Respuesta :
From the figure given,
[tex]\begin{gathered} j=\text{opposite}=\text{?} \\ k=adjacent=\text{?} \\ hypotenuse=94 \\ \theta=42^0 \end{gathered}[/tex]Let us solve for 'j'
To solve for j, we will employ the method of Sine of angles.
[tex]\begin{gathered} \text{ Sine of angles=}\frac{opposite}{\text{hypotenuse}} \\ \sin \theta=\frac{j}{hypotenuse} \end{gathered}[/tex][tex]\begin{gathered} \sin 42^0=\frac{j}{94} \\ \text{cross multiply} \\ j=94\sin 42^0 \\ j=94\times0.6691 \\ j=62.8954\approx62.9units(nearest\text{ tenth)} \end{gathered}[/tex]Let us solve for k
To solve for k, we will employ the method of Cosine of angles.
[tex]\begin{gathered} \text{ Cosine of angles=}\frac{k}{\text{hypotenuse}} \\ \cos \theta=\frac{k}{hypotenuse} \\ \cos 42^0=\frac{k}{94} \\ \text{cross multiply} \\ k=94\cos 42^0 \\ k=94\times0.7431 \\ k=69.8514\approx69.9units(nearest\text{ tenth)} \end{gathered}[/tex]Hence, the value of j=62.9units,
k=69.9units.
