Question

$△ABC$ is a triangle such that $B=\frac{\pi }{8}andC=\frac{5\pi }{8}$

then the length of altitude dropped from $A$ is?

• A. $\frac{a}{2}$
• B. $2a$
• C. $\frac{b+c}{2}$
• D. $\frac{\sqrt{3}}{2}(b+c)$

Answer 1

The correct option is A $\frac{a}{2}$

$△=\frac{1}{2}bc\sin \frac{\pi }{4}=\frac{1}{2}Pa\Rightarrow P=\frac{bc}{a\sqrt{2}}$
Again
by sine rule $\frac{a}{\sin \frac{\pi }{4}}=\frac{b}{\sin \frac{\pi }{4}}=\frac{c}{\sin \frac{5\pi }{8}}$
$\Rightarrow b=\sqrt{2}a\sin \frac{\pi }{8}andc=\sqrt{2}a\sin \frac{5\pi }{8}$
$\therefore P=\frac{2a^2\sin \frac{\pi }{8}\sin \frac{5\pi }{8}}{a\sqrt{2}}=\frac{a^2(\cos \frac{4\pi }{8}-\cos \frac{6\pi }{8})}{a\sqrt{2}}=\frac{a^2\times \frac{1}{\sqrt{2}}}{a\sqrt{2}}=\frac{a}{2}$