Introduction to Trigonometry ⇒⇒ Exercise 8.1

Question 1

In ΔABC right angled at B, AB = 24 cm, BC = 7 cm. Determine
(i) sin A, cos A
(ii) sin C, cos C

Solution :

In ΔABC For acute angle A

According to the question Given

$$\angle B =90°, AB = 24 cm, BC = 7 cm.$$

According to Pythagoras Theorem, In a right angled triangle

$$(Hypotenuse)^2 = (perpendicular)^2 + (base)^2 $$

$$(AC )^2 = (BC)^2 + (AB)^2 $$

$$ (AC )^2 = (7)^2 + (24)^2 $$

$$ (AC )^2 = 49 + 576 $$

$$ (AC ) = \sqrt{49 + 576} $$

$$ (AC ) = \sqrt{625} $$

$$ (AC ) = 25 cm $$

Therefore For acute angle A

The hypotenuse is AC = 25 cm.

The side adjacent to angle A is = Base = AB = 24 cm.

The side opposite to angle A is BC = 7 cm.

$$ sin A = {Perpendicular \over Hypotenuse} = {BC \over AC } = {7 \over 25} $$

$$ cos A = { Base \over Hypotenuse} = {AB \over AC }= { 24\over 25 }$$

(ii) Answer

In ΔABC For acute angle C

According to the question Given

$$\angle B =90°, AB = 24 cm, BC = 7 cm.$$

Hypotenuse = AC = 25 cm (as calculated)

Perpendicular = AB = 24 cm (given)

Base = BC = 7 cm (given)

Therefore For acute angle C

$$ sin C = { Base \over Hypotenuse} = {AB \over AC }= { 24\over 25 }$$

$$ cos C = {Perpendicular \over Hypotenuse} = {BC \over AC } = {7 \over 25} $$

Question 2

In Figure, find tan P – cot R.

Solution :

In ΔPQR For acute angle P

According to the question Given

$$\angle Q =90°, PQ = 12 cm, PR = 13 cm.$$

According to Pythagoras Theorem, In a right angled triangle

$$(Hypotenuse)^2 = (perpendicular)^2 + (base)^2 $$

$$(PR )^2 = (QR)^2 + (PQ)^2 $$

$$ (13 )^2 = (QR)^2 + (12)^2 $$

$$ 169 = (QR)^2 + 144 $$

$$ 169 - 144 = (QR)^2 $$

$$ \sqrt{25} = QR $$

$$ (QR ) = 5 cm $$

Therefore For acute angle P

The side opposite to angle P is QR = 5 cm.

The side adjacent to angle P is PQ = 12 cm.

tan P = $ Perpendicular \over Base$ = QR / PQ = $5 \over 12 $

In ΔPQR For acute angle R

According to the question Given

Hypotenuse = PR = 13 cm

Perpendicular = The side opposite to angle R is PQ = 12 cm.

Base = The side adjacent to angle R is QR = 5 cm.

Therefore For acute angle R

cot R = $ Base \over Perpendicular$ = QR / PQ = $5 \over 12 $

tan P – cot R = $$ {5 \over 12} - {5 \over 12 } = 0 $$

Thus, tanP−cotR = 0 .

Question 3

If sin A = $ 3\over 4 $ , calculate cos A and tan A.

Solution :

In ΔABC For acute angle A

According to the question suppose

$\angle $ B = 90°, AC = 4 unit, BC = 3 unit.

We have sin A =$ BC\over AC $ = $ 3\over 4 $

According to Pythagoras Theorem, In a right angled triangle

$$(Hypotenuse)^2 = (perpendicular)^2 + (base)^2 $$

$$⇒ (AC )^2 = (BC)^2 + (AB)^2 $$

$$⇒ (4 )^2 = (3)^2 + (AB)^2 $$

$$⇒ 16 = 9 + (AB)^2 $$

$$⇒ (AB)^2 = 16 - 9 $$

$$⇒ (AB ) = \sqrt{7} $$

$$⇒ (AB ) = \sqrt{7} unit $$

Therefore For acute angle A

The hypotenuse is AC = 4 cm.

The side adjacent to angle A is = Base = AB = $\sqrt{7}$ cm.

The side opposite to angle A is BC = 3 cm.

$$ cos A = { Base \over Hypotenuse} = {AB \over AC }= { \sqrt{7} \over 4 }$$

$$ tan A = { Perpendicular \over Base} = {BC \over AB }= { 3 \over \sqrt{7} }$$

Question 4

Given 15 cot A = 8, find sin A and sec A.

Solution :

In ΔABC For acute angle A

According to the question suppose

$\angle $ B = 90°

We have , 15 cot A = 8

cot A = $ 8\over 15 $

cot A = $ Base\over Perpendicular $ = $ AB\over BC $ ,
AB = 8 unit, BC = 15 unit.

