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

Question 1: Evaluate each of the following:

Answers:

$\displaystyle \text{i) } \mathrm{cosec} 30^{\circ} + \cot 45^{\circ} = 2 + 1 = 3$

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

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

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

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

$\displaystyle \text{iv) } \frac{\sin 30^{\circ} - \sin 90^{\circ} + 2 \cos 0^{\circ} }{\tan 30^{\circ} \tan 60^{\circ} } = \frac{\frac{1}{2} - 1 + 2 \times 1}{\frac{1}{\sqrt{3}} \times \sqrt{3}} = \frac{3}{2}$

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

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

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

$\displaystyle \$

Question 2: Find the value of $\displaystyle \theta$ in each of the following when $\displaystyle 0^{\circ} < \theta < 90^{\circ}$ :

Answers:

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

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

$\displaystyle \Rightarrow \sin 2 \theta = \sin 60^{\circ}$

$\displaystyle \Rightarrow 2 \theta = 60^{\circ} \Rightarrow \theta = 30^{\circ}$

$\displaystyle \text{ii) } 2 \cos 3\theta = 1$

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

$\displaystyle \Rightarrow \cos 3 \theta = \cos 60^{\circ}$

$\displaystyle \Rightarrow 3 \theta = 60^{\circ} \Rightarrow \theta = 20^{\circ}$

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

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

$\displaystyle \Rightarrow \tan 2 \theta = \tan 60^{\circ}$

$\displaystyle \Rightarrow 2 \theta = 60^{\circ} \Rightarrow \theta = 30^{\circ}$

$\displaystyle \text{iv) } 2 \cos \theta = 1$

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

$\displaystyle \Rightarrow \cos \theta = \cos 60^{\circ}$

$\displaystyle \Rightarrow \theta = 60^{\circ}$

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

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

$\displaystyle \Rightarrow \cos \theta = \cos 60^{\circ} \Rightarrow \theta = 60^{\circ}$

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

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

$\displaystyle \Rightarrow \sin \theta = \sin 30^{\circ} \Rightarrow \theta = 30^{\circ}$

$\displaystyle \text{vii) } 3 \tan^2 \theta - 1 = 0$

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

$\displaystyle \Rightarrow \tan \theta = \tan 30^{\circ} \Rightarrow \theta = 30^{\circ}$

$\displaystyle \$

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

Answers:

$\displaystyle \text{Given } \tan \theta + \cot \theta = 2$

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

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

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

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

$\displaystyle \Rightarrow \tan \theta = 1$

$\displaystyle \text{Therefore } \cot \theta = 1$

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

$\displaystyle \$

Question 4: if $\displaystyle x = 30^{\circ}$ , verify that

Answers:

$\displaystyle \text{i) } \sin 3x = 3 \sin x - 4 \sin^3 x$

$\displaystyle \text{Given } x = 30^{\circ}$

$\displaystyle \text{Therefore } \sin x = \frac{1}{2} \text{ ; } \sin 3x = \sin 90^{\circ} = 1$

Therefore $\displaystyle \text{LHS } = 3 \sin x - 4 \sin^3 x$

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

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

Therefore RHS = LHS. Hence proved.

$\displaystyle \text{ii) } \cos 3x = 4 \cos^3 x - 3 \cos x$

$\displaystyle \text{Given } x = 30^{\circ}$

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

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

$\displaystyle \text{LHS } = \cos 90^{\circ} = 0$

Therefore LHS = RHS. Hence proved.

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

$\displaystyle \text{LHS } = \tan 2x = \tan 60^{\circ} = \sqrt{3}$

$\displaystyle \text{RHS } = \frac{2 \tan x}{1 - \tan^2 x}$ = $\displaystyle \frac{2 \tan 30^{\circ} }{1 - \tan^2 30^{\circ} }$

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

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

$\displaystyle \text{LHS } = \sin x = \sin 30^{\circ} = \frac{1}{2}$

$\displaystyle \text{RHS } = \sqrt{ \frac{1- \cos 2x}{2} } = \sqrt{ \frac{1- \cos 60^{\circ} }{2} }$ $\displaystyle = \sqrt{ \frac{1- \frac{1}{2} }{2} } = \frac{1}{2}$

