Question

unit exam - right triangles and trigonometry
a surveyor stands at a corner of a triangular field. one side is 60 ft, the other 64 ft, and the third side opposite the corner is 100 ft. what is the angle at the surveyors position?
surveyors angle = ?°
round your answer to the nearest hundredth.

Explanation:

Step1: Apply the Law of Cosines

Let the angle at the surveyor be $\theta$, the sides adjacent to $\theta$ be $a=60$ ft, $b=64$ ft, and the side opposite $\theta$ be $c=100$ ft. The Law of Cosines states:
$$c^2 = a^2 + b^2 - 2ab\cos\theta$$

Step2: Rearrange to solve for $\cos\theta$

$$\cos\theta = \frac{a^2 + b^2 - c^2}{2ab}$$

Step3: Substitute the given values

$$\cos\theta = \frac{60^2 + 64^2 - 100^2}{2\times60\times64} = \frac{3600 + 4096 - 10000}{7680} = \frac{-2304}{7680}$$

Step4: Calculate $\cos\theta$ and find $\theta$

$$\cos\theta = -0.3$$
$$\theta = \arccos(-0.3)$$
Using a calculator, $\theta \approx 107.46^\circ$

Answer:

$107.46^\circ$