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? 36 feet 29 feet 24 feet 31 feet

Explanation:

Step1: Find the adjacent side and set up the tangent ratio.

We know the adjacent side (distance from student to tree base) is 20 feet, and we need to find the opposite side (height from eyes to tree top) using the tangent of the angle of elevation. The tangent of an angle in a right triangle is $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. First, we need to find $\sin50^{\circ}$ or $\tan50^{\circ}$. Wait, we are given $\cos50^{\circ}\approx0.64$. We know that $\sin^{2}\theta+\cos^{2}\theta = 1$, so $\sin50^{\circ}=\sqrt{1 - \cos^{2}50^{\circ}}$. But maybe easier to use $\tan50^{\circ}=\frac{\sin50^{\circ}}{\cos50^{\circ}}$. Alternatively, we can use the right triangle: let $h$ be the height from eyes to tree top. Then $\tan50^{\circ}=\frac{h}{20}$. But we can also find $\sin50^{\circ}$ first. Let's calculate $\sin50^{\circ}$: $\sin50^{\circ}=\sqrt{1 - 0.64^{2}}=\sqrt{1 - 0.4096}=\sqrt{0.5904}\approx0.768$. Wait, no, actually, the correct approach is: in the right triangle, the horizontal distance is 20 (adjacent), the vertical distance from eyes to tree top is $y$, and the angle is $50^{\circ}$. So $\tan50^{\circ}=\frac{y}{20}$, so $y = 20\times\tan50^{\circ}$. But we can also use $\sin$ and $\cos$. Wait, the hypotenuse $c$ can be found by $\cos50^{\circ}=\frac{20}{c}$, so $c=\frac{20}{\cos50^{\circ}}\approx\frac{20}{0.64}\approx31.25$. Then $\sin50^{\circ}=\frac{y}{c}$, so $y = c\times\sin50^{\circ}$. We know that $\sin50^{\circ}\approx0.766$ (since $\cos50^{\circ}\approx0.64$, $\sin50^{\circ}=\sqrt{1 - 0.64^{2}}=\sqrt{1 - 0.4096}=\sqrt{0.5904}\approx0.768$, close to 0.766). So $y\approx31.25\times0.768\approx24.0$. Wait, no, that's not right. Wait, actually, the angle of elevation is from the student's eyes to the top of the tree. So the right triangle has angle $50^{\circ}$, adjacent side 20 feet (horizontal distance), opposite side $h$ (height from eyes to tree top), and hypotenuse (line of sight). So $\tan(50^{\circ})=\frac{h}{20}$, so $h = 20\times\tan(50^{\circ})$. We know that $\tan(50^{\circ})\approx1.1918$. So $h\approx20\times1.1918\approx23.836$. Then the total height of the tree is $h + 5$ (since eyes are 5 feet above ground) $\approx23.836 + 5 = 28.836\approx29$ feet. Wait, but let's check with the given $\cos50^{\circ}\approx0.64$. Let's find $\sin50^{\circ}$ using $\sin^{2}+\cos^{2}=1$: $\sin50^{\circ}=\sqrt{1 - 0.64^{2}}=\sqrt{1 - 0.4096}=\sqrt{0.5904}\approx0.768$. Then, the hypotenuse $c=\frac{20}{\cos50^{\circ}}\approx\frac{20}{0.64}=31.25$. Then the opposite side (height from eyes to tree top) is $c\times\sin50^{\circ}\approx31.25\times0.768 = 24$. Then total height is $24 + 5 = 29$ feet. Yes, that matches. So:

Step1: Calculate the hypotenuse of the right triangle.

Using $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, so hypotenuse $c=\frac{\text{adjacent}}{\cos\theta}=\frac{20}{\cos50^{\circ}}\approx\frac{20}{0.64}=31.25$ feet.

Step2: Calculate the height from eyes to tree top.

Using $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, so opposite side (height from eyes to tree top) $y = c\times\sin50^{\circ}$. We know $\sin50^{\circ}=\sqrt{1 - \cos^{2}50^{\circ}}\approx\sqrt{1 - 0.64^{2}}\approx0.768$. So $y\approx31.25\times0.768 = 24$ feet.

Step3: Calculate the total height of the tree.

Add the height from eyes to ground (5 feet) to $y$: total height $= 24 + 5 = 29$ feet.

Answer:

29 feet