
Q) A triangle which does not contain right angles is called...?
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Q) Formula of sin (90°  A)=?
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Q) Define Pythagoras Theorem.
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Q) What is trigonometric identity?
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Q) What is adjacent to angle A?
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Q) What is trigonometric ratios?
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Q) Given tan A = 3 , find the other trigonometric ratios of the angle A?
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Q) In ABC, rightangled at B, AB = 5 cm and ∠ACB = 30° . Determine the lengths of the sides BC and AC.
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Q) In OPQ, rightangled OP = 7 cm and OQ  PQ = 1 cm Determine the values of sinQ and cosQ.
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Q) In PQR, right  angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠QPR and ∠PRQ.
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Q) Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
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Q) Explain Trigonometric Ratios of 30° and 60°?
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Q) If tan A = cot B, prove that A + B = 90°.
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Q) If sin3A = cos(A – 26°), where 3A is an acute angle, find the value of A.
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Q) Show that: (i) tan 48° tan 23° tan 42° tan 67° = 1 (ii) cos 38° cos 52°  sin 38° sin 52° = 0
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 You have already studied about triangles, and in particular, right triangles, in your earlier classes.
 Let us take some examples from our surroundings where right triangles can be imagined to be formed. For instance
 Suppose the students of a school are visiting Qutub Minar. Now, if a student is looking at the top of the Minar, a right triangle can be imagined to be made, Can the student find out the height of the Minar, without actually measuring it?
 Suppose a girl is sitting on the balcony of her house located on the bank of a river. She is looking down at a flower pot placed on a stair of a temple situated nearby on the other bank of the river. A right triangle is imagined to be made in this situation If you know the height at which the person is sitting, can you find the width of the river?.
 Suppose a hot air balloon is flying in the air. A girl happens to spot the balloon in the sky and runs to her mother to tell her about it. Her mother rushes out of the house to look at the balloon.Now when the girl had spotted the balloon initially it was at point A. When both the mother and daughter came out to see it, it had already travelled to another point B. Can you find the altitude of B from the ground?
 In all the situations given above, the distances or heights can be found by using some mathematical techniques, which come under a branch of mathematics called ‘trigonometry’.
 The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure).
 In fact, trigonometry is the study of relationships between the sides and angles of a triangle.
 The earliest known work on trigonometry was recorded in Egypt and Babylon.
 Early astronomers used it to find out the distances of the stars and planets from the Earth.
 Even today, most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts.
 In this chapter, we will study some ratios of the sides of a right triangle with respect to its acute angles, called trigonometric ratios of the angle.
 We will restrict our discussion to acute angles only. However, these ratios can be extended to other angles also.
 We will also define the trigonometric ratios for angles of measure 0° and 90°.
 We will calculate trigonometric ratios for some specific angles and establish some identities involving these ratios, called trigonometric identities.
We have seen triangles and their properties in previous classes.
There, we observed different daily life situations where we were using
triangles.
Let’s again look at some of the daily life examples.
Electric poles are present everywhere. They are usually erected
by using a metal wire.
The pole, wire and the ground form a
triangle.
But, if the length of the wire decreases, what will be
the shape of the triangle and what will be the angle of the wire
with the ground ?
A person is whitewashing a wall with the help of a ladder which
is kept as shown in the adjacent figure on
left.
If the person wants to paint at a higher position, what will the
person do? What will be the change in angle of the ladder with the
ground ?
In the temple at Jainath in Adilabad district, which was built in
13th century, the first rays of the Sun fall at the feet of the Idol of
Suryanarayana Swami in the month of December.
There is a relation
between distance of Idol from the door, height of the hole on the
door from which Sun rays are entering and angle of sun rays in that
month.
Is there any triangle forming in this context?
In a play ground, children like
to slide on slider and slider is
on a defined angle from earth.
What will happen to the slider if
we change the angle?
Will
children still be able to play on.
The above examples are geometrically showing the application part of triangles in our daily
life and we can measure the heights, distances and slopes by using the properties of triangles.
These types of problems are part of ‘trigonometry’ which is a branch of mathematics.
Now look at the example of a person who is white washing the wall with the help of a
ladder as shown in the previous figure.
Let us observe the following conditions.
We denote the foot of the ladder by A and top of it by C and the point of joining height
of the wall and base of the ladder as B.
Therefore, ΔABC is a right angle triangle with
right angle at B.
The angle between ladder and base is said to be θ.
If the person wants to white wash at a higher point on the
wall
What happens to the
angle made by the
ladder with the ground?
· What will be the change
in the distance AB?
2. If the person wants to white
wash at a lower point on the wall
· What happens to the angle made by the ladder with the ground?
· What will be the change in the distance AB?
We have observed in the above example of a person who was white washing.
