Question

to the nearest tenth, which is the perimeter of △abc?
a
61°
10
29°
b
c
a. 4.8
b. 8.7
c. 13.6
d. 23.6

Explanation:

Step1: Find the length of \(AC\)

Use the cosine function: \(\cos A=\frac{AC}{AB}\). Given \(A = 61^{\circ}\) and \(AB = 10\), then \(AC=AB\times\cos A\).
\(AC = 10\times\cos61^{\circ}\approx10\times0.485 = 4.85\)

Step2: Find the length of \(BC\)

Use the sine function: \(\sin A=\frac{BC}{AB}\). Given \(A = 61^{\circ}\) and \(AB = 10\), then \(BC=AB\times\sin A\).
\(BC = 10\times\sin61^{\circ}\approx10\times0.875=8.75\)

Step3: Calculate the perimeter of \(\triangle ABC\)

The perimeter \(P=AC + BC+AB\).
\(P\approx4.85 + 8.75+10=23.6\)

Answer:

D. 23.6