
Q) Definition of Oblique triangle?
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Q) State Pythagoras Theorem?
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Q) The sides of a triangle are of length 9 cm, 11 cm and 6 cm. Is this triangle a right triangle?
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Q) A triangle which does not contain right angles is called...?
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Q) Measures of two angles of a triangle are 55° and 30° . Find the measure of its third angle?
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Q) A tower stands vertically on the ground. From a point on the ground, which is 27 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 80°. Find the height of the tower?
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Q) An observer 3.5 m tall is 32.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 60°. What is the height of the chimney?
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Q) The shadow of a tower stantanding on a level ground is found to be 50 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower?
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Q) From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 6 m from the banks, find the width of the river?
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Q) What are the six trigonometrical ratios?
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Q) State Converse of Pythagorean Theorem.
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Q) If u_{n} = cos^{n} Θ + sin^{n} Θ then prove that, 2u_{4}  3u_{3} + 1 = 0?
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Q) Proof the theorems on properties of triangle p/(sinsin P) = q/(sinsin Q) = r/(sinsin R) = 2K.
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Q) Proof Pythagorean Theorem using Algebra?
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Q) If tanΘ + secΘ = 2/√3 and Θ is a positive acute angle, find the value of sinΘ.
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 Applications Of Trigonometry
 Introduction
 Theorem on Properties of Triangle
 Pythagorean Theorem
 Statement of Pythagoras theorem
 Verification of Pythagoras theorem by the method of dissection
 Proof of Pythagorean Theorem
 Proof of Pythagorean Theorem using Algebra
 Converse of Pythagorean Theorem
 Word problems using the Converse of Pythagorean Theorem
 Properties of Triangles
 Discuss About Some Of The Properties Of Triangles
 Examples of Properties of Triangle
 Worksheet on Area and Perimeter of Squares
 Worksheet on Circumference and Area of Circle
 Trig Ratios Proving Problems
 Examples on trig ratios proving problems
 Trig Ratio Problems
 Basic Trigonometric Ratios
 Definitions of Trigonometrical Ratios
 Surveyors have used trigonometry for centuries. One such large surveying project of the nineteenth century was the 'Great Trigonometric Survey' of British India for which the two largestever theodolites were built.
 During the survey in 1852, the highest mountain in the world was discovered. From a distance of over 160 km, the peak was observed from six different stations. In 1856, this peak was named after Sir George Everest, who had commissioned and first used the giant theodolites (see the figure alongside).
 The theodolites are now on display in the Museum of the Survey of India in Dehradun.
Proof the theorems on properties of triangle` p/(sin P) = q/(sin Q) = r/(sin R) = 2K`
Proof:
Let O be the circumcentre and K the circumradius of any triangle PQR.
since in triangle PQR, three angles are acute
In figure(i), then we observe that the triangle PQR is acuteangled
In figure (ii), the triangle PQR is obtuseangled (since its angle P is obtuse) and
In figure (iii), the triangle PQR is rightangled (since the angle P is right angle).
In figure (i) and figure (ii) we join QO and produce it to meet the circumference at S. Then join RS.
Clearly, QO = circumradius = K
Therefore, QS = 2 · QO = 2K and `/_`QRS = 90° (being the semicircular angle).
Now, from figure (i)we get,
`/_`QSR = `/_`QPR = P (being the angles on the same arc QR).
Therefore, from the triangle QRS we have,
`[QR]/[QS] = sin /_QSR`
`p/[2K] = sin P `
`p/sinP = 2K`
Again, from figure (ii) we get,
`/_`QSR = Π  P [since, `/_QSR + /_QPR = Π`]
Therefore, from the triangle QRS we get,
`[QR]/[QS] = sin /_QSR`
` p/[2K] = sin(Π  P)`
`p/[2K] = sin P`
`a/sin P = 2K`
Finally, for rightangled triangle, we get from figure (iii),
2K = p = `(p/sin( 90°))` = `p/sin P ` [since, P = 90°]
Therefore, for any triangle PQR (acuteangled, or obtuseangled or rightangled) we have,
Similarly, if we join PO and produce it to meet the circumference at T then joining RT and QE we can prove
`q/(sin Q) = 2K` and `r/(sin R )= 2K` …………………………….. (1)
Therefore, in any triangle PQR we have,
`p/(sin P) = q/(sin Q) = r/(sin R) = 2K`
Note:
(i) The relation `p/sin P` = `q/sin Q` = `r/sin R` is known as sine Rule.
(ii) since, p : q : r = `sin P` : `sin Q `: `sin R`
Therefore, in any triangle the lengths of sides are proportional to the sines of opposite angles.
(iii) From (1) we get, p = 2K `sin` P, q = 2K `sin` Q and r = 2K `sin` R. These relations give the sides in terms of sines of angles.
Again, from (1) we get, `sin` P = `p/2K`, `sin` Q = `q/2K` and `sin` R = `r/2K`
These relations give the sines of the angles in terms of the sides of any triangle.
Solved problems using theorem on properties of triangle:
1. In the triangle PQR, if P = 60°, show that,
` q + r = 2p cos [(Q−R)/2]`
Solution:
We have,
We know that
`p/[sinP] = q/[sinQ] = r/[sinR] = 2K.`
`p = 2K sin P, q = 2K sin Q and r = 2K sin R`.
`[q+r]/[2p] = [2KsinQ+2KsinR]/[2*2KsinP],` [since,`p = 2K sin P, q = 2K sin Q and r = 2K sin R`]
= `[sinQ+sin R]/[2sinP]`
=`(2sin[(Q+R)/2]cos[(Q−R)/2])/(2sin60°)`
= `[sin60°cos[(Q−R)/2]]/[sin60°]`,
[since, P + Q + R = 180°, and P = 60°
Therefore, Q + R = 180°  60° = 120°
` [(Q+R)/2 = 60°]`
`(q+r)/[2p] = cos [(Q−R)/2]`
Therefore, q + r = 2p `cos [(Q−R)/2]` proved.
2. In any triangle PQR, prove that,
`(q^2  r^2) `cot` P + (r^2  p^2) `cot` Q + (p^2  q^2) `cot` R = 0`.
Solution:
`p/sinP = q/sinQ = r/sinR = 2K`
⇒ p = 2K `sin` P, q = 2K `sin` Q and r = 2K `sin` R.
Now, `(q^2  r^2) `cot` P = (4K^2 `sin`2 Q  4K^2 `sin`2 R) `cot` P`
= `2K^2 (2 sin^2 Q  2 sin^2 R)`
= `2K^2 (1  cos^2Q  1 + cos^ 2R) cot P`
= `2K^2 [2 sin (Q + R) sin (Q  R)] cot P`
=`4K^2 sin (Π  P) sin (Q  R) cot p`, [since, P + Q + R = Π]
=` 4K^2 sin P sin (Q  R) [(cosP)/(sinP)]`
= `4K^2 `sin` (Q  R) `cos` {Π  (Q  R)}`
= ` 2K^2 . 2`sin` (Q  R) `cos` (Q + R)`
=`  2K^2 (`sin` 2Q  `sin` 2R)`
Similarly,
`(r^2  p^2) `cot` Q = 2K^2 (`sin` 2R  `sin` 2P)`
and `(p^2  q^2) `cot` R = 2K^2 (`sin` 2R  `sin` 2Q)`
Now L.H.S. = `(q^2  r^2) `cot`P + (r^2  p^2) `cot` Q + (p^2  q^2) `cot` R`
= ` 2K^2 (sin 2Q  sin 2R)  2K^2 (sin2R  sin 2P)  2K^2(sin 2P  sin 2Q)`
= ` 2K^2 ** 0`
= 0 = R.H.S. Proved.