When he
wants to paint at higher or lower points, he should change the position of ladder.
So, when ‘θ’ is
increased, the height also increases and the base decreases.
But, when θ is decreased, the height
also decreases and the base increases.
Do you agree with this statement?
Here, we have seen a right angle triangle ABC and have given ordinary names to all sides
and angles.
Now let’s name the sides again because trigonometric ratios of angles are based on
sides only.
Let’s take a right triangle ABC as show in the figure.
In triangle ABC, we can consider ∠CAB as A where angle A is an acute angle.
Since
AC is the longest side, it is called “hypotenuse”.
Here you observe the position
of side BC with respect to angle A.
It is
opposite to angle A and we can call it as
“opposite side of angle A”.
And the
remaining side AB can be called as
“Adjacent side of angle A”
AC = Hypotenuse
BC = Opposite side of angle A
AB = Adjacent side of angle A
What do you observe? Is there any relation between the opposite side of the angle A and
adjacent side of angle C? Like this, suppose you are erecting a pole by giving support of strong
ropes.
Is there any relationship between the length of the rope and the length of the pole? Here,
we have to understand the relationship between the sides and angles we will study this under the
section called trigonometric ratios.
We have seen the example problems in the beginning of the chapter which are related to our daily life situations. Let’s know about the trigonometric ratios and how they are defined.
You have seen some right triangles imagined to be formed in different situations. Let us take a right triangle ABC a below fig Here, ∠CAB (or, in brief, angle A) is an acute angle. Note that the position of the side BC with respect to angle A. It faces ∠A. We call it the side opposite to angle A. AC is the hypotenuse of the right triangle and the side AB is a part of ∠A. So, we call it the side adjacent to angle A.
Note that the position of sides change when you consider angle C in place of A
You have studied the concept of ‘ratio’ in your earlier classes. We now define certain ratios involving the sides of a right triangle, and call them trigonometric ratios.
The trigonometric ratios of the angle A in right triangle ABC (see above fig ) are defined as follows :
sine of `/_ A =(BC)/(AC)`=(side opposite to angle A )/(hypotenuse)
cosine of `/_ A =(AB)/(AC)`=(side adjacent to angle A)/( hypotenuse)
tangent of `/_ A =(BC)/(AB)` =(side opposite to angle A)/( hypotenuse)
cosec of `/_ A =(AC)/(BC)`=(hypotenuse)/(side opposite to angle A)=1/(sine of `/_`A)
secant of `/_ A =(AC)/(BC)`=(hypotenuse)/(side adjacent to angle A)=1/(cosine of `/_`A)
cosec of `/_ A =(AB)/(AC)`=(hypotenuse)/(side opposite to angle A)=1/(sine of `/_A`)
The ratios defined above are abbreviated as `sin A, cos A, tan A,csc A, sec A and cot A` respectively. Note that the ratios `csc A,sec A and cot A `are respectively, the reciprocals of the ratios `sin A, cos A and tan A`.
Also, observe that `tan A =(BC)/(AB)=((BC)/(AB))/((AB)/(AC))=sinA/cosA and cotA =cosA/sinA`So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.
Why don't you try to define the trigonometric ratios for angle C in the right triangle?
The first use of the idea of ‘sine’ in the way we use it today was in
the work Aryabhatiyam by Aryabhatta, in A.D. 500.
Aryabhatta used the word ardhajya
for the halfchord, which was shortened to jya or jiva in due course. When the Aryabhatiyam was
translated into Arabic, the word jiva was retained as it is.
The word jiva was translated into sinus,
which means curve, when the Arabic version was translated into Latin.
Soon the word sinus, also used as sine,
became common in mathematical texts throughout Europe. An English Professor of astronomy
Edmund Gunter (1581–1626), first used the abbreviated notation ‘sin’.
The origin of the terms ‘cosine’ and ‘tangent’ was much later. The cosine
function arose from the need to compute the sine of the complementary angle.
Aryabhatta called it kotijya. The name cosinus originated with Edmund Gunter. In 1674, the English Mathematician
Sir Jonas Moore first used the abbreviated notation ‘cos’.
Remark : Note that the symbol sin A is used as an
abbreviation for ‘the sine of the angle A’. sin A is not
the product of ‘sin’ and A.
‘sin’ separated from A
has no meaning. Similarly, cos A is not the product of
‘cos’ and A. Similar interpretations follow for other
trigonometric ratios also.
Now, if we take a point P on the hypotenuse
AC or a point Q on AC extended, of the right triangle
ABC and draw PM perpendicular to AB and QN
perpendicular to AB extended how
will the trigonometric ratios of ` /_`A in PAM differ
from those of `/_`A in CAB or from those of `/_ `A in
QAN?
To answer this, first look at these triangles. Is PAM similar to CAB? recall the AA similarity criterion. Using the criterion,
you will see that the triangles PAM and CAB are similar.
