Question

If the angles of a triangle are ${30}^{\circ }$ and ${45}^{\circ },$ and the included side is $(\sqrt{3}+1)$ cm, then

• A. area of the triangle is $\frac{1}{2}(\sqrt{3}+1)$ sq. units
• B. area of the triangle is $\frac{1}{2}(\sqrt{3}-1)$ sq. units
• C. ratio of greater side to smaller side is $\frac{\sqrt{3}+1}{\sqrt{2}}$
• D. $\frac{1}{4\sqrt{3}}$

Answer 1

Answer: (A), (C)

area of the triangle is $\frac{1}{2}(\sqrt{3}+1)$ sq. units
ratio of greater side to smaller side is $\frac{\sqrt{3}+1}{\sqrt{2}}$

Given that, $B={30}^{\circ }$; $C={45}^{\circ }$ and $a=\sqrt{3}+1$

$\Rightarrow A={180}^{\circ }-{30}^{\circ }-{45}^{\circ }={105}^{\circ }$
Now, from sine rule: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

$\Rightarrow \frac{\sqrt{3}+1}{\sin {105}^{\circ }}=\frac{b}{\sin {30}^{\circ }}=\frac{c}{\sin {45}^{\circ }}$

$\Rightarrow \frac{2\sqrt{2}(\sqrt{3}+1)}{\sqrt{3}+1}=2b=\sqrt{2}c$

$\Rightarrow b=\sqrt{2}$ and $c=2$

Therefore, $\Rightarrow \frac{a}{b}=\frac{\sqrt{3}+1}{\sqrt{2}}$
and $△=\frac{ab\sin C}{2}=\frac{(\sqrt{3}+1)\times \sqrt{2}\times \sin {45}^{\circ }}{2}=\frac{1}{2}(\sqrt{3}+1)$