# triangle proof

• March 9th 2009, 10:22 PM
soldatik21
triangle proof
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.
• March 9th 2009, 10:51 PM
u2_wa
Quote:

Originally Posted by soldatik21
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 $AB=AC$
$\angle{BAC}=x$
$\angle{ABC}=90-\frac{x}{2}$
Area $= \frac{1}{2}(s.sin(90-\frac{x}{2}))(2s.cos(90-\frac{x}{2}))$
$sin2x=2sinx.cosx$
$\frac{1}{2}sin(180-x)=sin(90-\frac{x}{2}).cos(90-\frac{x}{2})$
As $sin(180-x)=sinx
$

Hence, Area $= \frac{1}{2}.s^2.sinx$