Therefore, by the property of
similar triangles, the corresponding sides of the triangles are proportional.
So, we have `(AM)/(AB) =(AP)/(AC)= (MP)/(BC)`
From this, we find `(MP)/(AP)=(BC)/(AC)=sin A` .
Similarly, `(AM)/(AP)= (AB)/(AC)= cos A, (MP)/(AM) = (BC)/(AB) tan A` and so on.
those of angle A in CAB.
In the same way, you should check that the value of sin A (and also of other
trigonometric ratios) remains the same in QAN also.
From our observations, it is now clear that the values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.
Note : For the sake of convenience, we may write `sin^2A, cos^2A`, etc.,
in place of `(sin A)^2, (cos A)^2`, etc., respectively.
But `csc A = (sin A)^(–1) != sin^(–1) A` (it is called sine inverse A).
`sin^(–1)` A has a different meaning, which will be discussed in higher classes.
Similar conventions hold for the other trigonometric ratios as well.
Sometimes, the Greek letter ` theta` (theta) is also used to denote an angle.
We have defined six trigonometric ratios of an acute angle. If we know any one
of the ratios, can we obtain the other ratios? Let us see.
If in a right triangle ABC, `sin A = 1/3`,
then this means that `(BC)/(AC) = 1/3 ` , i.e.,
the
lengths of the sides BC and AC of the triangle
ABC are in the ratio 1 : 3 .
So if
BC is equal to k, then AC will be 3k, where k is any positive number.
To determine other
trigonometric ratios for the angle A, we need to find the length of the third side AB.
Do you remember the Pythagoras theorem? Let us use it to determine the required length AB.
`AB^2 = AC^2– BC^2`
=`(3k)^2 (k)^2 =8k^2`
Therefore, `AB= (2sqrt2 k)^2`
` AB=+2sqrt2 k`
So, we get AB =`2 sqrt2 k`(Why is AB not ` 2sqrt2 k` ?)
Now, `cos A =(AB)/(AC)= (2sqrt2k)/(3 k )=(2sqrt2)/3`
Similarly, you can obtain the other trigonometric ratios of the angle A.
Remark : Since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than 1 (or, in particular, equal to 1).
when we observe right angle triangles ABP, ACQ, ADR and AES,
∠ A is common, ∠ B, ∠ C, ∠ D and ∠ E are right angles and ∠ P, ∠ Q, ∠ R and ∠ S are
also equal.
Hence, we can say that triangles ABP, ACQ, ADR and AES are similar triangles.
When we observe the ratio of opposite side of angle A and hypotenuse in a right angle triangle
and the ratio of similar sides in another triangle, it is found to be constant in all the above right
angle triangles ABP, ACQ, ADR and AES.
And the ratios BP CQ DR , , AP AQ AR
and ES
AS
can be
named as “sine A” or simply “sin A” in those triangles.
If the value of angle A is “x” when it was
measured, then the ratio would be “sin x”.
Hence, we can conclude that the ratio of opposite side of an angle (measure of the angle)
and length of the hypotenuse is constant in all similar right angle triangles.
This ratio will
be named as “sine” of that angle.
Similarly, when we observe the ratios
AB AC AD ,AP AQ AR and
AE
AS , it is also found to be
constant.
And these are the ratios of the adjacent sides of the angle A and hypotenuses in right
angle triangles ABP, ACQ, ADR and AES.
So, the ratios
AB AC AD,AP AQ AR and
AE
AS will be
named as “cosine A” or simply “cos A” in those triangles.
If the value of the angle A is “x”, then
the ratio would be “cos x”
Hence, we can also conclude that the ratio of the adjacent side of an angle (measure of the
angle) and length of the hypotenuse is constant in all similar right triangles.
This ratio
will be named as “cosine” of that angle.
Similarly, the ratio of opposite side and adjacent side of an angle is constant and it can be
named as “tangent” of that angle.
Let's Define Ratios In A Right Angle Triangle
Consider a right angle triangle ABC having right angle at B as shown in the following figure.
Then, trigonometric ratios of the angle A in right angle triangle ABC are defined as follows :
sine of ∠ A = sinA = Length of the sideopposite toangle A / Length of hypotenuse
= `(BC)/(AC)`
cosine of ∠ A = cos A = Length of the sideadjacent to angle A / Length of hypotenuse
= `(AB)/(AC)`
tangent of ∠ A = tan A =
Length of the sideopposite to angle A/Length of the side adjacent to angle A
=
`(BC)/(AB)`
Try This
In a right angle triangle ABC, right angle is at C.
BC + CA = 23 cm and
BC − CA = 7cm, then find sin A and tan B.
Think Discuss
Discuss between your friends that
(i) sin x =
`4/
3` does exist for some value of angle x?
