Class 9: Trigonometric Ratios – Exercise 18.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Evaluate each of the following:

Answers:

i) $cosec \ 30^o + \cot 45^o = 2 + 1 = 3$

ii) $\cos 30^o \cos 45^o - \sin 30^o \sin 45^o =$ $\frac{\sqrt{3}}{2}$ $.$ $\frac{1}{\sqrt{2}}$ $-$ $\frac{1}{2}$ $.$ $\frac{1}{\sqrt{2}}$ $=$ $\frac{\sqrt{3} -1}{2 \sqrt{2}}$

iii) $\tan 30^o \sec 45^o + \tan 60^o \sec 30^o =$ $\frac{1}{\sqrt{3}}$ $. \sqrt{2} + \sqrt{3} .$ $\frac{2}{\sqrt{3}}$ $=$ $\frac{\sqrt{2}+ 2 \sqrt{3}}{\sqrt{3}}$

iv) $\sin 30^o \cos 45^o + \cos 30^o \sin 45^o =$ $\frac{1}{2}$ $.$ $\frac{1}{\sqrt{2}}$ $+$ $\frac{\sqrt{3}}{2}$ $.$ $\frac{1}{\sqrt{2}}$ $=$ $\frac{\sqrt{3}+1}{2 \sqrt{2}}$

v) $\frac{\sin^2 45^o + \cos^2 45^o}{\tan^2 60^o}$ $=$ $\frac{(\frac{1}{\sqrt{2}})^2 +(\frac{1}{\sqrt{2}})^2 }{(\sqrt{3})^2}$ $=$ $\frac{\frac{1}{2}+ \frac{1}{2}}{3}$ $=$ $\frac{1}{3}$

vi) $\frac{\sin 30^o - \sin 90^o + 2 \cos 0^o}{\tan 30^o \tan 60^o}$ $=$ $\frac{\frac{1}{2} - 1 + 2 \times 1}{\frac{1}{\sqrt{3}} \times \sqrt{3}}$ $=$ $\frac{3}{2}$

vii) $\frac{\sin 60^o}{\cos^2 45^o}$ $- \cot 30^o + 15 \cos 90^o =$ $\frac{\frac{\sqrt{3}}{2}}{(\frac{1}{\sqrt{2}})^2}$ $- \sqrt{3} + 15 \times 0 = \sqrt{3} -\sqrt{3} = 0$

viii) $\frac{5 \sin^2 30^o + \cos^2 45^o - 4 \tan^2 30^o}{2 \sin 30^o \cos 30^o + \tan 45^o}$

$=$ $\frac{5 \Big(\frac{1}{2}\Big)^2 + \Big(\frac{1}{\sqrt{2}}\Big)^2 - 4 \Big(\frac{1}{\sqrt{3}}\Big)^2 }{2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} + 1}$ $=$ $\frac{\frac{5}{4} + \frac{1}{2} - \frac{4}{3}}{\frac{\sqrt{3}}{2} +1}$ $=$ $\frac{\frac{5}{12}}{\Big(\frac{\sqrt{3}+2}{2}\Big)}$ $=$ $\frac{5}{6}$ $(2 - \sqrt{3})$

$\\$

Question 2: Find the value of $\theta$ in each of the following when $0^o < \theta < 90^o$ :

Answers:

i) $2 \sin 2 \theta = \sqrt{3}$

$\Rightarrow \sin 2 \theta =$ $\frac{\sqrt{3}}{2}$

$\Rightarrow \sin 2 \theta = \sin 60^o$

$\Rightarrow 2 \theta = 60^o$ $\Rightarrow \theta = 30^o$

ii) $2 \cos 3\theta = 1$

$\Rightarrow \cos 3 \theta =$ $\frac{1}{2}$

$\Rightarrow \cos 3 \theta = \cos 60^o$

$\Rightarrow 3 \theta = 60^o$ $\Rightarrow \theta = 20^o$

iii) $\sqrt{3} \tan 2 \theta - 3 = 0$

$\Rightarrow \tan 2 \theta =$ $\frac{3}{\sqrt{3}}$ $= \sqrt{3}$

$\Rightarrow \tan 2 \theta = \tan 60^o$

$\Rightarrow 2 \theta = 60^o$ $\Rightarrow \theta = 30^o$

iv) $2 \cos \theta = 1$

$\Rightarrow \cos \theta =$ $\frac{1}{2}$

$\Rightarrow \cos \theta = \cos 60^o$

$\Rightarrow \theta = 60^o$

v) $2 \cos^2 \theta =$ $\frac{1}{2}$

$\Rightarrow \cos^2 \theta =$ $\frac{1}{4}$ $\Rightarrow \cos \theta =$ $\frac{1}{2}$

