Question

If the base angles of a triangle are $22{\frac{1}{2}}^{\circ }$ and $112{\frac{1}{2}}^{\circ }$

then the altitude of the triangle is equal to $n$ times its base. The value of $n$ is:

• A. $\sqrt{2}$
• B. $2$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

In $\Delta ABC,\frac{BC}{\sin {45}^{\circ }}=\frac{AC}{\sin 22{\frac{1}{2}}^{\circ }}$

In $\Delta ALC,\frac{AL}{AC}=\sin 67{\frac{1}{2}}^{\circ }$

$\therefore AL=AC\cos 22{\frac{1}{2}}^{\circ }$ $=\frac{BC}{\sin {45}^{\circ }}\sin 22{\frac{1}{2}}^{\circ }\cos 22{\frac{1}{2}}^{\circ }$ [using Eq.(i)]

$\Rightarrow \text{Altitude},AL=\frac{1}{2}BC\frac{\sin {45}^{\circ }}{\sin {45}^{\circ }}=\frac{1}{2}BC=\frac{1}{2}\times \text{Base}$

$\therefore$ the altitude is $\frac{1}{2}$ times the base