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Let the Isosceles Triangle be ∆ ABC with sides AB = AC = 14', and BC = 17'

Draw a line bIsecting angle BAC. This line will be perpendicular to and bisect BC at point D.

Then ∆ DBA (or ∆ DCA) is a right angled triangle with AB the hypotenuse.

Angle ABD = Angle ABC is one of the two equal angles of the isosceles triangle.

Cos ABD = BD/AB = 8.5/14 = 0.607143, therefore Angle ABC = 52.62°

The third angle of the triangle is 180 - (2 x 52.62) = 180 - 105.24 = 74.76°

The angles are therefore 52.62° , 52.62° and 74.76° .

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