(ii) The value of sin A and cos A is always less than 1. Why?
(iii) tan A is product of tan and A.
There are three more ratios defined in trigonometry which are considered as multiplicative
inverse of the above three ratios.
Multiplicative inverse of “sine A” is “cosecant A”.
Simply written as “cosec A”
i.e., cosec A =
`1 /
(sin A)`
Similarly, multiplicative inverses of “cos A” is secant A” (simply written as “sec A”) and
that of “tan A” is “cotangent A (simply written as cot A)
i.e., sec A =
`1/
(cos A)`
and cot A =
`1/
(tan A)`
How can you define ‘cosec’ in terms of sides?
If sin A = Opposite sideof theangle A /Hypotenuse
then cosec A =
Hypotenuse/Opposite sideof theangle A
Try This
What will be the ratios of sides for sec A and cot A?
Let us see some examples
Example1
If tan A =`3/
4 `, then find the other trigonometric ratio of angle A.
Hence tan A = Opposite side/Adjacent side = `3/4`
Therefore, opposite side : adjacent side = 3:4
For angle A, opposite side = BC = 3k
Adjacent side = AB = 4k (where k is any positive number)
Now, we have in triangle ABC (by Pythagoras theorem)
= `(3k)^2` + `(4k)^2`
= `25k^2`
AC = `sqrt( 25k^2)`
= 5k = Hypotenuse
Now, we can easily write the other ratios of trigonometry
sin A =`(3k)/(5k)`
= `3/5`
and cos A =`(4k)/(5k)`
= `4/5`
And also cosec A =`1/(Sin A)`
= `5/3`
sec A =`1 / (cos A)` =`5/4`
cot A =`1/(tan A)` = `4/3`
Example2:
If ∠A and ∠P are acute angles such that sin A = sin P then prove that ∠A =∠P.
Solution:
Given sin A = sin P
we have sin A =
`(BC)/(AC)`
and sin P = `(QR)/(PQ)`
Then `(BC)/(AC)` = `(QR)/(PQ)`
By using Pythagoras theorem
`(AB)/(PR)` = `(sqrt((AC^2)(BC^2)))/(sqrt((PQ^2)(QR^2)))`
=`(sqrt((AC^2)(K^2)(BC^2)))/((sqrt((PQ^2)(K^2)(QR^2))))`
= `(AC)/(PQ)`
Hence,
`(AC)/(PQ)` =`(AB)/(PR)` = `(BC)/(QR)`
then ΔABC ∼ ΔPQR
Therefore, ∠A =∠P
Example:Consider a triangle PQR, right angled at P, in which PQ = 29 units, QR = 21 units
and ∠PQR = θ, then find the values of
(i) cos^{2}
θ + sin^{2}
θ
and
(ii) cos^{2}
θ − sin^{2}
θ
Solution:
ln PQR, we have,
PR = `(sqrt((PQ^2)(QR^2)))/((sqrt(((29)^2)(11)^2)))`
=`sqrt(400)`=20
cos θ =`(QR)/(PQ)`
= `(20)/(29)`
Sin θ= `(PR)/(PQ)`
=`(21)/(29)`
Now (i) cos^{2}
θ + sin^{2}
θ = `{((20)/(29))^2}` + `{((21)/(29))^2}`
= `{(441) + (400)}/{(841)}`
= 1.
(i) cos^{2}
θ  sin^{2}
θ = `{((20)/(29))^2}`  `{((21)/(29))^2}`
= `(41)/(841)`
Exercise
1.In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA
respectively.
Then, find out sin A, cos A and tan A.
2. The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25cm and ∠Q = 90o
respectively.
Then find, tan Q − tan R.
3. In a right angle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and
∠ BAC = θ.
Then, find cos θ and tan θ.
4. If cos A =
`(12)/
(13)` , then find sin A and tan A.
5. If 3 tan A = 4, then find sin A and cos A.
6. If ∠ A and ∠ X are acute angles such that cos A = cos X then show that
∠ A = ∠ X.
Example 1 : Given tan A = 3 , find the other trigonometric ratios of the angle A.
Solution :Let us first draw a right ABC .
Now, we know that `tan A = (BC)/(AB) = 4/3`.
Therefore, if ` BC = 4k`, then `AB = 3k`, where k is a
positive number.
Now, by using the Pythagoras Theorem, we have
`AC^2 = AB^2 + BC^2 = (4k)^2 + (3k)^2 = 25k^2`
So, AC = 5k
Now, we can write all the trigonometric ratios using their definitions.