 Pythagorean Theorem is also known as 'Pythagoras theorem' and is related to the sides of a right angled triangle.
 In a right triangle the area of the square on the hypotenuse is equal to the sum of the areas of the squares of its remaining two sides.
 In the given figure, ∆PQR is right angled at Q; PR is the hypotenuse and PQ, QR are
the remaining two sides, then
`(PR)^2 = (PQ)^2 + (QR)^2`
`(h)^2 = p^2 + b^2`
 [Here h  hypotenuse, p perpendicular, b  base]
 In the adjoining figure, `Delta` PQR is a right angled triangle where QR is its hypotenuse and PR > PQ.
 Square on QR is QRBA, square on PQ is PQST and the square on PR is PRUV.
 The point of intersection of the diagonal of the square PRUV is O.
 The straight line through the point O parallel to the QR intersects PV and RU at the point J and K respectively.
 Again the straight line through the point O perpendicular to JK intersects PR and VU at the point L and respectively.
 As a result, the square PRUV is divided into four parts which is marked as 1, 2, 3, 4 and the square PQST is marked 5.
 You can draw the same figure on a thick paper and cut it accordingly and now cut out the squares respectively from this figure. Cut the squares PRUV along JK and LM dividing it in four parts. Now, place the parts 1, 2, 3, 4and 5 properly on the square QRBA.
Note:
Thus, we find that `(QR)^2 = (PQ)^2 + (PR)^2`
(ii) Square drawn on side PQ, which means the area of a square of side PQ is denoted by `(PQ)^2`.
1. Find the value of x using Pythagorean theorem:
Solution:
Identify the sides and the hypotenuse of the right angle triangle.
 The one sides length = 8 m
and the other side length = 15.  'X' is the length of hypotenuse because it is opposite side of the right angle.
 Substitute the values into the Pythagorean formula (here 'x' is the hypotenuse)
`(h)^2 = p^2 + b^2`
[Here h  hypotenuse, p  perpendicular, b  base]
`x^2 = 8^2 + 15^2`
Solve to find the known value of 'x'
`x^2 = 64 + 225`
`x^2 = 289`
x = `sqrt(289)`
x = 17
Therefore, value of x (hypotenuse) = 17 m
2. Use the formula of Pythagorean theorem to determine the length of 'a'.
Solution:
 Identify the perpendicular, base and the hypotenuse of the right angle triangle.
 Length of perpendicular = 24 cm and
the length of base = a.  Length of hypotenuse = 25 cm. since hypotenuse is the opposite side of the right angle.
Substitute the values into the Pythagorean formula (here 'a' is the base)
`(h)^2 = p^2 + b^2`
[Here h  hypotenuse, p  perpendicular, b  base]
`25^2 = 24^2 + a^2`
Solve to find the known value of 'a'
`625 = 576 + a^2`
`625 – 576 = 576 – 576 + a^2`
`49 = a^2`
`a^2 = 49`
a = `sqrt(49)`
a = 7,
Therefore, length of a (base) = 7 cm
3. Solve to find the missing value of the triangle using the formula of Pythagorean Theorem:
Solution:
 Identify the perpendicular, base and the hypotenuse of the right angle triangle.
 Perpendicular = k and Base = 7.5
 Hypotenuse = 8.5,
since hypotenuse is the opposite side of the right angle.  Substitute the values into the Pythagorean formula (here 'k' is the perpendicular)
`(h)^2 = p^2 + b^2`
[Here h  hypotenuse, p  perpendicular, b  base]
`8.5^2 = k^2 + 7.5^2`
Solve to find the known value of 'k'
72.25 = `k^2` + 56.25
72.25 – 56.25 = `k^2` + 56.25 – 56.25
16 = `k^2`
`k^2 = 16`
k = `sqrt(16)`
k = 4
Therefore, missing value of the triangle 'k' (perpendicular) = 4
 The proof of Pythagorean Theorem in mathematics is very important.
 In a right angle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
 States that in a right triangle that, the square of a `(a^2)` plus the square of b `(b^2)` is equal to the square of c `(c^2)`.
In short it is written as: ` a^2 + b^2 = c^2`  Let QR = a, RP = b and PQ = c. Now, draw a square WXYZ of side (b + c). Take points E, F, G, H on sides WX, XY, YZ and ZW respectively such that WE = XF = YG = ZH = b.
 Then, we will get 4 rightangled triangle, hypotenuse of each of them is 'a': remaining sides of each of them are band c. Remaining part of the figure is the
 square EFGH, each of whose side is a, so area of the square EFGH is `a^2`.
Now, we are sure that square WXYZ = square EFGH + 4 `Delta` GYF
or, `(b + c)^2` = `a^2` + 4 * `1/2` b * c
or, `b^2 + c^2 + 2bc = a^2 + 2bc`
or, `b^2 + c^2 = a^2`
Given: A `Delta` XYZ in which `/_`XYZ = 90°.
To prove: `(XZ)^2 = (XY)^2 + (YZ)^2`
Construction: Draw YO `__` XZ
Proof: In `Delta`XOY and `Delta`XYZ, we have,
`/_`X = `/_`X > common
`/_`XOY = `/_`XYZ  >each equal to 90°
Therefore, `Delta` XOY `~` `Delta` XYZ > by AAsimilarity
`((XO)/(XY))` = `((XY)/(XZ))`
`XO ** XZ = (XY)^2 ` (i)
In `Delta`YOZ and `Delta`XYZ, we have,
`/_`Z = `/_`Z > common
`/_`YOZ = `/_`XYZ > each equal to 90°
Therefore, `Delta` YOZ `~` `Delta` XYZ > by AAsimilarity
⇒ `((OZ)/(YZ))` = `((YZ)/(XZ))`
⇒ `OZ × XZ = (YZ)^2 ` (ii)
From (i) and (ii) we get,
`XO * XZ + OZ * XZ = ((XY)^2 + (YZ)^2)`
⇒ `(XO + OZ) * XZ = ((XY)^2 + (YZ)^2)`
⇒` XZ * XZ = (XY)^2 + (YZ)^2)`
⇒ `(XZ)^2 = (XY)^2 + (YZ)^2`
Converse of Pythagorean Theorem states that:
 In a triangle, if the square of one side is equal to the sum of the squares of the other two sides then the angle opposite to the first side is a right angle.
