1. The triangle above describes the given scenario.
Right triangle: x is the hypotenuse, y is a leg (opposite to the angle of depression), and 65 in the measure of the other leg (adjacent to the angle of depression)
2. The vertical distance between the search boat and the treasure is y.
Use the trigonometric function that involves the two legs of a right triangle to find the value of y:
[tex]\begin{gathered} tan\theta=\frac{opposite}{adjacent} \\ \\ tan30º=\frac{y}{65} \\ \\ y=65*tan30º \\ \\ y=37.53 \end{gathered}[/tex]
The vertical distance between the search boat and the treasure is approximately 37.53 meters
3. The straight-like distance between the treasure and the shore is x.
Use the trigonometric function that involves the adjacent leg and hypotenuse of a right triangle to find x:
[tex]\begin{gathered} cos\theta=\frac{adjacent}{hypotenuse} \\ \\ cos30º=\frac{65}{x} \\ \\ x*cos30º=65 \\ \\ x=\frac{65}{cos30º} \\ \\ x=75.06 \end{gathered}[/tex]
The straight-like distance between the treasure and the shore is approximately 75.06 meters
4. Angle α is the angle that the treasure creates with search boat and the shore:
Use the inverse trigonometric function that involves adjacent leg (for this angle y) and hypotenuse (x): (use the calculator in radians mode)
[tex]\begin{gathered} cos\alpha=\frac{adjacent}{hypotenuse} \\ \\ cos\alpha=\frac{y}{x} \\ \\ cos\alpha=\frac{37.53}{75.06} \\ \\ \alpha=\cos^{-1}(\frac{37.53}{75.06}) \\ \\ \alpha=1.05 \end{gathered}[/tex]
Then, using arccos you can show that the measue of angle α is approximately 1.05 radians