$\Rightarrow \cos \theta = \cos 60^o$ $\Rightarrow \theta = 60^o$

vi) $2 \sin^2 \theta =$ $\frac{1}{2}$

$\Rightarrow \sin^2 \theta =$ $\frac{1}{4}$ $\Rightarrow \sin \theta =$ $\frac{1}{2}$

$\Rightarrow \sin \theta = \sin 30^o$ $\Rightarrow \theta = 30^o$

vii) $3 \tan^2 \theta - 1 = 0$

$\Rightarrow \tan^2 \theta =$ $\frac{1}{3}$ $\Rightarrow \tan \theta =$ $\frac{1}{\sqrt{3}}$

$\Rightarrow \tan \theta = \tan 30^o$ $\Rightarrow \theta = 30^o$

$\\$

Question 3: If $\theta$ is an acute angle and $\tan \theta + \cot \theta = 2$ , find the value of $\tan^7 \theta + \cot^7 \theta$ .

Answers:

Given $\tan \theta + \cot \theta = 2$

$\tan \theta +$ $\frac{1}{ \tan \theta }$ $= 2$

$\tan^2 \theta - 2 \tan \theta + 1 = 0$

$\Rightarrow (\tan \theta -1)^2 = 0$

$\Rightarrow \tan \theta -1 = 0$

$\Rightarrow \tan \theta = 1$

Therefore $\cot \theta = 1$

Hence $\tan^7 \theta + \cot^7 \theta = 1^7 + 1^7 = 2$

$\\$

Question 4: if $x = 30^o$ , verify that

Answers:

i) $\sin 3x = 3 \sin x - 4 \sin^3 x$

Given $x = 30^o$

Therefore $\sin x =$ $\frac{1}{2}$ ; $\sin 3x = \sin 90^o = 1$

Therefore LHS $= 3 \sin x - 4 \sin^3 x$

$= 3 \Big(\frac{1}{2}\Big) - 4\Big(\frac{1}{2}\Big)^3$

$=$ $\frac{3}{2}$ $-$ $\frac{1}{2}$ $= 1$

Therefore RHS = LHS. Hence proved.

ii) $\cos 3x = 4 \cos^3 x - 3 \cos x$

Given $x = 30^o$

Therefore $\cos x =$ $\frac{\sqrt{3}}{2}$ ; $\cos 3x = \cos 90^o = 0$

RHS $= 4 \Big($ $\frac{\sqrt{3}}{2}$ $\Big)^3 - 3 \Big($ $\frac{\sqrt{3}}{2}$ $\Big) =$ $\frac{3\sqrt{3}}{2}$ $-$ $\frac{3\sqrt{3}}{2}$

LHS $= \cos 90^o = 0$

Therefore LHS = RHS. Hence proved.

iii) $\tan 2x =$ $\frac{2 \tan x}{1 - \tan^2 x}$

LHS $= \tan 2x = \tan 60^o = \sqrt{3}$

RHS = $\frac{2 \tan x}{1 - \tan^2 x}$ = $\frac{2 \tan 30^o}{1 - \tan^2 30^o}$

$=$ $\frac{2. \frac{1}{\sqrt{3}}}{1- \frac{1}{3}}$ $= \sqrt{3}$

iv) $\sin x =$ $\sqrt{ \frac{1- \cos 2x}{2} }$

LHS $= \sin x = \sin 30^o =$ $\frac{1}{2}$

RHS $=$ $\sqrt{ \frac{1- \cos 2x}{2} }$ $=$ $\sqrt{ \frac{1- \cos 60^o}{2} }$ $=$ $\sqrt{ \frac{1- \frac{1}{2} }{2} }$ $=$ $\frac{1}{2}$

$\\$

Question 5: Find an acute angle $\theta$ when $\frac{\cos \theta - \sin \theta}{ \cos \theta + \sin \theta}$ $=$ $\frac{1 - \sqrt{3}}{1 + \sqrt{3}}$

Answers:

Given $\frac{\cos \theta - \sin \theta}{ \cos \theta + \sin \theta}$ $=$ $\frac{1 - \sqrt{3}}{1 + \sqrt{3}}$

