Question

In $\Delta$ABC, $B={45}^o$, $C={120}^o$, $a=40$cm, the length of the perpendicular from A on BC produced is?

• A. $40(\sqrt{3}+1)$
• B. $20(\sqrt{3}+1)$
• C. $40(3+\sqrt{3})$
• D. $20(3+\sqrt{3})$

Answer 1

The correct option is D $20(3+\sqrt{3})$

In $△ABC,B={45}^{\circ },C={120}^{\circ },a=40$

$A={180}^{\circ }-{120}^{\circ }-{45}^{\circ }={15}^{\circ }$

$\frac{a}{\sin A}=\frac{b}{\sin B}⟹b=a(\frac{\sin B}{\sin A})=40(\sqrt{3}+1)$

Lenth of altitude $=b\sin C=40(\sqrt{3}+1)\times \sin \left({120}^{\circ }\right)=20(3+\sqrt{3})$