Isosceles triangle

What are the angles of an isosceles triangle ABC if its base is long a=5 m and has an arm b=4 m.

Result

α =  77.4 °
β =  51.3 °
γ =  51.3 °

Solution:

\sin \dfrac{\alpha}{2} = \dfrac{a/2}{b} \ \\ \alpha = 2 \arcsin \dfrac{a/2}{b} = 2 \arcsin \dfrac{ 5/2}{ 4} = 77.4 \text{ ^\circ }
\beta = \gamma = (180- \alpha)/2 = 51.3 \text{ ^\circ }
\gamma =(180 - 2 \cdot \ \dfrac{ 180^\circ }{ \pi } \cdot a\sin(5/2/4))/2 = 51.3 \text{ ^\circ }



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