Question 5.6.7: Finding the height of an object from a distance The angle of...
Finding the height of an object from a distance
The angle of elevation of the top of a water tower from point A on the ground is 19.9°. From point B, 50.0 feet closer to the tower, the angle of elevation is 21.8°. What is the height of the tower?
Let y represent the height of the tower and x represent the distance from point B to the base of the tower, as shown in Fig. 5.88.
At point B, tan 21.8° = y/x or
x=\frac{y}{\tan 21.8^{\circ}}Since the distance to the base of the tower from point A is x + 50,
\tan 19.9^{\circ}=\frac{y}{x+50}Or
y = (x + 50) tan 19.9°.
To find the value of y we must write an equation that involves only y. Since x = y/tan 21.8°, we can substitute y/tan 21.8° for x in the last equation:
y=\left(\frac{y}{\tan 21.8^{\circ}}+50\right) \tan 19.9^{\circ} ,
y=\frac{y \cdot \tan 19.9^{\circ}}{\tan 21.8^{\circ}}+50 \tan 19.9^{\circ} Distributive property
y-\frac{y \cdot \tan 19.9^{\circ}}{\tan 21.8^{\circ}}=50 \tan 19.9^{\circ} ,
y\left(1-\frac{\tan 19.9^{\circ}}{\tan 21.8^{\circ}}\right)=50 \tan 19.9^{\circ} Factor out y.
y=\frac{50 \tan 19.9^{\circ}}{1-\frac{\tan 19.9^{\circ}}{\tan 21.8^{\circ}}} \approx 191 \text { feet }This computation is shown on a graphing calculator in Fig. 5.89.
