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 $\frac{1}{2}$ of its base.

• A. True
• B. False

Answer 1

The correct option is A True

In $\Delta$ABC

measure of angle A=180-112.5-22.5=45${}^{\circ }$,

Also $\angle ACD=\angle ABC+\angle BAC=22.5+45={67.5}^{\circ }$ ( exterior angle property)

using sine rule

$\frac{a}{sinA}=\frac{b}{sinB}$

$⟹\frac{BC}{sin45}=\frac{b}{sin22.5}-\left(1\right)$

Also In $\Delta ACD$,

$sin\left(\angle ACD\right)=\frac{h}{b}⟹b=\frac{h}{sin\left(\angle ACD\right)}-\left(2\right)$

Sunstituting Equation (2) in (1)-

$⟹\frac{BC}{sin45}=\frac{h}{sin22.5\times sin67.5}$

$⟹\frac{BC}{sin45}=\frac{2h}{Sin45}$

$⟹BC=2h⟹h=\frac{1}{2}BC$