Answer:
![x=64.1](https://img.qammunity.org/2023/formulas/mathematics/college/s98iarmgzxna8olgyd7gqmxyeb4yqp2zpc.png)
Step-by-step explanation:
Step 1. The first step is to find the missing side of the triangle using the Pythagorean theorem:
In this case:
![\begin{gathered} c=16 \\ b=7 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/y25ts3l0h36j7l9dv5o179ly9pzt3lxnwn.png)
and we need to find a.
Substituting c and b into the Pythagorean theorem formula:
![16^2=a^2+7^2](https://img.qammunity.org/2023/formulas/mathematics/college/t5ypz87xslsr1t8kyp9pk1nqwrn7a91q2d.png)
Solving for a:
![\begin{gathered} 16^2-7^2=a^2 \\ 256-49^{}=a^2 \\ 207=a^2 \\ \downarrow \\ \sqrt[]{207}=a \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/f73uu36ii5rnxtd9irt8xogrs5ih8d0c8r.png)
Step 2. The triangle now is:
And to find angle x we use:
![\begin{gathered} \\ \boxed{\sin x=\frac{\text{opposite side}}{hypotenuse}} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/e472mlrt50qmm846e98dyopuj7ehs3yd2s.png)
Substituting the known values:
![\begin{gathered} \sin x=\frac{\sqrt[]{207}}{16} \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7kxq3ype9h6fraxl211zrsbjhptk3aooio.png)
Step 3. Solving for x:
![\begin{gathered} \sin x=\frac{\sqrt[]{207}}{16} \\ \downarrow \\ x=\sin ^(-1)(\frac{\sqrt[]{207}}{16}) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4k8sb7zoau22mqfzkwlbv69ga6ogje1pj0.png)
Step 4. Solving the operations:
![\begin{gathered} x=\sin ^(-1)(0.899218) \\ \downarrow \\ \\ x=64.05552 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4gs6lb7h4q0xuzx52yczfxz8k9gg88woz8.png)
Rounding to the nearest tenth (1 decimal):
![x=64.1](https://img.qammunity.org/2023/formulas/mathematics/college/s98iarmgzxna8olgyd7gqmxyeb4yqp2zpc.png)
Answer:
![x=64.1](https://img.qammunity.org/2023/formulas/mathematics/college/s98iarmgzxna8olgyd7gqmxyeb4yqp2zpc.png)