Applying componendo and dividendo we get

$\frac{(\cos \theta - \sin \theta) + (\cos \theta + \sin \theta)}{(\cos \theta - \sin \theta) - (\cos \theta + \sin \theta)}$ $=$ $\frac{(1 - \sqrt{3}) + (1 + \sqrt{3})}{(1 - \sqrt{3}) - (1 + \sqrt{3})}$

$\frac{2 \cos \theta}{-2 \sin \theta}$ $=$ $\frac{2}{2\sqrt{3}}$

$\Rightarrow \cot \theta$ $=$ $\frac{1}{\sqrt{3}}$

$\Rightarrow \tan \theta = \sqrt{3} = \tan 60^o$

$\Rightarrow \theta = 60^o$

$\\$

Question 6: If $\sin (A+B) = 1$ and $\cos (A-B) =$ $\frac{\sqrt{3}}{2}$ $, 0^o < A + B \leq 90^o, A > B$ , then find $A, B$ .

Answers:

Given $\sin (A+B) = 1$

$\Rightarrow \sin (A+B) = \sin 90^o$

$\Rightarrow A + B = 90^o$ … … … … … i)

Similarly, $\cos (A-B) =$ $\frac{\sqrt{3}}{2}$

$\Rightarrow \cos (A-B) = \cos 30^o$

$\Rightarrow A - B = 30^o$ … … … … … ii)

Solving i) and ii) we get

$A = 60^o$ and $B = 30^o$

$\\$

Question 7: $ABC$ is a right triangle, right angled at $C$ . If $A = 30^o$ and $AB = 40$ units, find remaining two sides and $\angle B$ in $\triangle ABC$ .

Answers:

Given $\angle A = 30^o$ and $\angle C = 90^o$

Since $\angle A + \angle B + \angle C = 180^o$

$\Rightarrow \angle B = 60^o$

$\frac{BC}{AB}$ $= \sin 30^o$

$\Rightarrow BC = 40 \times$ $\frac{1}{2}$ $= 20$ units

Similarly, $\frac{AC}{40}$ $= \cos 30^o$

$\Rightarrow AC = 40 \times$ $\frac{\sqrt{3}}{2}$ $= 20\sqrt{3}$ units

Hence $AC = 20\sqrt{3}$ units, $BC = 20$ units and $\angle B = 60^o$

$\\$

Question 8: A rhombus of side $20$ cm has two angles of $60^o$ each. Find the length of the diagonals.

Answers:

Given $ABCD$ is a rhombus.

$\Rightarrow AB = BC = CD = DA = 20$ cm

Property of rhombus: Diagonals are perpendicular bisectors and $AC$ and $BD$ are bisectors of $\angle A$ and $\angle B$

$\cos \angle BAO =$ $\frac{OA}{AB}$

$\cos 30^o =$ $\frac{OA}{20}$ $\Rightarrow OA = 20 \times$ $\frac{\sqrt{3}}{2}$ $= 10\sqrt{3}$

Similarly, $\sin \angle BAO =$ $\frac{OB}{AB}$

$\sin 30^o =$ $\frac{OB}{20}$ $\Rightarrow OB =$ $\frac{20}{2}$ $= 10$

Therefore $AC = 2 \times OA = 20\sqrt{3}$ cm and $DB = 2 \times OB = 20$ cm

$\\$

Question 9: An equilateral triangle is inscribed in a circle of radius $latex 6 cm. Find the side.

Answers:

Given $OA = OB = OC = 6$ cm

$OD \perp BC$ , then $D$ is the mid point of $BC. OB$ and $OC$ are bisectors of $\angle B$ and $\angle C$ .

therefore $\angle OBD = 30^o$

In $\triangle OBD, \angle D = 90^o$ we have

$\angle OBD = 30^o$ and $OB = 6$ cm

Therefore $\cos \angle OBD =$ $\frac{BD}{OB}$

$\Rightarrow \cos 30^o =$ $\frac{BD}{OB}$

$\Rightarrow \cos 30^o =$ $\frac{BD}{6}$

$\Rightarrow BD = 6 \times$ $\frac{\sqrt{3}}{2}$ $= 3\sqrt{3}$ cm

$\Rightarrow BC = 2 BD = 2 \times 3\sqrt{3} = 6\sqrt{3}$ cm

$\\$

Question 10: If each of $\alpha, \beta$ and $\gamma$ are positive acute angle, such that $\sin (\alpha + \beta - \gamma) =$ $\frac{1}{2}$ , $\cos (\beta + \gamma - \alpha) =$ $\frac{1}{2}$ and $\tan (\gamma + \alpha - \beta) = 1$ , find the value of $\alpha, \beta$ and $\gamma$