In ΔABC For acute angle A

According to Pythagoras Theorem, In a right angled triangle

$$(Hypotenuse)^2 = (perpendicular)^2 + (base)^2 $$

$$⇒ (AC )^2 = (BC)^2 + (AB)^2 $$

$$⇒ (AC )^2 = (15)^2 + (8)^2 $$

$$⇒ (AC )^2 = (225 + 64 )$$

$$⇒ (AC )^2 = 289 $$

$$⇒ (AC ) = \sqrt{289} $$

$$⇒ (AC ) = 17 unit $$

Therefore For acute angle A

The hypotenuse is AC = 17 cm.

The side adjacent to angle A is = Base = AB = 8 cm.

The side opposite to angle A is BC = 15 cm.

$$ sin A = {Perpendicular \over Hypotenuse} = {BC \over AC } = {15 \over 17} $$

$$ sec A = { Hypotenusee \over Base} = {AC \over AB }= { 17 \over 8 }$$

Question 5

Given sec θ = $ 13\over 12 $, calculate all other trigonometric ratios.

Solution :

In ΔABC For acute angle A = θ°

According to the question suppose

$\angle $ B =90°

We have ,

sec θ = $ 13\over 12 $

sec θ = $ Hypotenuse\over Base$ = $ AC\over AB $ ,
AB = 12 unit, AC = 13 unit.

In ΔABC For acute angle A

According to Pythagoras Theorem, In a right angled triangle

$$(Hypotenuse)^2 = (perpendicular)^2 + (base)^2 $$

$$⇒ (AC )^2 = (BC)^2 + (AB)^2 $$

$$⇒ (AC )^2 = (BC)^2 + (AB)^2 $$

$$⇒ (13 )^2 = (BC)^2 + (12)^2 $$

$$⇒ 169 = (BC)^2 + 144 $$

$$⇒ (BC)^2 = 169 - 144 $$

$$⇒ (BC) = \sqrt{25} $$

$$⇒ (BC ) = 5 unit $$

Therefore For acute angle A

The hypotenuse is AC = 13 cm.

The side adjacent to angle A is = Base = AB = 12 cm.

The side opposite to angle A is BC = 5 cm.

$$ sin θ = {Perpendicular \over Hypotenuse} = {BC \over AC } = {5 \over 13} $$

$$ cos θ = { Base \over Hypotenuse} = {AB \over AC }= { 12 \over 13 }$$

$$ tan θ = { Perpendicular \over Base} = {BC \over AB }= { 5 \over 12 }$$

$$ cosec θ = { Hypotenuse\over Perpendicular} = {AC \over BC } = { 13 \over 5} $$

$$ cot θ = {$ Base \over Perpendicular} = {AB \over BC }= { 12 \over 5} $$

Question 6

If $\angle $ A and $\angle $ B are acute angles such that cos A = cos B, then show that $\angle $ A = $\angle $ B.

Solution :

Let Consider a right-angled triangle ABC, with the right angle at C.

In this triangle, the side opposite the right angle is AB

We have ,

for acute $\angle $ A

$$ cos A = { Base \over Hypotenuse} = {AC \over AB }$$

Now, for acute $\angle $ B

$$ cos B = { Base \over Hypotenuse} = {BC \over AB }$$

According to the question

cos A = cos B

⇒Therefore , $$ {AC \over AB } = {BC \over AB } $$

After cross multiplication

$$ ⇒ AC = BC $$

Since two sides AC and BC are equal. Thus, The given triangle is an isosceles triangle

Since the sides opposite to angles A and B are equal, the angles themselves must be equal.

Hence $\angle $ A = $\angle $ B

Question 7

If cot θ = = $ 7\over 8 $, evaluate
(i) $$ {{(1+sinθ)(1−sinθ)}\over {(1+cosθ)(1−cosθ)}}$$ (ii) $$ (cot)^2θ$$

Solution :

Suppose In Δ ABC For acute angle A = θ°

and $\angle $ B =90°

We have , cot θ = $ 7\over 8 $

cot θ = $ Base \over Perpendicular $ = ${AB \over BC } $
= AB = 7 unit, BC = 8 unit.

