Question

Solve the right triangle. Round decimal answers to the nearest tenth.find YZfind angle Xfind angle Z

Answer 1

Given the right triangle XYZ

`\begin{gathered} XZ^2=XY^2+YZ^2 \\ YZ^2=XZ^2-XY^2 \\ YZ^2=9^2-3^2=81-9=72 \\ YZ=\sqrt[]{72}=\sqrt[]{36\cdot2}=6\sqrt[]{2} \end{gathered}`

We will find the angles using the sine rule

`\begin{gathered} \angle Y=90,XZ=9 \\ \\ (XZ)/(\sin Y)=(YZ)/(\sin X)=(XY)/(\sin Z) \\ \\ (9)/(\sin90)=\frac{6\sqrt[]{2}}{\sin X}=(3)/(\sin Z) \end{gathered}`

So, the angle X will be:

`\begin{gathered} (9)/(\sin90)=\frac{6\sqrt[]{2}}{\sin X} \\ \\ \sin X=\frac{6\sqrt[]{2}}{9}\cdot\sin 90=0.9428 \\ \\ X=\sin ^(-1)0.9428\approx70.5^o^{} \end{gathered}`

The angle Z will be:

`\begin{gathered} (9)/(\sin90)=(3)/(\sin Z) \\ \\ \sin Z=(3)/(9)\sin 90=0.3333 \\ \\ Z=\sin ^(-1)0.3333\approx19.5^o \end{gathered}`

So, the answer is:

`\begin{gathered} YZ=6\sqrt[]{2}\approx8.5 \\ \\ \angle X=70.5^o \\ \\ \angle Z=19.5^o \end{gathered}`