Question

If ABC is a triangle in which $B={45}^o,C={120}^o$ and $a=40$, the length of the perpendicular from A on BC produced is

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

Answer 1

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

Referring to figure, we have from $\Delta$ ABC,
$AC/sin{45}^o=40/sin{15}^o$.
$\Rightarrow AC=\frac{40sin{45}^o}{sin{15}^o}$
$=\frac{40}{\sqrt{2}sin({45}^o-{30}^o)}=\frac{40.2\sqrt{2}}{\sqrt{2}(\sqrt{3}-1)}$
so that $AC=40(\sqrt{3}+1)$. Therefore, from right-angled triangle ADC, we get the length of the perpendicular
$AD=ACsin{60}^o=40(\sqrt{3}+1).\frac{\sqrt{3}}{2}=20(3+\sqrt{3})$.