Question

If the base angles of a triangle are ${22\frac{1}{2}}^0$ and ${112\frac{1}{2}}^0$ then the height of the triangle is equal to

• A. half the base
• B. the base
• C. twice the base
• D. four times the base

Answer 1

The correct option is A half the base

Let $A=22{\frac{1}{2}}^0\Rightarrow \therefore 4A={90}^0$
$A=90-3A$
$\sin A=\sin (90-3A)=\cos 3A$ .............$\left(1\right)$
$\therefore$ In $△ABD,$
we have $\sin (90-A)=\frac{H}{AB}$
or $\cos A=\frac{H}{AB}$
or $AB=\frac{H}{\cos A}$ ..............$\left(2\right)$
and in $△ABC,\frac{a}{\sin 2A}=\frac{AB}{\sin A}$
or $\frac{a}{2\sin A\cos A}=\frac{AB}{\sin A}$
or $\frac{a}{2\cos A}=AB$ ................$\left(3\right)$
From $\left(2\right)$ and $\left(3\right)$ we have
$\frac{H}{\cos A}=\frac{a}{2\cos A}$
$\therefore H=\frac{a}{2}=$half of the base of the triangle.