question_answer

If the angles of a triangle are \[30{}^\circ \] and \[45{}^\circ \] and the included side is \[(\sqrt{3}+1),\] then what is the area of the triangle?

A) \[\frac{\sqrt{3}+1}{2}\]

B) \[2(\sqrt{3}+1)\]

C) \[\frac{\sqrt{3}+1}{3}\]

D) \[\frac{\sqrt{3}-1}{2}\]

Correct Answer: A

Solution :

[a]

From \[\Delta ADB,\,\,AD=BD=x\]

In \[\Delta ADC,\]

\[\tan 30{}^\circ =\frac{x}{\sqrt{3}+1-x}\]

\[\Rightarrow \frac{1}{\sqrt{3}}=\frac{x}{\sqrt{3}+1-x}\Rightarrow \sqrt{3}x=\sqrt{3}+1-x\]

\[\Rightarrow (\sqrt{3}+1)x=\sqrt{3}+1\]

\[x=\frac{\sqrt{3}+1}{\sqrt{3}+1}\]

Area of \[\Delta ABC=\frac{1}{2}\times (\sqrt{3}+1)\times 1=\frac{\sqrt{3}+1}{2}\]