Question

1. use the primary trigonometric ratios to solve for x to the nearest centimetre, or degree.

(there are two right - triangle diagrams: the first has a 50° angle, a 10 cm side, and the side opposite the 50° angle is x; the second has legs 11 cm and x, and hypotenuse 16.7 cm)

Explanation:

Step1: Analyze the first triangle

We have a right - triangle with an angle of \(50^{\circ}\), the adjacent side to the \(50^{\circ}\) angle is \(10\) cm, and the side \(x\) is the opposite side to the \(50^{\circ}\) angle. We use the tangent function, where \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). So \(\tan(50^{\circ})=\frac{x}{10}\).
We know that \(\tan(50^{\circ})\approx1.1918\). Then \(x = 10\times\tan(50^{\circ})\approx10\times1.1918 = 11.918\approx12\) cm.

Step2: Analyze the second triangle

We have a right - triangle with hypotenuse \(c = 16.7\) cm and one leg \(a = 11\) cm. We want to find the angle \(x\). We can use the sine function \(\sin x=\frac{\text{opposite}}{\text{hypotenuse}}\), where the opposite side to angle \(x\) is \(11\) cm. So \(\sin x=\frac{11}{16.7}\approx0.6587\). Then \(x=\sin^{- 1}(0.6587)\approx41^{\circ}\) (since \(\sin(41^{\circ})\approx0.6561\) and \(\sin(42^{\circ})\approx0.6691\), and \(0.6587\) is closer to \(\sin(41^{\circ})\)).

Answer:

For the first triangle, \(x\approx12\) cm; for the second triangle, \(x\approx41^{\circ}\)