Answers:

Given $\sin (\alpha + \beta - \gamma) =$ $\frac{1}{2}$

$\Rightarrow$ $\sin (\alpha + \beta - \gamma) = \sin 30^o$

$\Rightarrow \alpha + \beta - \gamma = 30^o$ … … … … … i)

Similarly, $\cos (\beta + \gamma - \alpha) =$ $\frac{1}{2}$

$\Rightarrow$ $\cos (\beta + \gamma - \alpha) = \cos 60^o$

$\Rightarrow \beta + \gamma - \alpha = 60^o$ … … … … … ii)

and $\tan (\gamma + \alpha - \beta) = 1$

$\Rightarrow$ $\tan (\gamma + \alpha - \beta) = \tan 45^o$

$\Rightarrow \gamma + \alpha - \beta = 45^o$ … … … … … iii)

Adding i) , ii) and iii) we get

$\alpha + \beta + \gamma = 135^o$ … … … … … iv)

Now iv) – i) $\Rightarrow 2 \gamma = 105^o \Rightarrow \gamma = 52\frac{1}{2}^o$

iv) – ii) $\Rightarrow 2 \alpha = 75^o \Rightarrow \alpha = 37 \frac{1}{2}^o$

iv) – i) $\Rightarrow 2 \beta = 90^o \Rightarrow \beta = 45^o$

$\\$

Question 11: In an acute angles triangle $ABC$ , if $\tan (A + B - C) = 1$ and $\sec (B + C - A) = 2$ , find the value of $A, B$ and $C$ .

Answers:

Given $\tan (A + B - C) = 1$

$\Rightarrow \tan (A + B - C) = \tan 45^o$

$\Rightarrow A + B - C = 45^o$ … … … … … i)

Also $\sec (B + C - A) = 2$

$\Rightarrow \sec B +C - A = \sec 60^o$

$\Rightarrow B + C - A = 60^o$ … … … … … ii)

Adding i) and ii) we get $2 B = 105^o \Rightarrow B = 52\frac{1}{2}^o$

Substituting in ii) we get

$52\frac{1}{2}^o + C - A = 60^o \Rightarrow C - A = 7\frac{1}{2}^o$ … … … … … iii)

We know $A + B + C = 180^o$

$\Rightarrow A + 52\frac{1}{2}^o + C = 180^o$

$\Rightarrow A + C = 127\frac{1}{2}^o$ … … … … … iv)

Adding iii) and iv) we get

$2C = 135^o \Rightarrow C = 67\frac{1}{2}^o$

Therefore $A = 127\frac{1}{2}^o - 67\frac{1}{2}^o = 60^o$

Hence $A = 60^o, B = 52\frac{1}{2}^o$ and $C = 67\frac{1}{2}^o$

$\\$

Question 12: Evaluate

$\sin^2 30^o \cos 45^o + 4 \tan^2 30^o +$ $\frac{1}{2}$ $\sin^2 90^o - 2 \cos^2 90^o +$ $\frac{1}{24}$ $\cos^2 0^o$

Answers:

$\sin^2 30^o \cos 45^o + 4 \tan^2 30^o +$ $\frac{1}{2}$ $\sin^2 90^o - 2 \cos^2 90^o +$ $\frac{1}{24}$ $\cos^2 0^o$

$= \Big($ $\frac{1}{2}$ $\Big)^2 \times \Big($ $\frac{1}{\sqrt{2}}$ $\Big)^2 + 4 \times \Big($ $\frac{1}{\sqrt{3}}$ $\Big)^2 +$ $\frac{1}{2}$ $(1) - 2 (0)^2 +$ $\frac{1}{24}$ $(1)^2$

$=$ $\frac{1}{4}$ $\times$ $\frac{1}{2}$ $+$ $\frac{4}{3}$ $+$ $\frac{1}{2}$ $+$ $\frac{1}{24}$

