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? diagram of a tree, student, and right triangle options: 24 feet, 29 feet, 31 feet, 36 feet

Explanation:

Step1: Find the adjacent and opposite sides relation

We know that in the right triangle formed, the adjacent side (distance from student to tree base horizontally) is 20 feet, and we can use the tangent function for the angle of elevation. Wait, but we are given $\cos 50^{\circ}\approx0.64$. Wait, maybe first find the hypotenuse? No, wait, actually, we can use $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. But we can also find the opposite side (height from eyes to tree top) using $\tan 50^{\circ}$. But first, let's recall that $\tan\theta=\frac{\sin\theta}{\cos\theta}$, but maybe we can find $\sin 50^{\circ}$ from $\cos 50^{\circ}$. Since $\sin^{2}\theta+\cos^{2}\theta = 1$, so $\sin 50^{\circ}=\sqrt{1-\cos^{2}50^{\circ}}$. But maybe easier to use $\tan 50^{\circ}$. Alternatively, let's see: the horizontal distance is 20 feet (adjacent side for the angle of elevation). The height from the student's eyes to the tree top is the opposite side. Let's call that height $h$. Then $\tan 50^{\circ}=\frac{h}{20}$. But we can find $\tan 50^{\circ}$ using $\tan\theta=\frac{\sin\theta}{\cos\theta}$, and we know $\cos 50^{\circ}\approx0.64$, so $\sin 50^{\circ}=\sqrt{1 - 0.64^{2}}=\sqrt{1 - 0.4096}=\sqrt{0.5904}\approx0.768$. Then $\tan 50^{\circ}=\frac{0.768}{0.64}\approx1.2$. So $h = 20\times\tan 50^{\circ}\approx20\times1.2 = 24$? Wait, no, that can't be. Wait, maybe I made a mistake. Wait, actually, the angle of elevation is 50 degrees, adjacent side is 20 feet, so the opposite side (height from eyes to tree top) is $20\times\tan(50^{\circ})$. Let's calculate $\tan(50^{\circ})\approx1.1918$. So $20\times1.1918\approx23.836$. Then the total height of the tree is this height plus the 5 feet from the ground to the eyes. So $23.836 + 5\approx28.836$, which is approximately 29 feet. Wait, let's check again.

Wait, maybe the problem is using $\cos 50^{\circ}\approx0.64$ to find the hypotenuse first? Wait, no, the adjacent side is 20 feet, which is adjacent to the 50-degree angle. So $\cos 50^{\circ}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{20}{hypotenuse}$, so hypotenuse $=\frac{20}{\cos 50^{\circ}}\approx\frac{20}{0.64}=31.25$ feet. Then, the opposite side (height from eyes to tree top) can be found using $\sin 50^{\circ}=\frac{\text{opposite}}{\text{hypotenuse}}$, so opposite $=\text{hypotenuse}\times\sin 50^{\circ}$. We know $\sin 50^{\circ}=\sqrt{1 - 0.64^{2}}=\sqrt{1 - 0.4096}=\sqrt{0.5904}\approx0.768$. So opposite $=31.25\times0.768 = 24$ feet. Then total height is $24 + 5 = 29$ feet. Ah, that makes sense. So let's do that step by step.

Step1: Find the hypotenuse of the right triangle

We know that $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$. Here, $\theta = 50^{\circ}$, adjacent $= 20$ feet, $\cos 50^{\circ}\approx0.64$. So:
$$\cos 50^{\circ}=\frac{20}{\text{hypotenuse}}$$
Solving for hypotenuse:
$$\text{hypotenuse}=\frac{20}{\cos 50^{\circ}}\approx\frac{20}{0.64}=31.25\text{ feet}$$

Step2: Find the height from eyes to tree top (opposite side)

Using $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, where $\theta = 50^{\circ}$, hypotenuse $= 31.25$ feet. First, find $\sin 50^{\circ}$:
$$\sin 50^{\circ}=\sqrt{1 - \cos^{2}50^{\circ}}=\sqrt{1 - 0.64^{2}}=\sqrt{1 - 0.4096}=\sqrt{0.5904}\approx0.768$$
Then, opposite (height from eyes to tree top) $h_1$:
$$\sin 50^{\circ}=\frac{h_1}{\text{hypotenuse}}$$
$$h_1=\text{hypotenuse}\times\sin 50^{\circ}\approx31.25\times0.768 = 24\text{ feet}$$

Step3: Find the total height of the tree

The student's eyes are 5 feet above the ground, so total height $h$ is $h_1 + 5$:
$$h = 24 + 5 = 29\text{ feet}$$

Answer:

29 feet