$\displaystyle \$

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

Answers:

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

Applying componendo and dividendo we get

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

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

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

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

$\displaystyle \Rightarrow \theta = 60^{\circ}$

$\displaystyle \$

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

Answers:

$\displaystyle \text{Given } \sin (A+B) = 1$

$\displaystyle \Rightarrow \sin (A+B) = \sin 90^{\circ}$

$\displaystyle \Rightarrow A + B = 90^{\circ}$ … … … … … i)

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

$\displaystyle \Rightarrow \cos (A-B) = \cos 30^{\circ}$

$\displaystyle \Rightarrow A - B = 30^{\circ}$ … … … … … ii)

Solving i) and ii) we get

$\displaystyle A = 60^{\circ} \text{ and } B = 30^{\circ}$

$\displaystyle \$

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

Answers:

$\displaystyle \text{Given } \angle A = 30^{\circ} \text{ and } \angle C = 90^{\circ}$

Since $\displaystyle \angle A + \angle B + \angle C = 180^{\circ}$

$\displaystyle \Rightarrow \angle B = 60^{\circ}$

$\displaystyle \frac{BC}{AB} = \sin 30^{\circ}$

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

$\displaystyle \text{Similarly, } \frac{AC}{40} = \cos 30^{\circ}$

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

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

$\displaystyle \$

Question 8: A rhombus of side $\displaystyle 20 \text{ cm }$ has two angles of $\displaystyle 60^{\circ}$ each. Find the length of the diagonals.

Answers:

$\displaystyle \text{Given } ABCD$ is a rhombus.

$\displaystyle \Rightarrow AB = BC = CD = DA = 20 \text{ cm }$

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

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

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

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

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

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

$\displaystyle \$

Question 9: An equilateral triangle is inscribed in a circle of radius $\displaystyle 6 \text{ cm }$ . Find the side.

Answers:

$\displaystyle \text{Given } OA = OB = OC = 6 \text{ cm }$

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

$\displaystyle \text{Therefore } \angle OBD = 30^{\circ}$

In $\displaystyle \triangle OBD, \angle D = 90^{\circ}$ we have

$\displaystyle \angle OBD = 30^{\circ} \text{ and } OB = 6 \text{ cm }$

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

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

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

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

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

$\displaystyle \$

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

Answers:

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

$\displaystyle \Rightarrow \sin (\alpha + \beta - \gamma) = \sin 30^{\circ}$

$\displaystyle \Rightarrow \alpha + \beta - \gamma = 30^{\circ}$ … … … … … i)

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

$\displaystyle \Rightarrow \cos (\beta + \gamma - \alpha) = \cos 60^{\circ}$

$\displaystyle \Rightarrow \beta + \gamma - \alpha = 60^{\circ}$ … … … … … ii)

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

$\displaystyle \Rightarrow \tan (\gamma + \alpha - \beta) = \tan 45^{\circ}$

$\displaystyle \Rightarrow \gamma + \alpha - \beta = 45^{\circ}$ … … … … … iii)

Adding i) , ii) and iii) we get

$\displaystyle \alpha + \beta + \gamma = 135^{\circ}$ … … … … … iv)

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

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

iv) – $\displaystyle \text{i) } \Rightarrow 2 \beta = 90^{\circ} \Rightarrow \beta = 45^{\circ}$

$\displaystyle \$

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

Answers:

$\displaystyle \text{Given } \tan (A + B - C) = 1$

$\displaystyle \Rightarrow \tan (A + B - C) = \tan 45^{\circ}$

$\displaystyle \Rightarrow A + B - C = 45^{\circ}$ … … … … … i)

$\displaystyle \text{Also } \sec (B + C - A) = 2$

$\displaystyle \Rightarrow \sec B +C - A = \sec 60^{\circ}$

$\displaystyle \Rightarrow B + C - A = 60^{\circ}$ … … … … … ii)

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

Substituting in ii) we get

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

We know $\displaystyle A + B + C = 180^{\circ}$

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

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

Adding iii) and iv) we get

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

$\displaystyle \text{Therefore } A = 127\frac{1}{2}^{\circ} - 67\frac{1}{2}^{\circ} = 60^{\circ}$

