Question

1. find the length of side b.

(there is a triangle abc with angle at a being 70°, angle at b being 50°, side opposite angle b (side ac) is labeled b, side opposite angle a (side bc) is 3.5 cm, and side ab is labeled c. the multiple - choice options are: 3.33 cm, 0.05 cm, 2.85 cm, 4.29 cm)

Explanation:

Step1: Find angle at C

In a triangle, the sum of angles is \(180^\circ\). So, \(\angle C = 180^\circ - 70^\circ - 50^\circ = 60^\circ\).

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Here, \(a = 3.5\) cm, \(\angle A = 70^\circ\), \(\angle B = 50^\circ\), and we need to find \(b\). So, \(\frac{3.5}{\sin 70^\circ}=\frac{b}{\sin 50^\circ}\).

Step3: Solve for b

First, find \(\sin 70^\circ\approx0.9397\) and \(\sin 50^\circ\approx0.7660\). Then, \(b=\frac{3.5\times\sin 50^\circ}{\sin 70^\circ}=\frac{3.5\times0.7660}{0.9397}\approx\frac{2.681}{0.9397}\approx2.85\) cm.

Answer:

2.85 cm (corresponding to the option "2.85 cm")