Given: A `Delta`PQR in which `(PR)^2 = (PQ)^2 + (QR)^2`
To prove: `/_`Q = 90°
Construction: Draw a `Delta` XYZ such that XY = PQ, YZ = QR and `/_`Y = 90°
So, by Pythagoras theorem we get,
`(XZ)2 = (XY)^2 + (YZ)^2`
`(XZ)^2 = (PQ)^2 + (QR)^2` ……….. (i), [since XY = PQ and YZ = QR]
But, `(PR)^2 = (PQ)^2 + (QR)^2 `………… (ii), [given]
From (i) and (ii) we get,
`(PR)^2 = (XZ)^2 `
⇒ `PR = XZ`
Now, in `Delta`PQR and `Delta`XYZ, we get
PQ = XY,
QR = YZ and
PR = XZ
Therefore `Delta`PQR `~=` `Delta`XYZ
Hence `/_`Q = `/_`Y = 90°
1. The side of a triangle are of length 4.5 cm, 7.5 cm and 6 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?
Solution:
 We know that hypotenuse is the longest side.
If 4.5 cm, 7.5 cm and 6 cm are the lengths of angled triangle, then 7.5 cm will be the hypotenuse.  using the converse of Pythagoras theorem, we get
`(7.5)^2 = (6)^2 + (4.5)^2`
56.25 = 36 + 20.25
56.25 = 56.25  since, both the sides are equal therefore, 4.5 cm, 7.5 cm and 6 cm are the side of the right angled triangle having hypotenuse 7.5 cm.
2. The side of a triangle are of length 8 cm, 15 cm and 17 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?
Solution:
 We know that hypotenuse is the longest side. If 8 cm, 15 cm and 17 cm are the lengths of angled triangle, then 17 cm will be the hypotenuse.
 using the converse of Pythagoras theorem, we get
`(17)^2 = (15)^2 + (8)^2`
` 289 = 225 + 64`
` 289 = 289`  since, both the sides are equal therefore, 8 cm, 15 cm and 17 cm are the side of the right angled triangle having hypotenuse 17 cm.
3. The side of a triangle are of length 9 cm, 11 cm and 6 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?
Solution:
 We know that hypotenuse is the longest side. If 9 cm, 11 cm and 6 cm are the lengths of angled triangle, then 11 cm will be the hypotenuse.
 using the converse of Pythagoras theorem, we get
`(11)^2 = (9)^2 + (6)^2`
121 = 81 + 36
121 ` !=` 117  since, both the sides are not equal therefore 9 cm, 11 cm and 6 cm are not the side of the right angled triangle.
 The above examples of the converse of Pythagorean Theorem will help us to determine the right triangle when the sides of the triangles will be given in the questions.
 We know any triangle has six parts, the three sides and the three angles are generally called the elements of the triangle.
 Oblique triangle: A triangle which does not contain right angles is called an oblique triangle.
 These six parts are not independent of each other i.e., various relations exist among the six parts.
 We will learn how to deduce the relations between the sides and trigonometric ratios of the angles of a triangle.
 We know in any triangle ABC, the measures of the angles `/_`ABC, `/_`BCA and `/_`CAB at the vertices B, C and A are denoted by the letters B, C and A respectively.
 The measures of the sides AB, BC and CA opposite to angles C, A and B respectively are denoted by c, a and b.
The perimeter of the triangle is denoted by 2s and semiperimeter of the triangles denoted by `(a + b + c)/2` .  The area of the triangle is denoted by `Delta` or S. The radius of the circumcircle of the triangle is called the circumradius and is denoted by R.
 The radius of the incircle of the triangle is called the inradius and is denoted by r.
 The radius of an excircle of the triangle is called an exradius and the radii of the excircles opposite to the angles A, B, C are denoted by `r_1`, `r_2`, and `r_3` respectively.
 The six elements of a triangle are not independent and are connected by the relations
A + B + C = Π, a + b > c, b + c > a and c + a > b.  In addition to these relations, the elements of a triangle are connected by some trigonometric relations.
1. In the triangle, given write the names of its three sides, three angles and three vertices.
Solution:
 Three sides of `Delta`PQR are: PQ, QR and RP
 Three angles of `Delta`PQR are: `/_`PQR, `/_`QRP and `/_`RPQ
 Three vertices of `Delta`PQR are: P, Q and R
2. Measures of two angles of a triangle are 65° and 40° . Find the measure of its third angle.
Solution:
 Measures of two angles of a `Delta` are 65° and 40°
 Sum of the measures of two angles = 65° + 40° = 105°
 Sum of all three angles of `Delta` = 180°
 Therefore, measure of the third angle = 180°  105° = 75°
3. Is the construction of a triangle possible in which the lengths of sides are 5 cm, 4 cm and 9 cm?
Solution:
 The lengths of the sides are 5 cm, 4 cm, 9 cm.
5 cm + 4 cm = 9 cm.
Then the sum of two smaller sides is equal to the third side. But in a triangle, the sum of any two sides should be greater than the third side.
Hence, no triangle possible with sides 5 cm, 4 cm and 9 cm.
Students can practice the questions on area of squares and perimeter of squares.
1. Find the perimeter and area of the following squares whose dimensions are:
(a) 16 cm
(b) 5.3 m
(c) 2 m 37 cm
(d) 86 dm
2. Find the perimeter of a square whose area is 625 `cm^2`.
3. Find the area of the square whose perimeter is 440 cm.
4. The area of a square field is 100 ares. Find its perimeter.
5. How many square tiles of side 9 cm will be needed to fit in a square floor of a bathroom of side 720 cm. Find the cost of tiling at the rate of $75 per tile.
6. The areas of a square and rectangle are equal. If the side of the square is 20 cm and the breadth of the rectangle 10 cm, find the length of the rectangle and its perimeter.
7. If it costs $2400 to fence a square field at the rate of $6 per m, find the length of the side and the area of the field.
Perimeter = (Total cost)/(cost per `m^2`)
8. A wire in the shape of rectangle whose width is 22 cm is bent to form a square of side 31 cm. Find the length of the rectangle. Also, find which shape encloses more area.
9. The area of a square field is 81 hectares. Find the cost of fencing the field with a wire at the rate of $2.25 per m.
10. Convert the following units:
(a) 17 `m^2to cm^2`
(b) 18000 `m^2` to hectares
(c) 3000` m^2` to ares ,
(d) 6.3 `m^2` to `mm^2`
(e) 3.7 hectares to ares
Answers for worksheet on area and perimeter of squares are given below to check the exact answers of the above questions.
Answers:
1. (a) 64 cm, 256 `cm^2`
(b) 21.2 m, 28.09 `m^2`
(c) 9.48 m, 5.6169 `m^2`
(d) 344 dm, 7396 `dm^2`
2. 100 cm
3. 12100 `cm^2`
4. 400 m
5. $32000
6. 40 cm, 100 cm
7. 100 m, 10000 `m^2`
8. l = 40 `cm^2`
9. $2700
10. (a) 170000 `cm^2`
(b) 1.8 hectares
(c) 30 ares
(d) 6300000 `m^2`
(e) 370 ares
Students can practice the questions on area of squares and perimeter of squares.
