Question

In a

∆ A B C,

if

a = \sqrt{3} - 1, b = \sqrt{3} + 1

and

C = 60 °,

find the other side and other two angles.

Accepted Answer

Given that $a = \sqrt{3} - 1, b = \sqrt{3} + 1$

We know that $\cos C = \frac{(a^{2} + b^{2} - c^{2})}{2 a b}$ .

Now, $a^{2} = (\sqrt{3} - 1)^{2}$

$= 3 + 1 - 2 \sqrt{3}$

$= 4 - 2 \sqrt{3}$

And, $b^{2} = (\sqrt{3} + 1)^{2}$

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

$= 4 + 2 \sqrt{3}$

By applying those values in $cosine$ formula, we get:

$\Rightarrow \cos 60 ° = \frac{(4 - 2 \sqrt{3} + 4 + 2 \sqrt{3} - c^{2})}{2 (\sqrt{3} - 1) (\sqrt{3} + 1)}$

$\Rightarrow \frac{1}{2} = \frac{(8 - c^{2})}{2 (2)}$

$\Rightarrow 2 = 8 - c^{2}$

$\Rightarrow c^{2} = 8 - 2 = 6$

$\Rightarrow c = \sqrt{6}$

We know that $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} [using sine rule]$

$\Rightarrow \frac{(\sqrt{3} - 1)}{\sin A} = \frac{(\sqrt{3} + 1)}{\sin B} = \frac{\sqrt{6}}{\sin 60 °}$

Consider, $\frac{(\sqrt{3} + 1)}{\sin B} = \frac{\sqrt{6}}{(\frac{\sqrt{3}}{2})}$

$\Rightarrow \frac{(\sqrt{3} + 1)}{\sin B} = \frac{2 \sqrt{6}}{\sqrt{3}}$

$\Rightarrow \frac{(\sqrt{3} + 1)}{\sin B} = 2 \sqrt{2}$

$\Rightarrow \sin B = \frac{(\sqrt{3} + 1)}{2 \sqrt{2}}$

$\Rightarrow \sin B = (\frac{\sqrt{3}}{2}) (\frac{1}{\sqrt{2}}) + (\frac{1}{2}) (\frac{1}{\sqrt{2}})$

$= \sin 60 ° \cos 45 ° + \cos 60 ° \sin 45 °$

$= \sin (60 ° + 45 °) [∵ \sin (x + y) = \sin x \cos y + \cos x \sin y]$

$= \sin 105 °$
Hence, $∠ B = 105 °$

And, $A + B + C = 180 °$

$\Rightarrow A + 105 ° + 60 ° = 180$

$\Rightarrow A + 165 ° = 180 °$

$\Rightarrow A = 180 ° - 165 °$

Hence, $∠ A = 15 °$ .

Tamil Nadu Board Solutions for Chapter: Trigonometry, Exercise 10: Exercise

Tamil Nadu Board Mathematics Solutions for Exercise - Tamil Nadu Board Solutions for Chapter: Trigonometry, Exercise 10: Exercise

Questions from Tamil Nadu Board Solutions for Chapter: Trigonometry, Exercise 10: Exercise with Hints & Solutions

Learn and Prepare for any exam you want !