Question

The area of an isosceles triangle is $9c{m}^2$. If the equal sides are $6cm$ each in length, then the angle between them is

• A. ${60}^o$
• B. ${30}^o$
• C. ${90}^o$
• D. ${45}^o$

Answer 1

Answer: (B)

${30}^o$

Area $=9$, Sides $=6$

Area $=\frac{1}{2}AB\times AC\sin \theta$

$9=\frac{1}{\text{/}2}\times {\text{/}6}^3\times 6\times \sin \theta$

$\sin \theta =\frac{{\text{/}9}^1}{{\text{/}18}_2}=\frac{1}{2}$.