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

$\displaystyle \$

Question 12: Evaluate

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

Answers:

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

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

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

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

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

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

$\displaystyle \$

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

Answers:

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

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

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

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

$\displaystyle \$

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

Answers:

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

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

$\displaystyle \Rightarrow \sin 3x = \sin 60^{\circ}$

$\displaystyle \Rightarrow 3x = 60^{\circ}$

$\displaystyle \Rightarrow x = 20^{\circ}$

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

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

$\displaystyle \Rightarrow \tan x = \tan 30^{\circ}$

$\displaystyle \Rightarrow x = 30^{\circ}$

$\displaystyle \$

Question 15: If $\displaystyle A = B = 60^{\circ}$ , verify

$\displaystyle \text{i) } \cos (A - B) = \cos A \cos B + \sin A \sin B$

$\displaystyle \text{ii) } \sin (A - B) = \sin A \cos B - \cos A \sin B$

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

Answers:

$\displaystyle \text{Given } A = B = 60^{\circ}$ we have

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

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

$\displaystyle \cos 0^{\circ} = 1$ ; $\displaystyle \sin 0^{\circ} = 0$

$\displaystyle \text{i) } \text{LHS } = \cos (A - B) = \cos (60^{\circ} -60^{\circ} ) = \cos 0^{\circ} = 1$

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

Therefore LHS = RHS. Hence proved.

$\displaystyle \text{ii) } \text{LHS } = \sin (A - B) = \sin (60^{\circ} -60^{\circ} ) = \sin 0^{\circ} = 0$

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

Therefore LHS = RHS. Hence proved.

$\displaystyle \text{iii) } \text{LHS } = \tan (A-B) = \tan 0^{\circ} = 0$

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

Therefore LHS = RHS. Hence proved.

$\displaystyle \$

$\displaystyle \text{Question 16: If } \sin (A - B) = \sin A \cos B - \cos A \sin B \text{ and } \cos (A-B) = \cos A \cos B + \sin A \sin B \text{ , find the value of } \sin 15^{\circ} \text{ and } \cos 15^{\circ}$

Answers:

$\displaystyle \text{i) } \sin 15 = \sin (45-30) = \sin 45 \cos 30 - \cos 45 \sin 30$

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

$\displaystyle \text{ii) } \cos 15 = \cos (45-30) = \cos 45 \cos 30 + \sin 45 \sin 30$

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

$\displaystyle \$

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

Answers:

$\displaystyle \text{Given } \angle C = 90^{\circ} \text{ and } \angle B = 60^{\circ}$

$\displaystyle \text{Therefore } \angle A = 180^{\circ} - 90^{\circ} - 60^{\circ} = 30^{\circ}$

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

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

$\displaystyle \$

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

Answers:

$\displaystyle \text{Given } \sin (A+B) = 1$

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

$\displaystyle \Rightarrow A + B = 90^{\circ}$ … … … … … i)

$\displaystyle \text{Similarly, } \cos (A - B) = 1$

$\displaystyle \Rightarrow \cos (A - B) = \cos 0^{\circ}$

$\displaystyle \Rightarrow A = B$ … … … … … ii)

Solving i) and $\displaystyle \text{ii) } A = 45^{\circ} \text{ and } B = 45^{\circ}$

$\displaystyle \$

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

Answers:

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

$\displaystyle \Rightarrow \tan (A-B) = \tan 30^{\circ}$ … … … … … i)

$\displaystyle \Rightarrow A - B = 30^{\circ}$

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

$\displaystyle \Rightarrow \tan (A+B) = \tan 60^{\circ}$

$\displaystyle \Rightarrow A + B = 60^{\circ}$ … … … … … ii)

Solving i) and $\displaystyle \text{ii) } A = 45^{\circ} \text{ and } B = 15^{\circ}$

$\displaystyle \$

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

Answers:

$\displaystyle \text{We have } A = C = 45^{\circ} ( since \angle B = 90^{\circ} )$

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

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

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

$\displaystyle \$

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

Answers:

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

$\displaystyle \text{Therefore } \tan (A+B) = 1 = \tan 45^{\circ} \Rightarrow A = 45^{\circ}$

$\displaystyle \$

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