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.

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- Mar 9th 2009, 10:22 PMsoldatik21triangle 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. - Mar 9th 2009, 10:51 PMu2_wa
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**$\displaystyle = \frac{1}{2}.s^2.sinx$