Question

1. (5pts) find all angles and side lengths in the triangle diagram

Explanation:

Step1: Find the third angle

The sum of angles in a triangle is \(180^{\circ}\). Let the third angle be \(x\). Then \(x + 50^{\circ}+60^{\circ}=180^{\circ}\). So \(x=180^{\circ}-(50^{\circ} + 60^{\circ})=70^{\circ}\).

Step2: Use the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Let's assume the side opposite \(50^{\circ}\) is \(a\), opposite \(60^{\circ}\) is \(b\), and opposite \(70^{\circ}\) is \(c\). If we assume a side - angle pair (but since no side lengths are given initially, we can just express the ratios). Let's say we want to find the sides in terms of a common multiplier \(k\). Then \(a = k\sin50^{\circ}\), \(b=k\sin60^{\circ}\), \(c = k\sin70^{\circ}\). If we assume \(k = 1\) (for the sake of showing the relationship of side lengths relative to the angles), \(\sin50^{\circ}\approx0.766\), \(\sin60^{\circ}\approx0.866\), \(\sin70^{\circ}\approx0.940\). So the sides (relative) are approximately in the ratio \(0.766:0.866:0.940\)

Answer:

The angles are \(50^{\circ}\), \(60^{\circ}\), and \(70^{\circ}\). If we assume a common multiplier \(k\) for side - length calculation using the Law of Sines (\(\frac{a}{\sin50^{\circ}}=\frac{b}{\sin60^{\circ}}=\frac{c}{\sin70^{\circ}} = k\)), the sides (relative) are \(a\approx0.766k\), \(b\approx0.866k\), \(c\approx0.940k\)