Question

A person is watching a boat from the top of a lighthouse. The boat is approaching the lighthouse directly. When first noticed, the angle of depression to the boat is 14°52'. When the boat stops, the angle of depression is 45°10'. The lighthouse is 200 feet tall. How far did the boat travel from when it was first noticed until it stopped? Round your answer to the hundredths place

Answer 1

Answer:

Boat traveled 553.24 feet towards the lighthouse.

Step-by-step explanation:

In the figure attached AB is the light house of height 200 feet.

Angle of depression of the boat from the top of a lighthouse = angle of elevation of the lighthouse from the boat = 14°52'

so 1' = $\frac{1}{60}$ degree

so angle of elevation at point C = 14 + $\frac{52}{60}$

So angle of elevation from C = (14 + 0.87) = 14.87°

Similarly, when boat arrives at point D angle of elevation = 45°10' = 45 + $\frac{10}{60}$ = 45.17°

Now we have to calculate the distance CD, traveled by the boat.

In ΔABC

tan14.87 = $\frac{200}{BC}$

0.2655 =  $\frac{200}{BC}$

BC = $\frac{200}{0.26552}$

BC = 753.239 feet

Similarly in ΔABD

tan45.17 = $\frac{200}{BD}$

1 = $\frac{200}{BD}$

BD = 200 feet

So distance CD = BC - BD

CD = 753.239 - 200

     = 553.24 feet

Therefore, Boat traveled 553.24 feet towards the lighthouse.

Answer 2

The distance covered by the boat is $\boxed{556.71{\text{ feet}}}.$

Further explanation:

The Pythagorean formula can be expressed as,

$\boxed{{H^2} = {P^2} + {B^2}}.$

Here, H represents the hypotenuse, P represents the perpendicular and B represents the base.

The formula for tan of angle a can be expressed as

$\boxed{\tan a = \frac{P}{B}}$

Explanation:

The perpendicular AB. The length of AB is $200{\text{ feet}}.$

The angle of depression is $\angle ACB = {14^ \circ }52'.$

One degree has 60 minutes.

${1^ \circ } = 60'$

$\begin{aligned}\angle ACB&=14+\frac{{52}}{{60}}\\&= 14 + 0.87\\&= 14.87\\\end{aligned}$

The angle ABD is $\angle ADB={45^ \circ }10'.$

$\begin{aligned}\angle ADB &= {45^\circ }10' \\&= 45 + \frac{{10}}{{60}}\\&= 45 + 0.17 \\ &= {45.17^\circ }\\\end{aligned}$

In triangle ABC.

$\begin{aligned}\tan \left( {{{14.87}^\circ }} \right)&=\frac{{200}}{{BC}}\\0.26&= \frac{{200}}{{BC}}\\BC&= \frac{{200}}{{0.265}}\\BC &= 754.72{\text{ feet}}\\\end{aligned}$

In triangle ABD.

$\begin{aligned}\tan \left({48.217} \right)&= \frac{{200}}{{BD}}\\1.01&= \frac{{200}}{{BD}}\\BD&= \frac{{200}}{{1.01}}\\BD &= 198.01{\text{ feet}}\\\end{aligned}$

The distance boat can travel can be obtained as follows,

$\begin{aligned}DC &= BD- BC\\&= 754.72 - 198.01\\&= 556.71{\text{ feet}}\\\end{aligned}$

The distance covered by the boat is $\boxed{556.71{\text{ feet}}}.$

Kindly refer to the image attached.

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Trigonometry

Keywords: perpendicular, person watching boat, top, lighthouse, angle of depression, angle of elevation, 200 feet tall, travel, sides, right angle triangle, triangle, altitudes, hypotenuse, on the triangle, hypotenuse, trigonometric functions.