Richard is standing between two buildings in a townhouse development. The building on the left is 9m away and the angle of elevation to its security spotlight, A, is 68 degree. The building on the right is 6m away and the angle of elevation to its security spotlight, B, is 73degree. Which spotlight is farther away from Richard, and by how much?

Question

Richard is standing between two buildings in a townhouse development. The building on the left is 9m away and the angle of elevation to its security spotlight, A, is 68 degree. The building on the right is 6m away and the angle of elevation to its security spotlight, B, is 73degree. Which spotlight is farther away from Richard, and by how much?

Answer 1

Really don't understand need help.please

Answer 2

let the distances be a and b.

a/9 = sec 68°
b/6 = sec 73°

so,
a = 24.0
b = 20.5

so, a is farther away by 3.5m

Answer 3

To solve this problem, we can use trigonometry and the concept of similar triangles.

Let's denote the distance from Richard to the left building as x and the distance from Richard to the right building as y.

We can form two right-angled triangles, one for each building, with Richard as the common vertex. Let's call the angle between Richard's line of sight and the horizontal direction in both cases as angle C.

In the triangle formed with the left building, we have angle C = 68 degrees, and the distance to the left building is 9m. We can use the tangent function to find x:

tan(68°) = 9 / x

Rearranging the equation, we have:

x = 9 / tan(68°)

Now, let's consider the triangle formed with the right building. In this case, we have angle C = 73 degrees, and the distance to the right building is 6m. We can use the tangent function to find y:

tan(73°) = 6 / y

Rearranging the equation, we have:

y = 6 / tan(73°)

To determine which spotlight is farther away, we need to compare the values of x and y. We can calculate these values using a calculator:

x = 9 / tan(68°) ≈ 3.2591 meters
y = 6 / tan(73°) ≈ 2.53 meters

From the calculations, we can see that the spotlight on the left building is farther away from Richard by approximately 0.7291 meters.

Answer 4

To determine which spotlight is farther away from Richard and by how much, we need to find the distances between Richard and the two spotlights.

Let's start with the left building and spotlight A. We have the angle of elevation to spotlight A (68 degrees) and the distance between Richard and the left building (9m). To find the distance between Richard and spotlight A, we can use trigonometric functions.

We can use the tangent function since we have the angle and the opposite side length. The tangent function is defined as opposite/adjacent, so we have:

tan(68) = opposite/9

Rearranging the equation to solve for the opposite side length, we have:

opposite = tan(68) * 9

Using a calculator, we find that the opposite side length (distance between Richard and spotlight A) is approximately 22.50m.

Now, let's move on to the right building and spotlight B. We have the angle of elevation to spotlight B (73 degrees) and the distance between Richard and the right building (6m). Again, we can use the tangent function to find the distance between Richard and spotlight B.

tan(73) = opposite/6

Rearranging the equation to solve for the opposite side length, we have:

opposite = tan(73) * 6

Using a calculator, we find that the opposite side length (distance between Richard and spotlight B) is approximately 18.27m.

Comparing the two distances, we can see that the spotlight on the left building (A) is farther away from Richard by approximately 22.50m - 18.27m = 4.23m.