# an isosceles triange has a verticle angle of 108 degree and a base 20 cm long.calculate its altitude.

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You should remember that evaluating the sum of interior angles of triangle yields `180^o` such that:

`hatA + hatB + hatC = 180^o `

Notice that the triangle is an isosceles triangle, hence it has two equal angles such that:

`hatB= hatC`

The problem provides the angle `hat A = 108^o` , hence, substituting this value in relation `hatA + hatB + hatC = 180^o` yields:

`108^o + 2hatB = 180^o => 2 hat B = 180^o - 108^o => 2 hat B = 72^o => hat B = hat C = 36^o`

You should know that the height of an isosceles triangle that starts from the vertex A, as the equal legs, is also a median of the base side BC, opposed to the angle hatA.

`BD = DC = (BC)/2 = 10`

The height that starts from A falls to the side BC in the point D, hence, you may evaluate the length of the height using one of the two right angle triangles ABD or ACD.

Since in triangle ABD, the problem provides the length of the side BD, that is adjacent to the angle `hat(ABD) = 36^o` , you may use the tangent function such that:

`tan 36^o = (AD)/(BD) => tan 36^o = (AD)/(10)`

`AD = 10 tan 36^o => AD~~ 7.26cm`

**Hence, evaluating the length of the height of isosceles triangle ABC yields `AD ~~ 7.26 cm` .**