Question 1016339
By law of sines, *[tex \large \frac{200}{\sin 16^{\circ}} = \frac{50}{\sin \alpha} = \frac{x}{\sin \beta}] where alpha is opposite the 50 m side and beta is opposite the x m side. Then *[tex \large \sin \alpha = \frac{50}{200} \cdot \sin 16^{\circ} \approx 0.0689] so *[tex \large \alpha = \sin^{-1}(0.0689) \approx 4^{\circ}] (closer to 3.95 degrees)


Then *[tex \large \beta \approx 160^{\circ}], so *[tex \large x = \frac{200 \,\sin 160^{\circ}}{\sin \, 16^{\circ}} \approx 248] meters.


Alternatively, you can use the law of cosines:
*[tex \large 200^2 = 50^2 + x^2 - 2 \cdot 50 \cdot x \, \cos (16^{\circ})]
Then solve the remaining quadratic using your calculator. You get the same answer either way.