1. Find the perimeter and area of the following squares whose dimensions are:
(a) 16 cm
(b) 5.3 m
(c) 2 m 37 cm
(d) 86 dm
2. Find the perimeter of a square whose area is 625 `cm^2`.
3. Find the area of the square whose perimeter is 440 cm.
4. The area of a square field is 100 ares. Find its perimeter.
5. How many square tiles of side 9 cm will be needed to fit in a square floor of a bathroom of side 720 cm. Find the cost of tiling at the rate of $75 per tile.
6. The areas of a square and rectangle are equal. If the side of the square is 20 cm and the breadth of the rectangle 10 cm, find the length of the rectangle and its perimeter.
7. If it costs $2400 to fence a square field at the rate of $6 per m, find the length of the side and the area of the field.
Perimeter = (Total cost)/(cost per m^2)
8. A wire in the shape of rectangle whose width is 22 cm is bent to form a square of side 31 cm. Find the length of the rectangle. Also, find which shape encloses more area.
9. The area of a square field is 81 hectares. Find the cost of fencing the field with a wire at the rate of $2.25 per m.
10. Convert the following units:
(a) 17` m^2to cm^2`
(b) 18000 `m^2` to hectares
(c) 3000 `m^2` to ares
(d) 6.3 `m^2 to mm^2`
(e) 3.7 hectares to ares
Answers for worksheet on area and perimeter of squares are given below to check the exact answers of the above questions.
Answers:
1. 1. (a) 64 cm, 256 `cm^2 `
(b) 21.2 m, 28.09` m^2 `
(c) 9.48 m, 5.6169 `m^2`
(d) 344 dm, 7396 `dm^2`
2. 100 cm
3. 12100 `cm^2`
4. 400 m
5. $32000
6. 40 cm, 100 cm
7. 100 m, 10000 `m^2`
8. l = 40 `cm^2`
9. $2700
10. (a) 170000 `cm^2`
(b) 1.8 hectares
(c) 30 ares
(d) 6300000 `m^2`
(e) 370 ares
Students can recall the topic and practice the questions to get more ideas on how to find the circumference of the circle and the area of the circle.
1. Find the circumference of the circle whose radius is ………
(a) 12 cm
(b) 7.2 cm
(c) 15 mm (Take Π = `22/7`)
2. Find the area of the circle whose diameter is ………..
(a) 28 cm
(b) 10 cm
(c) 6.8 mm (Take Π = `22/7`)
3. If the circumference of a circular sheet is 176 m, find its area.
4. The area of a circle is 616 `cm^2`. Find its circumference.
5. From a circular sheet of a radius 5 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet.
6. Find the perimeter of the adjoining figure which is a semicircle including the diameter.
7. The diameter of a wheel is 70 cm. How many times the wheel will revolve in order to cover a distance of 110 m?
8. The ratio of the radii of two wheels is 4 : 5. Find the ratio of their circumference.
9. A well of diameter 150 cm has a stone parapet around it. If the length of the outer edge of the parapet is 616 cm, find the width of the parapet.
10. A thin wire is in the form of an equilateral triangle of side 11 cm. Find the area of a circle whose circumference is equal to the length of the wire.
11. Find the area of a circle whose circumference is same as the perimeter of square of side 22 cm.
12. From a rectangular metal sheet of size 20 cm by 30 cm, a circular sheet as big as possible is cut. Find the area of the remaining sheet.
13. Two circles have areas in the ratio 36 : 49. Find the ratio of their circumference.
14. Find the circumference of a wheel whose radius is 35 cm. Find the distance covered in 60 seconds, if it revolves 5 times per second.
15. The radius of a cycle wheel is 35 cm. Find the number of turns required to cover a distance of 1540 m.
16. A 3 m wide road runs around a circular park whose circumference is 132 m. Find the cost of fencing the outer boundary of the road at the rate of $10 per m.
17. A circular flowerbed is surrounded by a path 2.5 m wide. The diameter of the flowerbed is 40 m. Find the area of the path.
18. A square metallic frame has a perimeter 208 cm. It is bent in the shape of a circle. Find the area of the circle.
19. From a circular sheet of radius 18 cm, two circles of radii 4.5 cm and a rectangle of length 4 cm and breadth 1 cm are removed; find the area of the remaining sheet.
20. Find the area of a circle inscribed in a square of side 20 cm.
Answers for worksheet on circumference and area of circle are given below to check the exact answers of the above questions.
Answers:
1. 1. (a) `753/7` cm
(b) 45.25 cm
(c) `942/7` cm
2. (a) 616 `cm^2`
(b) `784/7` `cm^2`
(c) 36.33 `mm^2`
3. 2464` m^2`
4. 88 cm
5. `502/7` `cm^2`
6. 51.42
7. 50
8. 4 : 5
9. 23 cm
10. `865/8` `cm^2`
11. 616 `cm^2`
12. 285.71 `cm^2`
13. 6 : 7
14. 660 m
15. 700
16. $1508.57
17. 333.92` m^2`
18. `34415/11` `cm^2`
19. 887 `cm^2`
20. 314.2` cm^2`.
1. If (1 + `cos` A)( 1 + `cos` B)( 1 + `cos` C) = (1  `cos` A)( 1  `cos` B)( 1  `cos` C) then prove that each side = `+` `sin` A `sin` B `sin` C.
Solution:
Let, (1 + `cos` A) (1 + `cos` B) (1 + `cos` C) = k …. (i)
Therefore, according to the problem,
(1  `cos` A) (1  `cos` B) (1  `cos` C) = k ….. (ii)
Now multiplying both sides of (i) and (ii) we get,
(1 + `cos` A)(1 + `cos` B)(1 + `cos` C)(1  `cos` A)(1  `cos` B)(1  `cos` C) = `k^2`
`k^2` = (1  `cos^2 A`) (1  `cos^2 B`) (1  `cos^2 C`)
`k^2` = `sin^2 A sin^2 B sin^2 C`
k = `+ sin A sin B sin C`
Therefore, each side of the given condition
k = `+ sin A sin B sin C`
Proved.
2. If u_{n} = `cos`^{n} `theta` + `sin`^{n} `theta` then prove that, 2u_{6}  3u_{4} + 1 = 0`.
Solution:
since,u_{n} = `cos`^{n} `theta` + `sin`^{n} `theta`
Therefore, u_{6} = `cos^6 theta + sin^6 theta`
u_{6} = `(cos^2 theta)^3 + (sin^2 theta)^3`
u_{6} = `(cos^2 theta + sin^2 theta)^3  3 cos^2 theta sin^2 theta (cos^2 theta + sin^2 theta)`
u_{6} = `1  3cos^2 theta sin^2 theta` and
u_{4} = `cos^4 theta + sin^4 theta`
u_{4} = `(cos^2 theta)^2 + (sin^2 theta)^2`
u_{4} =` (cos^2 theta + sin^2 theta)^2  2 cos^2 theta sin^2 theta `
u_{4} = `1  2 cos^2 theta sin^2 theta`
Therefore,
2u_{6}  3u_{4} + 1
=` 2(1  3cos^2 theta sin^2 theta)  3(1  2 cos^2 theta sin^2 theta) + 1 `
= `2  6 cos^2 theta sin^2 theta  3 + 6 cos^2 theta sin^2 theta + 1 `
= 0.