`sin A =(BC)/(AC)=(4k)/(5k)=4/5`
`cos A = (AB)/(AC)=(3k)/(5k)=3/5 `
Therefore, `cot A = 1/(tanA)= 3/4` ,
`csc A =1/(sin A)=5/4` and `sec A =1/(cos A) =5/3`
Example 2 : If ∠ B and ∠ Q are acute angles such that `sin B = sin Q`,
then prove that ∠ B = ∠ Q.
Solution : Let us consider two right triangles ABC and PQR where sin B = sin Q .
We have `sin B =(AC)/(AB)`
and `sin Q =(PR)/(PQ)`
Then`(AC)/(AB)=(PR)/(PQ)`
Therefore,`(AC)/(PR)=(AB)/(PQ) = k`, say (1)
Now, using Pythagoras theorem,
`BC = sqrt(AB^2 − AC^2)`
`QR = sqrt(PQ^2 – PR^2)`
So,`(BC)/(QR)=sqrt(AB^2 − AC^2)/sqrt(PQ^2 − PR^2)`
=`sqrt(k^2 PQ^2 − k^2 PR^2)/sqrt(PQ ^2 − PR^2)`
=`k sqrt( PQ^2 − PR^2)/sqrt(PQ^2 − PR^2)= k` (2)
From (1) and (2), we have
`(AC)/(PR) = (AB)/(PQ) = (BC)/(QR)`
Then, by using Theorem 6.4, ACB `~` PRQ and therefore, `/_B = /_Q`.
Example 3 : Consider ACB, rightangled at C, in which AB = 29 units, BC = 21 units and `/_ ABC = theta` Determine the values of the below fig Determine the values of
(i) `cos^2theta + sin^2theta`,
(ii) `cos^2theta – sin^2theta`.
`AC =sqrt[AB^2 − BC^2]=sqrt[(29)^2 − (21)^2]`
= `sqrt[(29 − 21)(29+21)]` =`sqrt(p(8)(50))` =`sqrt[400]` = 20 units`
So, `sin theta =(AC)/(AB)=(20)/(29), `cos theta=(BC)/(AB)=(21)/(29)`.
Now, (i) `cos^2theta + sin^2theta=(20^ 2)/(29)=(21^2)/(29)=(20^2+21^2)/(29^2)=(400 +441)/(841)` =1
and (ii) `cos^2theta + sin^2theta =(21^ 2)/(29)=(20^2)/(29)=((21+20)(2120))/(29^2)=(41)/(841)`.
Example 4 : In a right triangle ABC, rightangled at B, if tan A = 1, then verify that 2`sin A cos A `= 1.
i.e., BC = AB
Let AB = BC = k, where k is a positive number.
Now,`AC = sqrt [AB^2 +BC^2]`
= `sqrt[(k )^2 +( k )^2] = ksqrt2 `
Therefore, `sinA=(BC)/(AC) =1/sqrt2 and cosA=(AB)/(AC)=1/sqrt2 `and
`cosA=(AB)/(AC)=1/sqrt2`
so, `2sinAcosA=2(1/sqrt2)(1/sqrt2)=1`, which is the required value.
Example 5 : In OPQ, rightangled OP = 7 cm and OQ – PQ = 1 cm Determine the values of sinQ and cosQ.
`OQ^2 = OP^2 + PQ^2`
i.e., `(1 + PQ)^2 = OP^2 + PQ^2` (Why?)
i.e., `1 + PQ^2 + 2PQ = OP^2 + PQ^2`
i.e., `1 + 2PQ = 7^2` (Why?)
i.e., `PQ = 24 cm and OQ = 1 + PQ = 25 cm`
So, `sinQ = 7/(25) and cosQ = (24)/(25)`
We already know about isosceles right angle triangle and right angle triangle with angles
30°, 60° and 90°.
Can we find sin 30°
or tan 60°
or cos 45°
etc.
with the help of these triangles?
Does sin 0°
or cos 0°
exist?
From geometry, you are already familiar with the construction of
angles of 30°, 45°, 60° and 90°. In this section, we will find
the values of the trigonometric ratios for these angles and, of course, for 0°.
Trigonometric Ratios of 45° In ABC, rightangled at B, if one angle is 45°, then the other angle is also 45°, i.e.,` /_A = /_C` = 45°
So, BC = AB (Why?)
Now, Suppose BC = AB = a.
Then by Pythagoras Theorem,
`AC^2 = AB^2 + BC^2 = a^2 + a^2 = 2a^2`,
and, therefore,
`AC = asqrt2`
Using the definitions of the trigonometric ratios, we have :
`sin`45° = side opposite to angle 45° / hypotenuse
`sin`45°=`(BC)/(AC)=a/(asqrt2)=1/(sqrt2)`
`cos`45° = side adjacent to angle 45° / hypotenuse
`cos`45°=`(AB)/(AC)=a/(asqrt2)=1/(sqrt2)`
`tan45°` = side opposite to angle 45° / side adjacent to angle 45°
`tan45°=(BC)/(AC)=a/(asqrt2)=1/(sqrt2)`
Also, `csc45°=1/ sin45°=sqrt2sec45°`
`csc45°=1/(cos45°)=sqrt2cot45°
=1/[tan45°] =1`
Trigonometric Ratios of 30° and 60°
Let us now calculate the trigonometric ratios of 30° and 60°. Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°,
therefore, `/_ A = /_ B = /_C` = 60°.
