Question

a right triangle has side lengths ac = 7 inches, bc = 24 inches, and ab = 25 inches. what are the measures of the angles in triangle abc? o m∠a≈46.2°, m∠b≈43.8°, m∠c≈90° o m∠a≈73.0°, m∠b≈17.0°, m∠c≈90° o m∠a≈73.7°, m∠b≈16.3°, m∠c≈90° o m∠a≈74.4°, m∠b≈15.6°, m∠c≈90°

Explanation:

Step1: Use the cosine - law for right - triangles

In right - triangle \(ABC\) with \(\angle C = 90^{\circ}\), \(\cos A=\frac{AC}{AB}\) and \(\cos B=\frac{BC}{AB}\).
We know that \(AC = 7\) inches, \(BC = 24\) inches, and \(AB = 25\) inches.
First, find \(\cos A=\frac{AC}{AB}=\frac{7}{25}=0.28\).
Then, find \(\angle A=\cos^{- 1}(0.28)\approx73.7^{\circ}\).

Step2: Find \(\angle B\)

Since the sum of the interior angles of a triangle is \(180^{\circ}\) and \(\angle C = 90^{\circ}\), \(\angle A+\angle B+\angle C=180^{\circ}\).
So, \(\angle B = 180^{\circ}-\angle A - \angle C\).
\(\angle B=180^{\circ}-90^{\circ}-\angle A\).
Substitute \(\angle A\approx73.7^{\circ}\), then \(\angle B\approx180^{\circ}-90^{\circ}-73.7^{\circ}=16.3^{\circ}\).

Answer:

C. \(m\angle A\approx73.7^{\circ},m\angle B\approx16.3^{\circ},m\angle C\approx90^{\circ}\)