$=$ $\frac{1}{8}$ $+$ $\frac{4}{3}$ $+$ $\frac{1}{2}$ $+$ $\frac{1}{24}$

$=$ $\frac{3+32+12+1}{24}$

$=$ $\frac{48}{24}$ $= 2$

$\\$

Question 13: Evaluate: $\cot^2 30^o - 2 \cos^2 60^o -$ $\frac{3}{4}$ $\sec^2 45^o - 4 \sec^2 30^o$

Answers:

$\cot^2 30^o - 2 \cos^2 60^o -$ $\frac{3}{4}$ $\sec^2 45^o - 4 \sec^2 30^o$

$= (\sqrt{3})^2 - 2 \Big($ $\frac{1}{2}$ $\Big)^2 -$ $\frac{3}{4}$ $(\sqrt{2})^2 - 4\Big($ $\frac{2}{\sqrt{3}}$ $\Big)^2$

$= 3 -$ $\frac{1}{2}$ $-$ $\frac{3}{2}$ $- \frac{16}{3}$

$= \frac{18-3-9-32}{6}$ $=$ $\frac{-26}{6}$ $=$ $\frac{-13}{3}$

$\\$

Question 14: Find the value of $x$ i) $2 \sin 3x = \sqrt{3}$ ii) $\sqrt{3} \sin x =\cos x$

Answers:

i) $2 \sin 3x = \sqrt{3}$

$\Rightarrow \sin 3x =$ $\frac{\sqrt{3}}{2}$

$\Rightarrow \sin 3x = \sin 60^o$

$\Rightarrow 3x = 60^o$

$\Rightarrow x = 20^o$

ii) $\sqrt{3} \sin x = \cos x$

$\Rightarrow \tan x =$ $\frac{1}{\sqrt{3}}$

$\Rightarrow \tan x = \tan 30^o$

$\Rightarrow x = 30^o$

$\\$

Question 15: If $A = B = 60^o$ , verify

i) $\cos (A - B) = \cos A \cos B + \sin A \sin B$

ii) $\sin (A - B) = \sin A \cos B - \cos A \sin B$

iii) $\tan (A - B) =$ $\frac{\tan A - \tan B}{1 + \tan A \tan B}$

Answers:

Given $A = B = 60^o$ we have

$\sin A =$ $\frac{\sqrt{3}}{2}$ ; $\cos A =$ $\frac{1}{2}$

$\sin B =$ $\frac{\sqrt{3}}{2}$ ; $\cos B =$ $\frac{1}{2}$

$\cos 0^o = 1$ ; $\sin 0^o = 0$

i) LHS $= \cos (A - B) = \cos (60^o-60^o) = \cos 0^o = 1$

RHS $= \cos A \cos B + \sin A \sin B =$ $\frac{1}{2}$ $\times$ $\frac{1}{2}$ $+$ $\frac{\sqrt{3}}{2}$ $\times$ $\frac{\sqrt{3}}{2}$ $=$ $\frac{1}{4}$ $+$ $\frac{3}{4}$ $= 1$

Therefore LHS = RHS. Hence proved.

ii) LHS $= \sin (A - B) = \sin (60^o-60^o) = \sin 0^o = 0$

RHS $= \sin A \cos B - \cos A \sin B =$ $\frac{\sqrt{3}}{2}$ $\times$ $\frac{1}{2}$ $-$ $\frac{\sqrt{3}}{2}$ $\times$ $\frac{1}{2}$ $=$ $\frac{\sqrt{3}}{4}$ $-$ $\frac{\sqrt{3}}{4}$ $= 0$

Therefore LHS = RHS. Hence proved.

iii) LHS $= \tan (A-B) = \tan 0^o = 0$

RHS $=$ $\frac{\tan A - \tan B}{1 + \tan A \tan B}$ $=$ $\frac{\sqrt{3} - \sqrt{3}}{1 + \sqrt{3} \times \sqrt{3}}$ $= 0$

Therefore LHS = RHS. Hence proved.

$\\$

Question 16: If $\sin (A - B) = \sin A \cos B - \cos A \sin B$ and $\cos (A-B) = \cos A \cos B + \sin A \sin B$ , find the value of $\sin 15^o$ and $\cos 15^o$

Answers:

i) $\sin 15 = \sin (45-30) = \sin 45 \cos 30 - \cos 45 \sin 30$

$=$ $\frac{1}{\sqrt{2}}$ $\times$ $\frac{\sqrt{3}}{2}$ $-$ $\frac{1}{\sqrt{2}}$ $\times$ $\frac{1}{2}$ $=$ $\frac{\sqrt{3}-1}{2\sqrt{2}}$

ii) $\cos 15 = \cos (45-30) = \cos 45 \cos 30 + \sin 45 \sin 30$

$=$ $\frac{1}{\sqrt{2}}$ $\times$ $\frac{\sqrt{3}}{2}$ $+$ $\frac{1}{\sqrt{2}}$ $\times$ $\frac{1}{2}$ $=$ $\frac{\sqrt{3}+1}{2\sqrt{2}}$

$\\$

Question 17: In a right angled triangle, right angled at $C$ , if $\angle B = 60^o$ , $AB = 15$ units, find the remaining angles and sides.

