Question

using the figure, find angles in degrees and minutes and sides to the nearest whole number.

1. a = 28, b = 45
2. ∠b = 30°, a = 12
3. ∠a = 35°40, c = 20

Explanation:

Step1: Recall trigonometric relations in right - triangle

In right - triangle \(ABC\) with \(\angle C = 90^{\circ}\), \(\tan A=\frac{a}{b}\), \(\tan B=\frac{b}{a}\), \(\sin A=\frac{a}{c}\), \(\sin B=\frac{b}{c}\), \(\cos A=\frac{b}{c}\), \(\cos B=\frac{a}{c}\).

Step2: Solve for problem 5

Given \(a = 28\) and \(b = 45\). First, find \(\angle A\): \(\tan A=\frac{a}{b}=\frac{28}{45}\approx0.6222\). Then \(A=\arctan(0.6222)\approx32^{\circ}\). Since \(A + B=90^{\circ}\), \(B = 90^{\circ}-A\approx90^{\circ}-32^{\circ}=58^{\circ}\).

Step3: Solve for problem 6

Given \(\angle B = 30^{\circ}\) and \(a = 12\). Since \(\sin B=\frac{b}{c}\) and \(\cos B=\frac{a}{c}\), and \(\cos B=\cos30^{\circ}=\frac{\sqrt{3}}{2}=\frac{a}{c}\), \(c=\frac{a}{\cos30^{\circ}}=\frac{12}{\frac{\sqrt{3}}{2}} = 8\sqrt{3}\approx13.86\). Also, \(\sin B=\frac{b}{c}\), \(b = c\sin B=8\sqrt{3}\times\frac{1}{2}=4\sqrt{3}\approx6.93\). And \(\angle A=90^{\circ}-\angle B = 60^{\circ}\).

Step4: Solve for problem 7

Given \(\angle A=35^{\circ}40'\). First, convert \(40'\) to degrees: \(40'=\frac{40}{60}\approx0.67^{\circ}\), so \(\angle A\approx35.67^{\circ}\). Then \(\angle B = 90^{\circ}-\angle A=90^{\circ}-35.67^{\circ}=54.33^{\circ}=54^{\circ}20'\) (since \(0.33\times60 = 20'\)). Also, \(\sin A=\frac{a}{c}\), \(a = c\sin A=20\times\sin(35.67^{\circ})\approx20\times0.583=11.66\), \(\cos A=\frac{b}{c}\), \(b = c\cos A=20\times\cos(35.67^{\circ})\approx20\times0.813 = 16.26\).

Answer:

Problem 5: \(\angle A\approx32^{\circ}\), \(\angle B\approx58^{\circ}\)
Problem 6: \(\angle A = 60^{\circ}\), \(b\approx6.93\), \(c\approx13.86\)
Problem 7: \(\angle B = 54^{\circ}20'\), \(a\approx11.66\), \(b\approx16.26\)