Question

kayson is looking at two buildings, building a and building b, at an angle of elevation of 59°. building a is 40 feet away, and building b is 60 feet away. which building is taller and by approximately how many feet?
building a; it is around 66.57 feet taller than building b
building a; it is around 33.29 feet taller than building b
building b; it is around 66.57 feet taller than building a
building b; it is around 33.29 feet taller than building a

Explanation:

Step1: Recall the tangent function formula

The tangent of an angle of elevation in a right triangle (where the height of the building is the opposite side and the distance from Kayson to the building is the adjacent side) is given by $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}=\frac{h}{d}$, so the height $h$ of a building is $h = d\times\tan(\theta)$.

Step2: Calculate the height of Building A

For Building A, the distance $d_A = 40$ feet and the angle of elevation $\theta = 59^\circ$. Using the formula $h_A = d_A\times\tan(59^\circ)$. We know that $\tan(59^\circ)\approx1.6643$. So $h_A = 40\times1.6643 = 66.572$ feet.

Step3: Calculate the height of Building B

For Building B, the distance $d_B = 60$ feet and the angle of elevation $\theta = 59^\circ$. Using the formula $h_B = d_B\times\tan(59^\circ)$. So $h_B = 60\times1.6643 = 99.858$ feet. Wait, no, wait, maybe I misread the problem. Wait, the problem says "at an angle of elevation of 59°" for both? Wait, no, maybe the angle is the same, but the distances are 40 and 60? Wait, no, let's re - read. The problem says "Kayson is looking at two buildings, building A and building B, at an angle of elevation of 59°. Building A is 40 feet away, and building B is 60 feet away. Which building is taller and by approximately how many feet?" Wait, no, that can't be, because if the angle is the same, the taller distance (adjacent) would lead to taller height? Wait, no, wait, maybe I made a mistake. Wait, no, the formula is $h = d\times\tan(\theta)$. So if $\theta$ is the same, larger $d$ means larger $h$. But wait, the options are about which is taller. Wait, but let's check the options again. Wait, the options are:

1. Building A; it is around 66.57 feet taller than building B
2. Building A; it is around 33.29 feet taller than building B
3. Building B; it is around 66.57 feet taller than building A
4. Building B; it is around 33.29 feet taller than building A

Wait, maybe I misread the distance. Wait, maybe Building A is 60 feet away and Building B is 40? No, the problem says "Building A is 40 feet away, and building B is 60 feet away". Wait, no, let's recalculate. Wait, $h_A=40\times\tan(59^\circ)\approx40\times1.6643 = 66.57$ feet. $h_B = 60\times\tan(59^\circ)\approx60\times1.6643 = 99.858$ feet. Then the difference is $h_B - h_A=(60 - 40)\times\tan(59^\circ)=20\times1.6643 = 33.286\approx33.29$ feet. So Building B is taller than Building A by approximately 33.29 feet.

Answer:

Building B; it is around 33.29 feet taller than building A