Question

15. Find the missing sides/angles.i=94jk=42k

Answer 1

From the figure given,

`\begin{gathered} j=\text{opposite}=\text{?} \\ k=adjacent=\text{?} \\ hypotenuse=94 \\ \theta=42^0 \end{gathered}`

Let us solve for 'j'

To solve for j, we will employ the method of Sine of angles.

`\begin{gathered} \text{ Sine of angles=}\frac{opposite}{\text{hypotenuse}} \\ \sin \theta=(j)/(hypotenuse) \end{gathered}`
`\begin{gathered} \sin 42^0=(j)/(94) \\ \text{cross multiply} \\ j=94\sin 42^0 \\ j=94*0.6691 \\ j=62.8954\approx62.9units(nearest\text{ tenth)} \end{gathered}`

Let us solve for k

To solve for k, we will employ the method of Cosine of angles.

`\begin{gathered} \text{ Cosine of angles=}\frac{k}{\text{hypotenuse}} \\ \cos \theta=(k)/(hypotenuse) \\ \cos 42^0=(k)/(94) \\ \text{cross multiply} \\ k=94\cos 42^0 \\ k=94*0.7431 \\ k=69.8514\approx69.9units(nearest\text{ tenth)} \end{gathered}`

Hence, the value of j=62.9units,

k=69.9units.