Question

Let area of a triangle $ABC$ is $\frac{\sqrt{3}-1}{2},b=1$ and $c=(\sqrt{3}-1)$ and $\angle A$ is acute. The measure of the angle $C$ is

Answer 1

REF.Image

We know

$\Delta ABC$

area of $\Delta ABC=\frac{1}{2}$bc sin A

$\frac{\sqrt{3}-1}{2}=\frac{1}{2}\times 1\times (\sqrt{3}-1)sinA$

sin A =1

sin A = sin 90

$A+{90}^{\circ }$

$a^2=b^2+c^2$

$=1+(\sqrt{3}-1{)}^2$

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

$a^2=5-2\sqrt{3}$

$a=\sqrt{5-2\sqrt{3}}$

$\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=k$

$\frac{sin90}{\sqrt{5-2\sqrt{3}}}=\frac{sinC}{\sqrt{3}-1}$

$sinC=\frac{\sqrt{3}-1}{\sqrt{5-2\sqrt{3}}}$