Question

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

Answer 1

Answer:

$m\angle A=m\angle B=75.5^{\circ},\\ \\m\angle C=29^{\circ}$

Step-by-step explanation:

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

Use the cosine rule:

$AB^2=AC^2+BC^2-2\cdot AC\cdot BC\cdot \cos \angle C,\\ \\4^2=8^2+8^2-2\cdot 8\cdot 8\cdot \cos \angle C,\\ \\16=64+64-128\cos \angle C,\\ \\16-64-64=-128\cos \angle C,\\ \\\cos \angle C=\dfrac{112}{128}=\dfrac{7}{8},\\ \\m\angle C\approx 29^{\circ}.$

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

$m\angle A=m\angle B=\dfrac{180^{\circ}-29^{\circ}}{2}=75.5^{\circ}.$