Therefore, 2u_{6}  3u_{4} + 1 = 0.
Proved.
3. If a `sin theta`  b `cos theta` = c then prove that, a `cos theta` + b `sin theta` = `+sqrt(a^2 + b^2  c^2)`.
Solution:
Given: a `sin` `theta`  b `cos` `theta` = c
`(a sin theta  b cos theta)^2 = c^2` [Squaring both sides]
`a^2 sin^2 theta + b^2 cos^2 theta  2ab sin theta cos theta = c^2`
` a^2 sin^2 theta  b^2 cos^2 theta + 2ab sin theta cos theta =  c^2 `
`a^2  a^2 sin^2 theta + b^2  b^2 cos^2 theta + 2ab sin theta cos theta = a^2 + b^2  c^2`
`a^2(1  sin^2 theta) + b^2(1  cos^2 theta) + 2ab sin theta cos theta = a^2 + b^2  c^2`
`a^2 cos^2 theta + b^2 sin^2 theta + 2 · a cos theta· b sin theta = a^2 + b^2  c^2`
`(a cos theta + b sin theta)^2 = a^2 + b^2  c^2`
Now taking square root on both the sides we get,
a `cos theta` + b `sin theta` = `+ sqrt(a^2 + b^2  c^2)`.
Proved.
1. If `sec` `theta` = `17/8` and `theta` is a positive acute angle, find the value of `csc theta` using Pythagoras theorem.
Solution:
Draw a rightangled `Delta ABC` such that `/_`ABC = `theta`,
Hypotenuse = BA = 17, and Adjacent side (or base) = BC = 8.
Then we get,
`sec` `theta` = `17/8`
Now, from the rightangled `Delta` ABC we get,
`AC^2 + BC^2 = BA^2`
` AC^2 = BA^2  BC^2`
`AC^2 = (17)^2  8^2`
`AC^2 = 289  64`
`AC^2 = 225`
Therefore, AC = 15 (since `theta` is a positive acute angle so, AC is also positive)
Therefore, `csc theta` = `((BA)/(AC))` ,
`csc theta` = `17/15`
In this question on Trig ratio problems we will learn how to find the value of `sin` `theta` when `theta` is a positive acute angle.
2. If `tan theta` + `sec theta` = `2/(sqrt(3))` and `theta` is a positive acute angle, find the value of `sin theta`.
Solution:
Given, `tan theta` + `sec theta` = `2/(sqrt(3))`,
`(sin(theta)/(cos (theta)))` + `(1/(cos (theta)))` = `2/(sqrt(3))`,( since `tan theta` = `(sin (theta)/cos (theta))` and `sec theta` = `(1/cos (theta))`
`((sin (theta) + 1)/(cos(theta)))` = `2/(sqrt(3))`
`sqrt(3)` (`sin` `theta` + 1) = 2 `cos` `theta`
3`(sin theta + 1)^2` = 4 `cos^2` `theta`, (Squaring both sides)
3(`sin^2` `theta` + 2 `sin` `theta` + 1) = 4(1  `sin^2 theta`)
3 `sin^2` `theta` + 6 `sin` `theta` + 3 = 4  4 `sin^2 theta`
3 `sin^2` `theta` + 6 `sin` `theta` + 3  4 + 4 `sin^2 theta` = 0
7 `sin^2` `theta` + 6 `sin` `theta`  1 = 0
7 `sin^2` `theta` + 7 `sin` `theta`  `sin` `theta`  1 =0
7 `sin` `theta` (`sin` `theta` + 1)  1 (`sin` `theta` + 1) =0
(7 `sin` `theta`  1)(`sin` `theta` + 1) = 0
Therefore,
Either, 7 `sin` `theta`  1 = 0
7 `sin` `theta` = 1
`sin` `theta` = `1/7`
or, `sin` `theta` + 1 = 0,
`sin` `theta` =  1
According to the problem, `theta` is a positive acute angle; so, we neglect, `sin` `theta` = 1.
Therefore, `sin` `theta` = `1/7`
The below solved Trig ratio problems will help us to find the values of the ratio using trigonometric identity.
3. If `theta` is a positive acute angle and `sec` `theta` = `25/7`, find the value of csc`theta` using trigonometric identity.
Solution:
Given, `sec` `theta` = `25/7`
Therefore, `cos` `theta` = `1/(sec(theta))`
`cos``theta` = `1/(25/7)`
`cos``theta` = `7/25`
We know that, `sin^2 theta` + `cos^2 theta` = 1
`sin^2 theta` = 1  `cos^2 theta`
`sin^2 theta` = 1  `(7/25)^2`
`sin^2 theta` = 1  `(49/625) `
`sin^2 theta` = `(625 – 49)/625`
`sin^2 theta` = `576/625`
Now, taking square root on both the sides we get,
`sin``theta` =` 24/25` (since `theta` is a positive acute angle so, `sin` `theta` is also positive)
Therefore, csc`theta` = `1/sin(theta)`
`csc theta` = `1/(24/25)`
`csc theta` = `25/24` .
In this question on Trig ratio problems we will learn how to find the minimum value of the given Tratio.
4. Find the minimum value of `cos^2 theta` + `sec^2 theta`
Solution:
`cos^2 theta` + `sec^2 theta`
= `(cos theta)^2 + (sec theta)^2  2 cos theta· sec theta + 2 cos theta sec theta`
= `(cos theta  sec theta)^2 + 2 · 1`(since, `cos` `theta`· `sec` `theta` = 1)
= `(cos theta  sec theta)^2 + 2`
`(cos theta  sec theta)^2` `>=` 0
Therefore, `(cos theta  sec theta)^2 + 2 >= 2 `(since, adding 2 on both the sides)
i.e., `cos^2 theta + sec^2 theta >= 2`
Therefore, the minimum value of `cos^2 theta + sec^2 theta` is 2.
Problems on Trigonometric Ratios
1. If `sin` `theta` = `8/17`, find other trigonometric ratios of `theta`.
Solution:Let us draw a `Delta` OMP in which `/_`M = 90°.
Then `sin` `theta` = `(MP)/(OP)` = `8/17`.
Let MP = 8k and OP = 17k, where k is positive.
By Pythagoras theorem, we get
`OP^2 = OM^2 + MP^2`
`OM^2 = OP^2 – MP^2`
`OM^2 = [(17k)^2 – (8k)^2]`
`OM^2 = [289k^2 – 64k^2]`
`OM^2 = 225k^2`
`OM = sqrt(225k^2)`
OM = 15k
Therefore, `sin` `theta` =` (MP)/(OP) `= `(8k)/(17k)` = `(8)/(17)`
`cos theta` = `(OM)/(OP)` = `(15k)/(17k)` =` 15/17`
`tan theta = sin (theta)/cos (theta)`
= `(8/(17) * (17)/(15)) = (8/(15))`
`csc theta = 1/sin theta = [17]/8`
`sec theta = 1/cos theta = [17]/[15]` and
`cot theta = 1/tan theta = [15]/8`.
