Question

solve the right triangle abc, where c = 90°. a = 76.3 yd, b = 41.8 yd c = □ yd (simplify your answer. type an integer or a decimal. round to the nearest tenth as needed.) a = □°□ (simplify your answers. type integers. round to the nearest ten minutes as needed.) b = □°□ (simplify your answers. type integers. round to the nearest ten minutes as needed.)

Explanation:

Step1: Find side c using Pythagorean theorem

By the Pythagorean theorem \(c = \sqrt{a^{2}+b^{2}}\), where \(a = 76.3\) and \(b = 41.8\).
\[c=\sqrt{(76.3)^{2}+(41.8)^{2}}=\sqrt{5821.69 + 1747.24}=\sqrt{7568.93}\approx87.0\]

Step2: Find angle A using tangent function

\(\tan(A)=\frac{a}{b}\), so \(A=\arctan(\frac{a}{b})\). Substituting \(a = 76.3\) and \(b = 41.8\), we have \(A=\arctan(\frac{76.3}{41.8})\approx\arctan(1.82536)\).
\(A\approx61.3^{\circ}\). Converting the decimal part to minutes, \(0.3\times60 = 18\), so \(A = 61^{\circ}18'\).

Step3: Find angle B

Since the sum of angles in a triangle is \(180^{\circ}\) and \(C = 90^{\circ}\), \(B=180^{\circ}-A - C\).
\(B = 180^{\circ}- 90^{\circ}-61^{\circ}18'=28^{\circ}42'\)

Answer:

c = 87.0 yd
A = 61°18'
B = 28°42'