pq is a tower standing vertically on the horizontal ground. from a, the angle of elevation of top p of the tower is 45°. on moving 50 m. up a slope of 15°, the angle of elevation becomes 75° from b. find height of the tower
so, where does the slope begin, and does it drop lower when moving away from the tower, or what?
To find the height of the tower, we can use the trigonometric ratios of sine and tangent.
Let's break down the problem and label the given information:
1. Angle of elevation of the top of the tower from A (denoted as angle α): 45°
2. Angle of elevation of the top of the tower from B (denoted as angle β): 75°
3. Distance from point A to the base of the tower: Unknown
4. Distance from point B to the base of the tower: 50 meters
5. Angle of the slope from point B (denoted as angle θ): 15°
First, let's find the distance from point A to the base of the tower (denoted as h).
We can use the trigonometric ratio of tangent for this:
tangent(angle α) = opposite/adjacent
tan(45°) = h/x
Since tangent(45°) = 1, we have:
1 = h/x
Therefore, x = h, so the distance from point A to the base of the tower is h.
Now, let's find the height of the tower (denoted as t).
We know that the distance from point B to the base of the tower is 50 meters. Using the trigonometric ratio of sine, we can write:
sine(angle β) = opposite/hypotenuse
sin(75°) = t/50
Rearranging the equation, we have:
t = 50 * sin(75°)
Using a calculator, we can compute the value of sin(75°) ≈ 0.9659, therefore:
t ≈ 50 * 0.9659 ≈ 48.29
So, the height of the tower is approximately 48.29 meters.