Question

a student is standing 20 feet away from the base of a tree. he looks to the top of the tree at a 50° angle of elevation. his eyes are 5 feet above the ground. using cos 50° ≈ 0.64, what is the height of the tree to the nearest foot?

Explanation:

Step1: Find vertical distance above eye - level

We use the tangent function. In a right - triangle, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. Here, $\theta = 50^{\circ}$ and the adjacent side (distance from student to tree) is $20$ feet. So the vertical distance $h_1$ from the student's eye - level to the top of the tree is $h_1 = 20\times\tan50^{\circ}$. Since $\tan50^{\circ}\approx1.19$, then $h_1=20\times1.19 = 23.8$ feet.

Step2: Calculate total height of the tree

The student's eyes are $5$ feet above the ground. Let the total height of the tree be $h$. Then $h=h_1 + 5$. Substituting the value of $h_1$, we get $h=23.8+5=28.8\approx29$ feet.

Answer:

29 feet