Draw the perpendicular AD from A to the side BC
Now
Therefore, BD =DC
and ` /_BAD = /_CAD (CPCT)`
Now observe that:
ABD is a right triangle, right angled at D with `/_BAD` = 30° and `/_ABD` = 60°
As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle. So, let us suppose that AB = 2a.
Then, `BD=1/3 BC =a`
and `AD^2 = AB^2 – BD^2 = (2a)^2 – (a)^2 = 3a^2`,
Therefore, `AD =asqrt3`
Now, we have : `sin`30° =`(BD)/(AB)`=`a/(2a)=1/2`,
`cos`30°=`(AD)/(AB)=(asqrt3)/(2a)=sqrt3/2`
`tan`30° =`(BD)/(AD)=a/(asqrt3)=1/sqrt3`.
Also, `csc30°=1/[sin30°]=2,sec30°=1/ [cos30°]=2/sqrt3`
`cot30°=1/[tan30°]=sqrt3`.
Similarly, `sin60° =(AD)/(AB)=(asqrt3)/(2a)=sqrt3/2`,`cos60°=1/2=tan30°=sqrt3`
`csc60° =2/sqrt3,sec60°= 2` and `cot60°=1/sqrt3`
Let us see what happens to the trigonometric ratios of angle A, if it is made smaller and smaller in the right triangle ABC ( ) fig1 till it becomes zero. As `/_A` gets smaller and smaller, the length of the side BC decreases.The point C gets closer to point B, and finally when `/_A ` becomes very close to 0°, AC becomes almost the same as AB fig2
When `/_A` is very close to 0°, BC gets very close to 0 and so the value of
BC
`sin`A = AC is very close to 0. Also, when `/_A ` is very close to 0°, AC is nearly the
AB
same as AB and so the value of `cos`A = AC is very close to 1.
This helps us to see how we can define the values of `sin`A and ` cos` A when A = 0°.
We define : `sin`0° = 0 and ` cos`0° = 1.
Using these, we have :
`tan0°=[sin0°]/ [cos0°]=0,cot0°=1/[tan0°]`,which is not defined. (Why?)
`sec 0°=1/ [cos 0°]=1`,and `csc0°=1/[sin0°]`, which is again not defined.(Why?)
Now, let us see what happens to the trigonometric ratios of `/_` A, when it is made larger and larger in ABC till it becomes 90°. As `/_`A gets larger and larger, `/_`C gets smaller and smaller. Therefore, as in the case above, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when `/_`A is very close to 90°, `/_`C becomes very close to 0° and the side AC almost coincides with side BC .
When `/_`C is very close to 0°, `/_`A is very close to 90°, side AC is nearly the same as side BC, and so sin A is very close to 1. Also when `/_` A is very close to 90°, `/_`C is very close to 0°, and the side AB is nearly zero, so cos A is very close to 0.
So, we define : `sin`90° = 1 and `cos `90° = 0.
Now, why don’t you find the other trigonometric ratios of 90°?
We shall now give the values of all the trigonometric ratios of 0°, 30°, 45°, 60° and 90° in this below fig for ready reference.
Now, let us see the values of trigonometric ratios of all the above discussed angles in the
form of a table.
Think  Discuss
What can you say about the values of sin A and cos A, as the value of angle A
increases from 0°
to 90°
? (observe the above table)
If A > B, then sin A > sin B. Is it true ?
If A > B, then cos A > cos B. Is it true ? Discuss.
Example:
In ∆ABC, right angle is at B, AB = 5 cm and ∠ACB = 30°
.
Determine the lengths
of the sides BC and AC.
Solution:
Given AB=5 cm and
∠ACB=30°
.
To find the length of side BC,
we will choose the trignometric ratio involving
BC and the given side AB.
Since BC is the
side adjacent to angle C and AB is the side
opposite to angle C.
Therefore,
`(AB)/(BC)` = tan C
i.e. `5 /(BC)` = tan 30°
= `1/sqrt(3)`
which gives BC = 5 `(sqrt(3))` cm
Now, by using the Pythagoras theorem
`(AC)^2`
= `(AB)^2`
+ `(BC)^2`
`(AC)^2`
= `5^2`
+ 5 `sqrt(3)^ 2`
`(AC)^2`
= 25 + 75
AC = 100 = 10 cm
Remark : From the table above you can observe that as `/_`A increases from 0° to 90°, sin A increases from 0 to 1 and cos A decreases from 1 to 0. Let us illustrate the use of the values in the table above through some examples.
