you have an isosceles triangle whose angle is x and the two equal sides are s. show that the area of the triangle is given by A=1/2 s^2 sinx.
I got 1/2 s sinx but i cant get s^2 can someone help? i greatly appreciate it.
Suppose sides of triangle as A,B&C
Side $\displaystyle AB=AC$
$\displaystyle \angle{BAC}=x$
$\displaystyle \angle{ABC}=90-\frac{x}{2}$
Area $\displaystyle = \frac{1}{2}(s.sin(90-\frac{x}{2}))(2s.cos(90-\frac{x}{2}))$
$\displaystyle sin2x=2sinx.cosx$
$\displaystyle \frac{1}{2}sin(180-x)=sin(90-\frac{x}{2}).cos(90-\frac{x}{2})$
As $\displaystyle sin(180-x)=sinx
$
Hence, Area $\displaystyle = \frac{1}{2}.s^2.sinx$