In ΔABC For acute angle A

According to Pythagoras Theorem, In a right angled ΔABC

$$(Hypotenuse)^2 = (perpendicular)^2 + (base)^2 $$

$$⇒ (AC )^2 = (BC)^2 + (AB)^2 $$

$$⇒ (AC )^2 = (8)^2 + (7)^2 $$

$$⇒ (AC )^2 = 64 + 49 $$

$$⇒ (AC )^2 = 113 $$

$$⇒ AC = \sqrt{113} $$

$$⇒ AC = \sqrt{113} unit $$

Therefore

$$ sin θ = {Perpendicular \over Hypotenuse} = {BC \over AC } = {8 \over \sqrt{113}} $$

$$ cos θ = { Base \over Hypotenuse} = {AB \over AC }= { 7 \over \sqrt{113} }$$

(i) $ {{(1+sinθ)(1−sinθ)}\over {(1+cosθ)(1−cosθ)}}$

$$ {{(1+sinθ)(1−sinθ)}\over {(1+cosθ)(1−cosθ)}}$$

$$ {{(1-sin^2θ)}\over {(1−cos^2θ)}}$$

We can use the algebraic identity$$ [∵{{(a+b)(a−b)} = (a^2-b^2)}]$$

Put the values of above trigonometric ratios, we get

$$ ⇒ {{(1-{8 \over \sqrt{113}}^2)}\over {(1−{ 7 \over \sqrt{113} }^2)}}$$

$$⇒ {{(1-{64 \over 113})}\over {(1−{ 49 \over 113 })}}$$

$$ ⇒ {49 \over 113} \over { 64 \over 113 }$$

$$ ⇒ {49 \over 64}$$

Therefore , $ {{(1+sinθ)(1−sinθ)}\over {(1+cosθ)(1−cosθ)}}$ = $ ⇒ {49 \over 64}$

(ii) $ (cot)^2θ$

$$ ⇒ (cot)^2θ$$

We are given that cotθ $ 7\over 8 $

$$⇒ (cot)^2θ = ({ 7 \over 8 })^2$$

$$ ⇒ (cot)^2θ = { 49 \over 64 }$$

Question 8

If 3cotA = 4, check whether
(i) $$ {{(1 - tan^2A )}\over {(1 + tan^2A )}} = cos^2A - sin^2A $$ or not.

Solution :

In ΔABC For acute angle A

According to the question suppose

$\angle $ B =90°

We have ,

3cotA = 4

$$ cotA = { 4\over 3} $$

cotA = $ {Base \over Perpendicular} = {AB \over BC } $ ,

AB = 4 unit, BC = 3 unit.

In ΔABC For acute angle A

According to Pythagoras Theorem, In a right angled ΔABC

$$(Hypotenuse)^2 = (perpendicular)^2 + (base)^2 $$

$$⇒ (AC )^2 = (BC)^2 + (AB)^2 $$

$$⇒ (AC )^2 = (3)^2 + (4)^2 $$

$$⇒ (AC )^2 = 9 + 16 $$

$$⇒ (AC )^2 = 25 $$

$$⇒ AC = \sqrt{25} $$

$$⇒ AC = 5 unit $$

Therefore For acute angle A

$$ sin A = {Perpendicular \over Hypotenuse} = {BC \over AC } = {3 \over 5} $$

$$ cos A = { Base \over Hypotenuse} = {AB \over AC }= { 4 \over 5 }$$

$$ tan A = { Perpendicular \over Base} = {BC \over AB }= { 3 \over 4 }$$

L.H.S.

$$ {{(1 - tan^2A )}\over {(1 + tan^2A )}} $$

Put the values of above trigonometric ratios, we get

$$⇒ {{1 - ({ 3 \over 4 })^2 }\over {1 + ({ 3 \over 4 })^2 }} $$

$$⇒ {{1 - ({ 9 \over 16 }) }\over {1 + ({ 9 \over 16}) }} $$

$$⇒ {{16 - 9 }\over {16 + 9 }} $$

$$⇒ L.H.S = {7 \over 25 } $$

R.H.S.

$$ cos^2A - sin^2A $$

Put the values of above trigonometric ratios, we get

$$⇒ {({4 \over 5})^2 } - {({3 \over 5})^2 } $$

$$⇒ {16 \over 25} - {9 \over 25} $$

$$⇒ {{16 - 9 } \over 25} $$

$$⇒ R.H.S. = {7 \over 25 } $$

Hence, proved that $ {{(1 - tan^2A )}\over {(1 + tan^2A )}} = cos^2A - sin^2A $

Question 9

In ΔABC, right angled at B. If tan A = = $ 1\over \sqrt3 $, find the value of
(i) sin A cos C + cos A sin C
(ii) cos A cos C − sin A sin C

Solution :

In ΔABC For acute angle A

According to the question suppose

$\angle $ B =90°

tan A = = $ 1\over \sqrt3 $

tan A = $ Perpendicular \over Base $ = ${BC \over AB } $= AB = $ \sqrt 3 $ unit, BC = 1 unit.

