Respuesta :
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.
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.

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.
