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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 16°23'. When the boat stops, the angle of depression is 49°29' . 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.

check the picture below.

now, bear in mind that, there are 60minutes in 1 degree, so 23' is just 23/60 degrees or about 0.38 degrees.

and 29' is just 29/60 degrees or about 0.48 degrees.

so 16°23' is about 16.38°, and 49°29' is about 49.48°

[tex]\bf tan(16.38^o)=\cfrac{200}{a+b}\implies a+b=\cfrac{200}{tan(16.38^o)} \\\\\\ \boxed{b=\cfrac{200}{tan(16.38^o)}-a} \\\\\\ tan(49.48^o)=\cfrac{200}{b}\implies \boxed{b=\cfrac{200}{tan(49.48^o)}}\\\\ -------------------------------\\\\ \cfrac{200}{tan(49.48^o)}=\cfrac{200}{tan(16.38^o)}-a\\\\\\a=\cfrac{200}{tan(16.38^o)}-\cfrac{200}{tan(49.48^o)}[/tex]

make sure your calculator is in Degree mode.

AkshayG AkshayG · Answer 2 · 2019-11-04T05:23:35+01:00

The distance covered by the boat is [tex]\boxed{489.67{\text{ feet}}}.[/tex]

Further explanation:

The Pythagorean formula can be expressed as,

[tex]\boxed{{H^2} = {P^2} + {B^2}}.[/tex]

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

[tex]\boxed{\tan a = \frac{P}{B}}[/tex]

Explanation:

The perpendicular AB. The length of AB is [tex]200{\text{ feet}}.[/tex]

The angle of depression is [tex]\angle ACB = {14^ \circ }52'.[/tex]

One degree has 60 minutes.

[tex]{1^ \circ } = 60'[/tex]

[tex]\begin{aligned}\angle ACB &= 49 + \frac{{29}}{{60}}\\&= 49 + 0.48\\&= 49.48\\\end{aligned}[/tex]

The angle ADB is [tex]\angle ADB = {16^ \circ }23'.[/tex]

[tex]\begin{aligned}\angle ADB&= {16^ \circ }23' \\&= 16 + \frac{{23}}{{60}}\\&= {16.38^ \circ }\\\end{aligned}[/tex]

In triangle ABC.

[tex]\begin{aligned}\tan\left( {{{49.48}^\circ }} \right)&=\frac{{200}}{{BC}}\\1.13&= \frac{{200}}{{BC}}\\BC &= \frac{{200}}{{1.13}}\\BC &= 177{\text{ feet}}\\\end{aligned}[/tex]

In triangle ABD.

[tex]\begin{aligned}\tan \left( {{{16.38}^ \circ }} \right)&= \frac{{200}}{{BD}}\\0.30&= \frac{{200}}{{BD}}\\BD&= \frac{{200}}{{0.30}}\\BD &= 666.67{\text{ feet}}\\\end{aligned}[/tex]

The distance boat can travel can be obtained as follows,

[tex]\begin{aligned}DC& = BD - BC\\&= 666.67 - 177\\&= 489.67{\text{ feet}}\\\end{aligned}[/tex]

The distance covered by the boat is [tex]\boxed{489.67{\text{ feet}}}.[/tex]

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 https://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.