Example 6 : In ABC, rightangled at B,
AB = 5 cm and `/_`ACB = 30° .
Determine the lengths of the sides BC and AC.
Solution : To find the length of the side BC, we will
choose the trigonometric ratio involving BC and the
given side AB. Since BC is the side adjacent to angle
C and AB is the side opposite to angle C,
therefore `(AB)/(BC)=tanC`
i.e, `5/(BC)=tan30°=1/sqrt3`
which gives BC = `5sqrt3` cm
To find the length of the side AC, we consider
`sin30° =(AB)/(AC)`
(Why?)
]
i.e, `1/2=5/(AC)`
i.e., AC = 10 cm
Note that alternatively we could have used Pythagoras theorem to determine the third side in the example above,
i.e., `AC = sqrt(AB ^2 +BC^2) =sqrt [5^2 + (5 sqrt3)^2] cm = 10cm`.
Example 7 : In PQR, right  angled at Q, PQ = 3 cm and PR = 6 cm.
Determine `/_`QPR and `/_`PRQ.
Solution : Given PQ = 3 cm and PR = 6 cm.
Therefore, `(PQ)/(PR )= sin R`
or `sinR =3/6=1/2 `
So, `/_`PRQ = 30°
and therefore, `/_`QPR = 60°. (Why?)
You may note that if one of the sides and any other part (either an acute angle or any side) of a right triangle is known, the remaining sides and angles of the triangle can be determined.
Example 8 :If `sin(A – B) = 1/2,cos(A + B) =1/2`, 0° `<` A + B `<=` 90° , A `>` B, find A and B.
Solution : Since, `sin`(A – B) =`1/2`, therefore, (A – B )= 30° (Why?) (1)
Also, since ,`cos `(A + B) =` 1/2 `, therefore, (A + B) = 60° (Why?) (2)
Solving (1) and (2), we get : A = 45° and B = 15°.
Recall that two angles are said to be complementary if their sum equals 90°. In ABC, rightangled at B, do you see any pair of complementary angles?
Since `/_`A + `/_` C = 90°, they form such a pair. We have:
`sinA = (BC)/(AC) , cosA = (AB)/(AC) ,tanA = (BC)/(AB)`,
`csc A =(AC)/(BC) , sec A = (AC)/(AB) ,cot A =(AB)/(BC)` (1)
Now let us write the trigonometric ratios for `/_`C = 90° – `/_` A.
For convenience, we shall write 90° – A instead of 90° – `/_` A.
What would be the side opposite and the side adjacent to the angle 90° – A?
You will find that AB is the side opposite and BC is the side adjacent to the angle 90° – A. Therefore,
`csc`(90°A)=`(AC)/(AB)` , `sec`(90°A)= `(AC)/(BC)` `cot`(90°A) =`(BC)/(AB)` (2)
Now, compare the ratios in (1) and (2). Observe that :
`sin`(90°A) = `(AB)/(AC)`=`cos`A and `cos`(90°A)= `(BC)/(AC)`=`sin`A
`tan`(90°A) = `(AB)/(BC)`=`cot`A and `cot`(90°A)=`(BC)/(AB)`=`tan`A
Also, `csc`(90°A)=`(AC)/(AB)`=`sec`A and `sec`(90°A)=`(AC)/(BC)`=`csc`A
so,`sin(90°A) = cosA`, `cos(90°A)=sinA`
`tan(90°A) = cotA` ,`cot (90°A)= tanA`
`sec(90°A)= csc A` ` csc(90°A)= secA`
for all values of angle A lying between 0° and 90°. Check whether this holds for A = 0° or A = 90°.
Note : tan 0° = 0 = cot 90°, sec 0° = 1 = cosec 90° and sec 90°, cosec 0°, tan 90° and cot 0° are not defined.
Example:If cos 7A = sin(A − 6°), where 7A is an acute angle, find the value of A.
Solution:Given cos 7A = sin(A − 6°) ...(1)
sin (90 − 7A) = sin (A − 6°)
since (90 − 7A) and (A − 6°)
are both acute angles,
therefore
90° − 7A = A − 6°
8A = 96°
which gives A = 12°
Example:If sin A = cos B, then prove that A + B = 90°.
Solution:
Given that sin A = cos B ...(1)
We know cos B = sin (90° − B), we can write (1) as
sin A = sin (90° − B)
If A, B are acute angles, then A = 90° − B
⇒ A + B = 90°
.
