Question

1. solve the triangles below.

a) a
25 cm
b 48°
20 cm
c

Explanation:

Step1: Use the Law of Cosines to find side AC

Let \(a = 20\), \(c=25\), and \(\angle B = 48^{\circ}\). The Law of Cosines formula is \(b^{2}=a^{2}+c^{2}-2ac\cos B\).
\[

$$\begin{align*} b^{2}&=20^{2}+25^{2}-2\times20\times25\times\cos(48^{\circ})\\ b^{2}&=400 + 625-1000\times0.669131\\ b^{2}&=1025- 669.131\\ b^{2}&=355.869\\ b&=\sqrt{355.869}\approx18.86\mathrm{cm} \end{align*}$$

\]

Step2: Use the Law of Sines to find \(\angle A\)

The Law of Sines is \(\frac{\sin A}{a}=\frac{\sin B}{b}\). We know \(a = 20\), \(b\approx18.86\), and \(\angle B = 48^{\circ}\).
\[

$$\begin{align*} \sin A&=\frac{a\sin B}{b}\\ \sin A&=\frac{20\times\sin(48^{\circ})}{18.86}\\ \sin A&=\frac{20\times0.743145}{18.86}\\ \sin A&=\frac{14.8629}{18.86}\approx0.7881 \end{align*}$$

\]
\(\angle A=\sin^{- 1}(0.7881)\approx52.0^{\circ}\)

Step3: Find \(\angle C\)

Since the sum of angles in a triangle is \(180^{\circ}\), \(\angle C=180^{\circ}-\angle A - \angle B\).
\(\angle C=180^{\circ}-52.0^{\circ}-48^{\circ}=80^{\circ}\)

Answer:

Side \(AC\approx18.86\mathrm{cm}\), \(\angle A\approx52.0^{\circ}\), \(\angle C = 80^{\circ}\)