Question

1. determine the indicated side lengths to the nearest tenth of a unit and the indicated angle measures to the nearest degree.

a)
b)
c)
d)
e)

Explanation:

Step1: Recall the Law of Sines

The Law of Sines states that for a triangle with sides \(a\), \(b\), \(c\) and opposite - angles \(A\), \(B\), \(C\) respectively, \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\).

Step2: Solve part a

Let the triangle be \(\triangle DEF\) with \(\angle D = 53^{\circ}\), \(d = 22.5\mathrm{cm}\), \(\angle E=68^{\circ}\). First, find \(\angle F=180^{\circ}-(53^{\circ}+68^{\circ}) = 59^{\circ}\). Using the Law of Sines \(\frac{e}{\sin E}=\frac{d}{\sin D}\), so \(e=\frac{d\sin E}{\sin D}=\frac{22.5\times\sin68^{\circ}}{\sin53^{\circ}}\approx\frac{22.5\times0.9272}{0.7986}\approx26.1\mathrm{cm}\).

Step3: Solve part b

In \(\triangle ABC\), \(\angle B = 60^{\circ}\), \(b = 40.0\mathrm{cm}\), \(\angle A = 40^{\circ}\). Then \(\angle C=180^{\circ}-(40^{\circ}+60^{\circ}) = 80^{\circ}\). Using the Law of Sines \(\frac{a}{\sin A}=\frac{b}{\sin B}\), so \(a=\frac{b\sin A}{\sin B}=\frac{40\times\sin40^{\circ}}{\sin60^{\circ}}\approx\frac{40\times0.6428}{0.8660}\approx29.7\mathrm{cm}\).

Step4: Solve part c

In \(\triangle XYZ\), \(\angle X = 88^{\circ}\), \(\angle Z = 25^{\circ}\), \(x = 3.0\mathrm{cm}\). First, \(\angle Y=180^{\circ}-(88^{\circ}+25^{\circ}) = 67^{\circ}\). Using the Law of Sines \(\frac{y}{\sin Y}=\frac{x}{\sin X}\), so \(y=\frac{x\sin Y}{\sin X}=\frac{3\times\sin67^{\circ}}{\sin88^{\circ}}\approx\frac{3\times0.9205}{0.9994}\approx2.8\mathrm{cm}\).

Step5: Solve part d

In \(\triangle LMN\), \(l = 45.2\mathrm{cm}\), \(n = 24.4\mathrm{cm}\), \(\angle L = 29^{\circ}\). Using the Law of Sines \(\frac{\sin N}{n}=\frac{\sin L}{l}\), so \(\sin N=\frac{n\sin L}{l}=\frac{24.4\times\sin29^{\circ}}{45.2}\approx\frac{24.4\times0.4848}{45.2}\approx0.261\). Then \(N=\sin^{- 1}(0.261)\approx15^{\circ}\).

Step6: Solve part e

In \(\triangle PQR\), \(p = 50\mathrm{cm}\), \(q = 65\mathrm{cm}\), \(\angle Q = 50^{\circ}\). Using the Law of Sines \(\frac{\sin P}{p}=\frac{\sin Q}{q}\), so \(\sin P=\frac{p\sin Q}{q}=\frac{50\times\sin50^{\circ}}{65}\approx\frac{50\times0.7660}{65}\approx0.5892\). Then \(P=\sin^{-1}(0.5892)\approx36^{\circ}\).

Answer:

a. Side length \(e\approx26.1\mathrm{cm}\)
b. Side length \(a\approx29.7\mathrm{cm}\)
c. Side length \(y\approx2.8\mathrm{cm}\)
d. Angle \(N\approx15^{\circ}\)
e. Angle \(P\approx36^{\circ}\)