2. If `cos` A =` 9/41`, find other trigonometric ratios of `/_`A.
Solution:
Let us draw a `Delta` ABC in which `/_`B = 90°.
Then `cos` `theta` = `(AB)/(AC)` =`( 9)/(41)`.
Let AB = 9k and AC = 41k, where k is positive.
By Pythagoras theorem, we get
`AC^2 = AB^2 + BC^2`
`BC^2 = AC^2 – AB^2 `
`BC^2 = [(41k)^2 – (9k)^2]`
`BC^2 = [1681k^2 – 81k^2]`
` BC^2 = 1600k^2`
`BC = sqrt(1600k^2)`
BC = 40k
Therefore, `sin A = [BC]/[AC] = [40k]/[41k] = [40]/[41]`
`cos A = [AB]/[AC] = = [9k]/[41k] = 9/[41]`
`tan` A = `sin A/cos A = ((40/41) * (41/9)) = ([40]/9)`
`csc A = 1/sin A = [41]/[40]`
`sec A = 1/cos A = [41]/9` and
`cot A = 1/tan A = 9/[40]`.
3. Show that the value of `sin` `theta` and `cos` `theta` cannot be more than 1.
Solution:
We know, in a right angle triangle the hypotenuse is the longest side
`sin``theta` = perpendicular/hypotenuse = `(MP)/(OP)`< 1 since perpendicular cannot be greater than hypotenuse; `sin` `theta` cannot be more than 1.
Similarly, `cos` `theta` = base/hypotenuse = `(OM)/(OP)`< 1 since base cannot be greater than hypotenuse; `cos` `theta` cannot be more than 1.
4. Is that possible when A and B be acute angles, `sin` A = 0.3 and `cos` B = 0.7?
Solution:
since A and B are acute angles, 0 `<= sin A <= 1 and 0 <= cos B <=1`, that means the value of `sin A `and `cos B` lies between 0 to 1. So, it is possible that `sin A = 0.3` and
`cos B = 0.7`
5. If 0° `<=` A `<=` 90° can `sin` A = 0.4 and `cos` A = 0.5 be possible?
Solution:
We know that `sin^2A + cos^2A` = 1
Now put the value of `sin` A and `cos` A in the above equation we get;
`(0.4)^2 + (0.5)^2 = 0.41` which is `!=` 1, `sin` A = 0.4 and `cos` A = 0.5 cannot be possible.
6. If `sin theta` = `1/2`, show that (3`cos theta`  4 `cos^3 theta`) =0.
Solution:
Let us draw a `Delta` ABC in which `/_`B = 90° and `/_`BAC = `theta`.
Then `sin` `theta` = `(BC)/(AC)` = `1/2`.
Let BC = k and AC = 2k, where k is positive.
By Pythagoras theorem, we get
`AC^2 = AB^2 + BC^2`
` AB^2 = AC^2 – BC^2`
` AB^2 = [(2k)^2 – k^2]`
` AB^2 = [4k^2 – k^2]`
` AB^2 = 3k^2`
` AB = sqrt(3k^2)`
AB = `sqrt(3)`k.
Therefore, `cos` `theta` = `(AB)/(AC)` = `sqrt(3k)/(2k)` = `sqrt(3)/2`
Now, (3`cos` `theta`  4 `cos^3 theta`)
= `3(sqrt(3))/2  4 **((sqrt(3))/2)^3`
=`3(sqrt(3))/2  4 ** 3(sqrt(3))/8`
= `3(sqrt(3))/2  3(sqrt(3))/2`
= 0
Hence, (3`cos theta`  4`cos^3 theta`) = 0
7. Show that `sin alpha + cos alpha > 1` when 0° `<=` `alpha` `<=` 90°
Solution:
From the right triangle MOP,
`sin alpha` = perpendicular/ hypotenuse
`cos alpha` = base/hypotenuse
Now, `sin alpha + cos alpha`
= perpendicular/ hypotenuse + base/ hypotenuse
= (perpendicular + base)/hypotenuse, which is > 1, since we know that the sum of two sides of a triangle is always greater than the third side.
8. If `cos` `theta` = `3/5`, find the value of `(5cosec (theta)  4 tan (theta))/(sec (theta) + cot(theta))`
Solution:
Let us draw a ∆ ABC in which `/_`B = 90°.
Let `/_`A = `theta`°
Then `cos` `theta` = `(AB)/(AC)` = `3/5`.
Let AB = 3k and AC = 5k, where k is positive.
By Pythagoras theorem, we get
`AC^2 = AB^2 + BC^2`
`BC^2 = AC^2 – AB^2`
`BC^2 = [(5k)^2 – (3k)^2]`
`BC^2 = [25k^2 – 9k^2]`
`BC^2 = 16k^2`
`BC = sqrt(16k^2)`
`BC = 4k`
Therefore, `sec` `theta` = `1/cos(theta)` = `5/3`
`tan theta = (BC)/(AB) =(4k)/(3k) = 4/3`
`cot``theta` = `1/tan(theta)` = `3/4` and
csc`theta` = `(AC)/(BC)` = `(5k)/(4k)` = `5/4`
Now `{5csc (theta) 4 tan (theta)}/{sec (theta) + cot (theta)}`
= `{5 * (5/4)  4 * (4/3)}/{(5/3) + (3/4)}`
= `{[(25)/(4) (16)/3]}/{(5/3) +(3/4)}`
= `(11)/(12) * (12)/(29)`
= `(11)/(29)`
9. Express 1 + 2 `sin` A `cos` A as a perfect square.
Solution:
1 + 2 `sin` A `cos` A
= `sin`2 A + `cos`2 A + 2`sin` A `cos` A, [since we know that `sin^2` `theta` + `cos^2` `theta` = 1]
= `(sin A + cos A)^2`
10. If 3 `tan` `theta` = 4, evaluate `(3 sin(theta) + 2 cos(theta))/(3 sin(theta)  2 cos(theta))`.
Solution: Given,
3 `tan` `theta` = 4
`tan``theta` = `4/3`
Now,
`(3sin (theta) + 2 cos(theta))/((3 sin(theta)  2 cos(theta))`
= `(3 tan (theta) + 2)/(3 tan(theta)  2)`, [dividing both numerator and denominator by `cos` `theta`]
= `(3 * (4/3) + 2)/(3 * (4/3) 2)`, putting the value of `tan` `theta` = `4/3`
= `6/2`
= 3.
