Question

In triangle abc, the sides are ac=bc=8 inches and ab=4. find the angles

Answer 1

`m\angle A=m\angle B=75.5^(\circ),\\ \\m\angle C=29^(\circ)`

Explanation:

In triangle ABC the sides AC=BC=8 in and AB=4 in.

Use the cosine rule:

`AB^2=AC^2+BC^2-2\cdot AC\cdot BC\cdot \cos \angle C,\\ \\4^2=8^2+8^2-2\cdot 8\cdot 8\cdot \cos \angle C,\\ \\16=64+64-128\cos \angle C,\\ \\16-64-64=-128\cos \angle C,\\ \\\cos \angle C=(112)/(128)=(7)/(8),\\ \\m\angle C\approx 29^(\circ).`

Since triangle ABC is isosceles, angles A and B have the same measures. Then

`m\angle A=m\angle B=(180^(\circ)-29^(\circ))/(2)=75.5^(\circ).`