Question

a boat is heading towards a lighthouse, whose beacon - light is 143 feet above the water. from point a, the boats crew measures the angle of elevation to the beacon, 5°, before they draw closer. they measure the angle of elevation a second time from point b to be 10°. find the distance from point a to point b. round your answer to the nearest tenth of a foot if necessary.

Explanation:

Step1: Find distance from A to L

We know that in right triangle \( \triangle ALP \) (where \( P \) is the top of the lighthouse), \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \). For angle of elevation \( 5^\circ \) at point \( A \), the opposite side is the height of the lighthouse \( 143 \) feet and adjacent side is \( AL \). So, \( \tan(5^\circ)=\frac{143}{AL} \). Solving for \( AL \), we get \( AL = \frac{143}{\tan(5^\circ)} \).

Step2: Find distance from B to L

For angle of elevation \( 10^\circ \) at point \( B \), using the same right triangle concept, \( \tan(10^\circ)=\frac{143}{BL} \). Solving for \( BL \), we get \( BL=\frac{143}{\tan(10^\circ)} \).

Step3: Find distance from A to B

The distance \( AB = AL - BL \). Substituting the values of \( AL \) and \( BL \), we have \( AB=\frac{143}{\tan(5^\circ)}-\frac{143}{\tan(10^\circ)} \).

Calculating the values:
\( \tan(5^\circ)\approx0.0875 \), \( \tan(10^\circ)\approx0.1763 \)
\( AL=\frac{143}{0.0875}\approx1634.29 \)
\( BL=\frac{143}{0.1763}\approx811.12 \)
\( AB = 1634.29 - 811.12 = 823.17\approx823.2 \)

Answer:

\( 823.2 \) feet