Question

If the base angles of a triangle are $22{\frac{1}{2}}^o,112{\frac{1}{2}}^o$, then height and base are in the ratio

• A. $1:2$
• B. $1:3$
• C. $3:1$
• D. $2:1$

Answer 1

Answer: (A)

$1:2$

$\angle BAC=180-112.5-22.5={45}^0$
$\angle ACL=180-112.2={67.5}^0$
now in $\Delta ABC$
$\frac{BC}{\sin 45}=\frac{AC}{\sin 22.5}$
$\Rightarrow AC=\frac{BC\sin 22.5}{\sin 45}$
In $\Delta ALC$
$AL=AC\sin 67.5=AC\cos 22.5$
$\Rightarrow AL=\frac{BC\sin 22.5}{\sin 45}\cos 22.5$
$=\sin 22.5\cos 22.5\frac{BC}{\sin 45}$
$=\frac{\sin 45}{2}\frac{BC}{\sin 45}$
$=\frac{BC}{2}$
Hence altitude is half of its base.