11. If `(sec (theta) + tan (theta))/(sec(theta)  tan(theta))` = `209/79`, find the value of `theta`.
Solution:` (sec(theta) + tan(theta))/(sec(theta)  tan(theta))` = `209/79`
⇒ `[(sec(theta) + tan(theta))  (sec(theta)  tan(theta))]/[(sec(theta) + tan(theta)) + (sec(theta)  tan(theta))]` = `[(209) – (79)]/[(209 )+ (79)]`, (Applying componendo and dividendo)
⇒ `(2 tan(theta))/(2 sec(theta))` =`(130)/(288)`
⇒ `(sin(theta)/cos(theta))` × `cos(theta)` = `65/144`
⇒`sin``theta` = `65/144`.
12. If `(5 cot(theta))` = 3, find the value of `(5 sin(theta)  3 cos(theta))/(4 (sin(theta)) + 3 (cos(theta)))`.
Solution:
Given 5 `cot` `theta` = 3
⇒`cot(theta)` = `(3/5)`
Now` (5 (sin(theta) ) 3 (cos(theta)))/(4 (sin(theta) + 3 (cos(theta)))`
= `(5  3 cot(theta))/(4 sin(theta) + 3 cot(theta))`, [dividing both numerator and denominator by `sin` `theta`]
= `(5  3 × (3/5))/(4 + 3 × (3/5))`
= `(5  (9/5))/(4 + (9/5))`
= `(16/5) × (5/29)`
= `(16)/(29)`.
13. Find the value of `theta` (0° ≤`theta`≤ 90°), when `sin^2 ``theta`  3 `sin` `theta` + 2 = 0
Solution:
⇒ `sin^2` `theta` 3 `sin` `theta` + 2 = 0
⇒ `sin^2` `theta` – 2 `sin` `theta` – `sin` `theta` + 2 = 0
⇒`sin``theta`(`sin` `theta`  2)  1(`sin` `theta`  2) = 0
⇒ (`sin` `theta`  2)(`sin` `theta`  1) = 0
⇒ (`sin` `theta`  2) = 0 or, (`sin` `theta`  1) = 0
⇒`sin``theta` = 2 or, `sin` `theta` = 1
So, value of `sin` `theta` can’t be greater than 1,
Therefore `sin` `theta` = 1
⇒`theta` = 90°
To know about the basic trigonometric ratios with respect to a rightangled triangle
let a ray OA revolve in the anticlockwise direction and assume the position `(OA)_1`, so that an angle `/_AOA_1` = `theta` is formed.
Now any number of points P, Q, R, .......... are taken on `OA_1`, and perpendiculars PX, QY, RZ, ........ are drawn on OA from those points respectively.
All the rightangled triangles POX, QOY, ROZ, ......... are similar to each other.
Now from the properties of similar triangles we know,
(i) `(PX)/(OP)` = `(QY)/(OQ)` = `(RZ)/(OR)` = .....
(ii) `(OX)/(OP)` =`(QY)/(OQ)` = `(OZ)/(OR)` = .....
(iii) `(PX)/(OX)` = `(QY)/ (OQ)` = `(RZ)/(OZ)` = .....
(v) `(OP)/(OX)` = `(OQ)/(OX)` = `(OR)/(OZ)` = .....
(iv) `(OP)/(PX)` = `(OQ)/(QY)` = `(OR)/(RZ)` = .....
(vi) `(OX)/(PX)` = `(OY)/(QY)` = `(OZ)/(RZ)` = .....
Thus we see in a set of similar rightangled triangles with respect to the same acute angle
(i) perpendicular : hypotenuse i.e., perpendicular/hypotenuse remains same.
(ii) base : hypotenuse
(iii) perpendicular : base do not change for the aforesaid similar rightangled triangles. So we can say that the values of these ratios do not depend on the size of triangles or the length of their sides. The values entirely depend on the magnitude of the acute angle `theta`.
It is so because all the triangles are right angled triangles having a common acute angle `theta`. Similar relations will hold whatever be the measure of the acute angle `theta`.
So we see that in similar rightangled triangles the ratio of any two sides, with reference to a common acute angle, give a definite value. This is the concept on the basis trigonometric ratios.
Again we have shown that the ratio of any two sides of a rightangled triangle, have six different ratios.
These six ratios are identified by six different names, one for each.
Now we will define trigonometrical ratios of positive acute angles and their relations.
Let a revolving line OY rotates about O in the anticlockwise sense and starting from the initial position OX comes in the final position OY and traces out an angle `/_`XOY = `theta` where ϴ is acute. Take a any point P on OY and draw PM perpendicular to OX. Clearly, POM is a rightangled triangle. With respect to the angle `theta` we shall call the sides, OP, PM and OM of the `Delta`POM as the hypotenuse, opposite side is also known as the perpendicular and adjacent side is also known as the base.
Now, the six trigonometrical ratios of the angle `theta` are defined as follows:
What are the six trigonometrical ratios?
Perpendicular/Hypotenuse = `(PM)/(OP)` = sine of the angle `theta`;
or, `sin theta` = `(PM)/(OP)`
Adjacent/Hypotenuse = `(OM)/(OP)` = cosine of the angle `theta`;
or, `cos theta` = `(OM)/(OP)`
Perpendicular/Adjacent = `(PM)/(OM)` = tangent of the angle `theta`;
or, `tan` θ = `(PM)/(OM)`
Hypotenuse/Perpendicular = `(OP)/(PM)` = cosecant of the angle `theta`;
or, `cosec (theta)` = `(OP)/(PM)`
Hypotenuse/Adjacent = `(OP)/(OM)`= secant of the angle `theta`;
or, `sec (theta)` = `(OP)/(OM)`
and Adjacent/Perpendicular = `(OM)/(PM )`= tangent of the angle `theta`;
or, `cot (theta)` = `(OM)/(PM)`
The six ratios `sin` `theta`, `cos` `theta`, `tan` `theta`, `cosec` `theta`, `sec` `theta` and `cot` `theta` are called Trigonometrical Ratios of the angle `theta`.
Sometimes there are two other ratios in addition. They are known as Conversed `sin` and Conversed `sin`.
These two ratios are defined as follows:
conversed `sin` of angle `theta` or Converse `theta` = 1  `cos` `theta`
and Conversed `sin` of angle `theta` or Converse `theta` = 1  `sin` `theta`.
 since each trigonometrical ratio is defined as the ratio of two lengths hence each of them is a pure number.
 Note that `sin` `theta` does not imply `sin` `theta`; in fact, it represents the ratio of perpendicular and hypotenuse with respect to the angle `theta` of a rightangled triangle.
 In a rightangled triangle the side opposite to rightangle is the hypotenuse, the side opposite to given angle `theta` is the perpendicular and the remaining side is the adjacent side.
Example 1 : A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.
Solution : First let us draw a simple diagram to represent the problem (see Fig. 9.4). Here AB represents the tower, CB is the distance of the point from the tower and `/_` ACB is the angle of elevation. We need to determine the height of the tower, i.e., AB. Also, ACB is a triangle, rightangled at B. To solve the problem, we choose the trigonometric ratio `tan` 60° (or `cot` 60°), as the ratio involves AB and BC.
Now, `tan` 60° =`(AB)/(BC)`
`sqrt(3)`=`(AB)/(15)`
i.e., AB = `15/(sqrt(3))`.
