Question

in \\( \triangle tuv, u = 750 \mathrm { cm }, m \angle t = 105 ^ { \circ } \\) and \\( m \angle u = 42 ^ { \circ } \\). find the length of \\( t \\), to the nearest centimeter.

Explanation:

Step1: Find the measure of angle \(V\)

The sum of angles in a triangle is \(180^{\circ}\). So, \(m\angle V=180^{\circ}-(m\angle T + m\angle U)\)
\(m\angle V=180^{\circ}-(105^{\circ}+ 42^{\circ})=33^{\circ}\)

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{t}{\sin T}=\frac{u}{\sin U}\)
We know \(u = 750\), \(m\angle T = 105^{\circ}\), \(m\angle U=42^{\circ}\)
Substitute into the formula: \(t=\frac{u\sin T}{\sin U}\)
\(\sin105^{\circ}=\sin(60^{\circ}+ 45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}+\frac{1}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.9659\), \(\sin42^{\circ}\approx0.6691\)
\(t=\frac{750\times\sin105^{\circ}}{\sin42^{\circ}}\)
\(t=\frac{750\times0.9659}{0.6691}\)
\(t=\frac{724.425}{0.6691}\approx1083\)

Answer:

\(1083\)