Chapter 7.6, Problem 99AYU

To determine

To show: That the area $A$ of an isosceles triangle whose equal sides are of length $s$, and where $\theta$ is the angle between them, is $A=\frac{1}{2}{s}^2\sin \theta$. [Hint: See the illustration. The height $h$ bisects the angle $\theta$ and is the perpendicular bisector of the base.]

Answer to Problem 99AYU

$A=\frac{1}{2}{s}^2\sin \theta$

Explanation of Solution

Area of the triangle $=\frac{1}{2}\text{bh}$.

Here $\sin \frac{\theta }{2}=\frac{\frac{\text{base}}{2}}{s}$,

$\text{base }=2s\sin \frac{\theta }{2}$

$\cos \frac{\theta }{2}=\frac{h}{s}$

$h=s\cos \frac{\theta }{2}$

Area of the triangle $=\frac{1}{2}\text{bh}$,

$=\frac{1}{2}2s\sin \frac{\theta }{2}s\cos \frac{\theta }{2}$

$=\frac{s^2\sin \theta }{2}$ By using $\left[\sin 2\theta =2\sin \theta \cos \theta \right]$.

$A=\frac{1}{2}{s}^2\sin \theta$