Question

1. find the length of the indicated side in each triangle, to the nearest tenth of a unit.

(first triangle: triangle with side x, side 16cm, angles 30° and 45° at the base vertices)
(second triangle: triangle xsb with side xs=7 km, side sb=10 km, angle at s is 54°, side xb is s=?)

Explanation:

Step1: Analyze the first triangle

First, find the third angle of the triangle. The sum of angles in a triangle is \(180^\circ\). So the third angle \(= 180^\circ - 30^\circ - 45^\circ = 105^\circ\). We can use the Law of Sines, which states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Here, we have a side of \(16\) cm opposite \(30^\circ\), and we want to find \(x\) opposite \(45^\circ\). So \(\frac{x}{\sin 45^\circ}=\frac{16}{\sin 30^\circ}\).

Step2: Solve for \(x\) in the first triangle

We know that \(\sin 30^\circ = 0.5\) and \(\sin 45^\circ=\frac{\sqrt{2}}{2}\approx0.7071\). So \(x=\frac{16\times\sin 45^\circ}{\sin 30^\circ}=\frac{16\times0.7071}{0.5}= 16\times1.4142 = 22.6272\approx22.6\) cm.

Step3: Analyze the second triangle

For the triangle with sides \(7\) km, \(10\) km and angle \(54^\circ\) between them (wait, no, angle at \(S\) is \(54^\circ\), sides \(SX = 7\) km, \(SB = 10\) km, and we need to find \(s = XB\). Use the Law of Cosines, which is \(c^{2}=a^{2}+b^{2}-2ab\cos C\). Here, \(a = 7\), \(b = 10\), \(C = 54^\circ\), so \(s^{2}=7^{2}+10^{2}-2\times7\times10\times\cos 54^\circ\).

Step4: Calculate \(s\) in the second triangle

First, calculate \(\cos 54^\circ\approx0.5878\). Then \(s^{2}=49 + 100-140\times0.5878=149 - 82.292 = 66.708\). Then \(s=\sqrt{66.708}\approx8.2\) km.

Answer:

First triangle: \(x\approx22.6\) cm; Second triangle: \(s\approx8.2\) km