Question

1. find x to the nearest tenth of a centimetre.

a)
b)

Explanation:

Step1: Use sine - ratio for right - triangle a

In right - triangle a, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. Here, $\theta = 51^{\circ}$ and the hypotenuse is $33$ cm, and the side opposite to the angle $\theta$ is $x$. So, $\sin(51^{\circ})=\frac{x}{33}$.
$x = 33\times\sin(51^{\circ})$

Step2: Calculate the value of $x$ for triangle a

We know that $\sin(51^{\circ})\approx0.777$. Then $x = 33\times0.777=25.641\approx25.6$ cm.

Step3: Use sine - ratio for right - triangle b

In right - triangle b, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. Here, $\theta = 42^{\circ}$ and the hypotenuse is $12$ cm, and the side opposite to the angle $\theta$ is $x$. So, $\sin(42^{\circ})=\frac{x}{12}$.
$x = 12\times\sin(42^{\circ})$

Step4: Calculate the value of $x$ for triangle b

We know that $\sin(42^{\circ})\approx0.669$. Then $x = 12\times0.669 = 8.028\approx8.0$ cm.

Answer:

a) $25.6$ cm
b) $8.0$ cm