Answers:

Given $\angle C = 90^o$ and $\angle B = 60^o$

Therefore $\angle A = 180^o - 90^o - 60^o = 30^o$

$\frac{BC}{15}$ $= \sin 30^o \Rightarrow BC =$ $\frac{1}{2}$ $\times 15 = 7$ $\frac{1}{2}$ units

$\frac{AC}{15}$ $= \cos 30^o \Rightarrow AC =$ $\frac{\sqrt{3}}{2}$ $\times 15$ units

$\\$

Question 18: If $\sin (A + B) = 1$ and $\cos (A - B) = 1, 0^o < A + B \leq 90^o$ and $A \geq B$ , find $A$ and $B$ .

Answers:

Given $\sin (A+B) = 1$

$\Rightarrow \sin (A + B) = \sin 90$

$\Rightarrow A + B = 90^o$ … … … … … i)

Similarly, $\cos (A - B) = 1$

$\Rightarrow \cos (A - B) = \cos 0^o$

$\Rightarrow A = B$ … … … … … ii)

Solving i) and ii) $A = 45^o$ and $B = 45^o$

$\\$

Question 19: If $\tan (A-B) =$ $\frac{1}{\sqrt{3}}$ and $\tan (A + B) = \sqrt{3}, 0^o < A + B \leq 90^o$ and $A > B$ . Find $A$ and $B$ .

Answers:

Given $\tan (A-B) =$ $\frac{1}{\sqrt{3}}$

$\Rightarrow \tan (A-B) = \tan 30^o$ … … … … … i)

$\Rightarrow A - B = 30^o$

Also $\tan (A+B) = \sqrt{3}$

$\Rightarrow \tan (A+B) = \tan 60^o$

$\Rightarrow A + B = 60^o$ … … … … … ii)

Solving i) and ii) $A = 45^o$ and $B = 15^o$

$\\$

Question 20: In $\triangle ABC$ , right angled at $B, \angle A = \angle C$ . Find values of i) $\sin A \cos C + \cos A \sin C$ ii) $\sin A \sin B + \cos A \cos B$

Answers:

We have $A = C = 45^o ( \ since \ \angle B = 90^o)$

$\Rightarrow \sin A = \sin C = \cos A = \cos C =$ $\frac{1}{\sqrt{2}}$

i) $\sin A \cos C + \cos A \sin C =$ $\frac{1}{\sqrt{2}}$ $+$ $\times$ $\frac{1}{\sqrt{2}}$ $+$ $+$ $\frac{1}{\sqrt{2}}$ $+$ $\times$ $\frac{1}{\sqrt{2}}$ $+$ $=$ $\frac{1}{2}$ $+$ $+$ $\frac{1}{2}$ $+$ $= 1$

ii) $\sin A \sin B + \cos A \cos B =$ $\frac{1}{\sqrt{2}}$ $+$ $\times 1 +$ $\frac{1}{\sqrt{2}}$ $+$ $\times 0 =$ $\frac{1}{\sqrt{2}}$

$\\$

Question 21: If $A$ and $B$ are acute angles such that $tan A =$ $\frac{1}{2}$ and $tan B =$ $\frac{1}{3}$ and $tan A =$ $\frac{tan A + tan B}{1 - tan A tan B}$ . Find $A + B$ .

Answers:

RHS $=$ $\frac{tan A + tan B}{1 - tan A tan B}$ $=$ $\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}}$ $=$ $\frac{5}{5}$ $= 1$

Therefore $\tan (A+B) = 1 = \tan 45^o \Rightarrow A = 45^o$

$\\$

Question 22: In adjoining figure $ABC$ is a right angled triangle at $B$ and $\triangle ABD$ is a right angled triangle at $A$ . If $BD \perp AC$ and $BC = 2\sqrt{3}$ , find $AD$ .