Hence, the height of the tower is 15`sqrt(3)` m.
Example 2 : An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3m below the top of the pole to undertake the repair work (see Fig. 9.5). What should be the length of the ladder that she should use which, when inclined at an angle of 60° to the horizontal, would enable to her reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (You may take 3 = 1.73).
Solution : In Fig. 9.5, the electrician is required to reach the point B on the pole AD. So, BD = AD – AB = (5 – 1.3)m = 3.7 m.
Here, BC represents the ladder. We need to find its length, i.e., the hypotenuse of the right triangle BDC.
Now, can you think which trigonometric ratio should we consider? It should be `sin` 60°.
`(BD)/(BC)`= `sin` 60°
or `(3.7)/ (BC)`= `(sqrt(3)/2)`
BC=`((3.7 × 2)/sqrt(3))`=4.28 m (approx.)
e., the length of the ladder should be 4.28 m.
`(DC)/(BD)` = `cot` 60° =`1/(sqrt(3))`
DC = `(3.7)/(sqrt(3))`= 2.14 m (approx.)
Therefore, she should place the foot of the ladder at a distance of 2.14 m from the pole.
Example 3: An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney?
Solution: Here, AB is the chimney, CD the observer and `/_` ADE the angle of elevation (see Fig. 9.6). In this case, ADE is a triangle, rightangled at E and we are required to find the height of the chimney.
We have AB = AE + BE = AE + 1.5
And DE = CB = 28.5 m
To determine AE, we choose a trigonometric ratio, which involves both AE and DE. Let us choose the tangent of the angle of elevation
`tan` 45° = `(AE)/( DE)`
1= `(AE)/ (28.5)`
Therefore, AE = 28.5
So the height of the chimney (AB) = (28.5 + 1.5) m = 30 m
Example 4 : From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (You may take 3 = 1.732)
Solution :
In Fig. 9.7, AB denotes the height of the building, BD the flagstaff and P the given point. Note that there are two right triangles PAB and PAD. We are required to find the length of the flagstaff, i.e., DB and the distance of the building from the point P, i.e., PA.
since, we know the height of the building AB, we will first consider the right `Delta` PAB.
We have `tan` 30° = `(AB) /(AP)`
i.e., 1/`sqrt(3)` = `(10)/(AP)`
Therefore, AP = 10 `sqrt(3)` i.e., the distance of the building from P is 10`sqrt(3)` m = 17.32 m.
Next, let us suppose DB = x m. Then AD = (10 + x) m.
Now, in right `Delta` PAD, `tan` 45deg; = `(AD)/(AP)`= `(10 + x) /(10sqrt(3))`
Therefore, 1 = `(10+x)/10(sqrt(3))`
x = 10 (`sqrt(3)`1 ) = 7.32
So, the length of the flagstaff is 7.32 m.
Example 5 : The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.
Solution : In Fig. 9.8, AB is the tower and BC is the length of the shadow when the Sun's altitude is 60°, i.e., the angle of elevation of the top of the tower from the tip of the shadow is 60° and DB is the length of the shadow, when the angle of elevation is 30°.
Now, let AB be h m and BC be x m. According to the question, DB is 40 m longer than BC.
So, DB = (40 + x) m
Now, we have two right triangles ABC and ABD.
In `Delta` ABC, `tan` 60° = `(AB)/ (BC)` or, `sqrt(3)` = `(h)/ (x............. (1))`
In `Delta` ABD, `tan` 30° = `(AB)/ (BD)` i.e., `(1)/sqrt(3)` =`(h)/(x+40........(2) )`
From (1), we have h = x`sqrt(3)`
Putting this value in (2), we get ( x`sqrt(3)`)`sqrt(3)` = x + 40, i.e., 3x = x + 40
i.e., x = 20
So, h = 20 `sqrt(3)`.......[From (1)]
Therefore, the height of the tower is 20 `sqrt(3)` m.
Example 6 : The angles of depression of the top and the bottom of an 8 m tall building from the top of a multistored building are 30° and 45°, respectively. Find the height of the multistored building and the distance between the two buildings.
Solution : In Fig. 9.9, PC denotes the multistorey building and AB denotes the 8 m tall building. We are interested to determine the height of the multistored building, i.e., PC and the distance between the two buildings, i.e., AC.
Look at the figure carefully. Observe that PB is a transversal to the parallel lines PQ and BD. Therefore, `/_` QPB and `/_` PBD are alternate angles, and so are equal. So `/_` PBD = 30°. Similarly, `/_` PAC = 45°. In right `Delta` PBD,
we have `(PD)/(BD)` = `tan` 30° = `1/sqrt(3)` or BD = PD`sqrt(3)`
In right `Delta` PAC, we have `(PC)/ (AC)` = `tan` 45° = 1 i.e.,
PC = AC
Also, PC = PD + DC, therefore, PD + DC = AC.
since, AC = BD and DC = AB = 8 m, we get PD + 8 = BD = PD`sqrt(3)` (Why?)
`PD= 8/(sqrt(3)1)`= `(8(sqrt(3)+1))/[(sqrt(3)+1)(sqrt(3)1)]`
=4(`sqrt(3)`+1)m
So, the height of the multistored building is `{4(sqrt(3)+1)+8 }m=4(3+sqrt3) m`
and the distance between the two buildings is also 4(3+`sqrt (3)`) m.
Example 7 : From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.
Solution : In Fig 9.10, A and B represent points on the bank on opposite sides of the river, so that AB is the width of the river.
P is a point on the bridge at a height of 3 m, i.e., DP = 3 m.
We are interested to determine the width of the river, which is the length of the side AB of the `Delta` APB.
AB = AD + DB
In right `Delta` APD, `/_` A = 30°.
So, `tan` 30° = `(PD)/( AD)`
i.e., `(1/ sqrt(3))` = `3/ (AD)` or AD = 3 `sqrt(3)` m
Also, in right `Delta` PBD, `/_` B = 45°. So, BD = PD = 3 m. Now, AB = BD + AD
= 3 + 3`sqrt(3)` = 3 (1 + `sqrt(3)` ) m.
Therefore, the width of the river is 3 (1 + `sqrt(3)` ) m..
(i) The relation` p/(sin P) = q/(sin Q )= r/(sin R)` is known as sine Rule.
(ii) since, p : q : r = `sin` P : `sin` Q : `sin` R
Therefore, in any triangle the lengths of sides are proportional to the sin's of opposite angles.
Pythagorean Theorem is also known as 'Pythagoras theorem' and is related to the sides of a right angled triangle.
(i) these parts together exactly fit the square.
Thus, we find that `(QR)^2 = (PQ)^2 + (PR)^2`
(ii) Square drawn on side PQ, which means the area of a square of side PQ is denoted by `(PQ)^2`.
Oblique triangle:
A triangle which does not contain right angles is called an oblique triangle.
These six parts are not independent of each other i.e., various relations exist among the six parts.
Perimeter = (Total cost)/(cost per `m^2`)