Example: Express sin 81o
+ tan 81o
in terms of trigonometric ratios of angles between
0o
and 45o
Solution:
We can write sin 81°
= cos(90° − 81°
) = cos 9°
tan 81°
= tan(90° − 81°
) = cot 9°
Then, sin 81°
+ tan 81°
= cos 9°
+ cot 9°
Example 9 : Evaluate `[tan65°]/[ cot25°]`
Solution : We know :`cotA = tan(90° – A)`
So, `cot25° =tan(90° – 25°) = tan65°`
i.e.,`[tan 65°]/[cot25°]=[tan65°]/[tan65°]=1`
Example 10 : If `sin`3A = `cos`(A – 26°), where 3A is an acute angle, find the value of A.
Solution : We are given that `sin`3A = `cos`(A – 26°). (1)
Since`sin`3A = `cos`(90° – 3A), we can write (1) as
`cos`(90° – 3A) = `cos`(A – 26°)
Since 90° – 3A and (A – 26°) are both acute angles, therefore,
(90° – 3A) = (A – 26°)
which gives A = 29°
Example 11 : Express `cot`85° + `cos`75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution : `cot`85° +`cos`75° = `cot`(90° – 5°) + `cos`(90° – 15°) = `tan`5° +`sin`15°.
You may recall that an equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved.
In this section, we will prove one trigonometric
identity, and use it further to prove other useful
trigonometric identities.
`AB^2 + BC^2 = AC^2` ______________(1)
Dividing each term of (1) by `AC^2`, we get
`(AB^2)/(AC^2) = (BC^2)/(AC^2)= (AC^2)/(AC^2)`
i.e., `(AB^2)/(AC)=(BC^2)/(AC)=(AC^2)/(AC)`
i.e.,`(cosA)^2 + (sinA)^2 = 1`
i.e.,`cos^2 A + sin^2 A = 1 ` _____________________ (2)
This is true for all A such that 0°` <=` A `<=` 90°. So, this is a trigonometric identity. Let us now divide (1) by ` (AB)^2`. We get
or`(AB^2)/(AB) = (BC^2)/(AB) = (AC^2)/(AB)`
i.e.,`1 + tan^2 A = sec^2 A` ______________(3)
Let us see what we get on dividing (1) by `BC^2`. We get
i.e.,`(AB^2)/(BC) = (BC^2)/(BC) = (AC^2)/(BC)`
i.e.,`cot^2 A + 1 = csc^2 A`_________________(4)
Note that `csc `A and `cot` A are not defined for A = 0°. Therefore (4) is true for all A such that 0° `< `A `<=` 90°.
Using these identities, we can express each trigonometric ratio in terms of other trigonometric ratios, i.e., if any one of the ratios is known, we can also determine the values of other trigonometric ratios.
Let us see how we can do this using these identities. Suppose we know that
`tanA = 1/sqrt3 Then, cotA =sqrt3`.
Since, `sec^2 A = 1 + tan^2 A = 1+1/3=4/3, sec A = 2/sqrt3 , and cosA =sqrt3/2`,
Again, `sin A = sqrt( 1 − cos^ 2 A)= sqrt(1 −3/4)=1/2`. Therefore, `csc A = 2`.
Solution : Since `cos^2 A + sin ^2 A = 1`, therefore,
`cos^2 A = 1 – sin ^2 A`,
i.e., `cos A =+sqrt (1 − sin ^2 A)`
This gives `cosA = sqrt(1 − sin ^2 A)` (Why?)
Hence, `tan A = sin A/cosA = sin A/(1 – sin ^2 A) and sec A = 1/cos A =1/sqrt(1 − sin ^2 A)`
In this chapter, you have studied the following points :
1 In a right triangle ABC, rightangled at B,
(i)`sinA `= side opposite to angle A /A hypotenuse,
`cos A` = side adjacent to angle/ hypotenuse
(ii)`tan A` = side opposite to angle A/side adjacent to angle A ?
21231213434
 `csc A =1/sin A ; secA =1/cosA;
 If one of the trigonometric ratios of an acute angle is known, the remaining trigonometric ratios of the angle can be easily determined.
 The values of trigonometric ratios for angles 0°, 30°, 45°, 60° and 90°. The value of `sin`A or `cos`A never exceeds 1, whereas the value of `sec`A or `csc` A is always greater than or equal to 1.
 `sin`(90° – A) = `cosA`,
 `cos`(90° – A) = `sinA`;
 `tan` (90° – A) =`cotA`,
 ` cot`(90° – A) = `tanA`;
 `sec`(90° – A) = `cscA `,
 `csc`(90° – A) = `secA`.
 `sin^2` A + `cos^2` A = 1,
 `sec^2` A – `tan^2` A = 1 for 0° `<=` A `<` 90°, `csc^2` A = 1 + `cot^2` A for 0° ` <` A `<=`90°.
tanA =1/ cotA,
tanA =sinA/ cos A`.