In ΔABC For acute angle A

According to Pythagoras Theorem, In a right angled ΔABC

$$(Hypotenuse)^2 = (perpendicular)^2 + (base)^2 $$

$$⇒ (AC )^2 = (BC)^2 + (AB)^2 $$

$$⇒ (AC )^2 = (1)^2 + (\sqrt 3)^2 $$

$$⇒ (AC )^2 = 1 + 3 $$

$$⇒ (AC )^2 = 4 $$

$$⇒ AC = \sqrt{4} $$

$$⇒ AC = 2 unit $$

Therefore For acute angle A

$$ sin A = {Perpendicular \over Hypotenuse} = {BC \over AC } = {1 \over 2} $$

$$ sin C = { Perpendicular \over Hypotenuse} = {AB \over AC }= { \sqrt 3 \over 2 }$$

$$ cos A = {Base \over Hypotenuse} = { AB \over AC } = {\sqrt 3\over 2} $$

$$ cos C = { Base \over Hypotenuse} = {BC \over AC }= { 1 \over 2 }$$

(i) sin A cos C + cos A sin C

$$ sin A cos C + cos A sin C $$

Put the values of above trigonometric ratios, we get

$$ ⇒ { {1 \over 2} ×{ 1 \over 2 } } + { {\sqrt 3\over 2} × {\sqrt 3\over 2}} $$

$$ ⇒ { {1 \over 4} } + { 3\over 4} $$

$$ ⇒ {4 \over 4 }$$

$$ ⇒ 1 $$

Hence, sin A cos C + cos A sin C = 1.

(ii) cos A cos C − sin A sin C

$$ ⇒ { {\sqrt 3\over 2} ×{ 1 \over 2 } } - { { 1 \over 2 } × {\sqrt 3\over 2}} $$

$$ ⇒ { {\sqrt 3\over 4} } - { {\sqrt 3\over 4}} $$

$$ ⇒ {0} $$

Hence, cos A cos C - sin A sin C = 0.

Question 10

In ΔPQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Solution :

In ΔPQR For acute angle P

According to the question Given

$$ \angle Q =90°, PR + QR = 25 cm, PQ = 5 cm.$$

Let PR = x cm

Therefore, We know that PR + QR=25, which can be rearranged to express QR in terms of PR

QR = 25 - PR

QR = 25 - x ... (i)

According to Pythagoras Theorem, In a right angled triangle

$$(Hypotenuse)^2 = (perpendicular)^2 + (base)^2 $$

$$(PR )^2 = (QR)^2 + (PQ)^2 $$

$$ (x )^2 = (25 - x)^2 + (5)^2 $$

$$ (x )^2 = (25)^2 + (x)^2 - 50x + 25 $$

$$ (x )^2 = 625 + (x)^2 - 50x + 25 $$

$$ (x )^2 - (x)^2 = 650 - 50x $$

$$ 50x = 650 $$

$$ x = { 650 \over 50} $$

$$ x = 13 $$

$$ PR = x = 13 cm $$

Now,Put the values of PR in expression (i)

$$ QR = 25 - PR ... (i)$$

$$ QR = 25 - 13 $$

$$ QR = 12 cm $$

So, the lengths of the sides are PQ = 5 cm, QR = 12 cm, and PR = 13 cm.

Therefore For acute angle P

$$ sin P = { Perpendicular \over Hypotenuse } = {QR \over PR } = {12 \over 13 } $$

$$ cos P = {Base \over Hypotenuse} = {PQ \over PR } = {5 \over 13 }$$

$$ tan P = { Perpendicular \over Base} = {QR \over PQ } = {12 \over 5} $$

Question 11

State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii)sec A = $ 12\over 5$ , for some value of $ \angle$ A .
(iii)cos A is the abbreviation used for the cosecant of $ \angle$ A
(iv) cot A is the product of cot and A.
(v) sinθ = $ 4\over 3 $ , for some $ \angle$ θ

Solution :

(i) False,

The value of tanA can be equal to, less than, or greater than 1.

The tangent of an acute angle in a right-angled triangle is the ratio of the side opposite to the angle to the side adjacent to it .

This ratio can have any positive value depending on the lengths of the sides.
In a right-angled triangle where the opposite side is longer than the adjacent side, tanA will be greater than 1 .

(ii) True, Since, 12/5 is greater than 1

In a right-angled triangle, the hypotenuse is always the longest side. This means the ratio of the hypotenuse to any other side must always be greater than 1.

(iii) False

cosA is the abbreviation used for the cosine of angle A. The abbreviation for the cosecant of angle A is cosec A.

(iv) False

cotA is not a product. It is a single term representing the cotangent of angle A.
The letters 'c', 'o', and 't' together define the trigonometric ratio (cotangent),

and 'A' specifies the angle to which the ratio applies. The term has no meaning if 'A' is separated from 'cot'.

(v) False

The sine of an angle is the ratio of the side opposite to the angle to the hypotenuse (sinθ = hypotenuse /opposite ).

In any right-angled triangle, the hypotenuse is always the longest side. Therefore, the length of the opposite side can never be greater than the length of the hypotenuse.

This means the value of the ratio hypotenuse /opposite can never be greater than 1.

Since 4/3 = 1.33, which is greater than 1, this value is impossible for sinθ.