Question

Two angles of a triangle are $\frac{\pi }{6}$ and $\frac{\pi }{4},$ and the length of the included side is $(\sqrt{3}+1)cm.$ The area of the triangle is

• A. $\frac{\sqrt{3}-1}{2}c{m}^2$
• B. $\frac{\sqrt{3}}{2}c{m}^2$
• C. $\frac{\sqrt{3}+1}{2}c{m}^2$
• D. none of these

Answer 1

Answer: (C)

$\frac{\sqrt{3}+1}{2}c{m}^2$

As shown in the figure, let $AD=a$

Thus $CD=a[\because \tan C=\frac{AD}{CD}=1]$
$\Rightarrow BD=\sqrt{3}+1-a$

$\tan B=\frac{1}{\sqrt{3}}=\frac{a}{\sqrt{3}+1-a}$

$(\sqrt{3}+1)a=\sqrt{3}+1\Rightarrow a=1$

$\therefore \Delta =\frac{1}{2}(\sqrt{3}+1)$