Trigonometry Word problem with angles of elevation

[jmgXD6]

Homework Statement

An Antenna that is 90 feet high is on top of a hill. From a point at the base of the hill, the angles of elevation to the top and bottom of the antenna are 28.5° and 25°, respectively. To the nearest whole number of feet, how high is the hill?

A, 189 ft
B, 213 ft
C, 548 ft
D, 623 ft
E, 697 ft

Homework Equations

well I know Sinθ=o/h Cosθ=a/h Tanθ=o/a

The Attempt at a Solution

I don't know the length of the base so I couldn't figure it out like the textbook examples, but from angle 25° I figured the hypotenuse to the base of the antenna was 212.5, which would round to 213 so I know A and B can't be it.

 

[Mark44]

Homework Statement

An Antenna that is 90 feet high is on top of a hill. From a point at the base of the hill, the angles of elevation to the top and bottom of the antenna are 28.5° and 25°, respectively. To the nearest whole number of feet, how high is the hill?

A, 189 ft
B, 213 ft
C, 548 ft
D, 623 ft
E, 697 ft

Homework Equations

well I know Sinθ=o/h Cosθ=a/h Tanθ=o/a

The Attempt at a Solution

I don't know the length of the base so I couldn't figure it out like the textbook examples, but from angle 25° I figured the hypotenuse to the base of the antenna was 212.5, which would round to 213 so I know A and B can't be it.

I hope you have a drawing of the two triangles that represent this situation. You have the hypotenuse of the right triangle whose acute angle is 25°. That makes the other angle of this triangle 65°. You are given the length of the side opposite the 3.5° angle, and you should be able to figure out the obtuse angle of the upper triangle. Knowing two angles and a side, you should be able to use the Law of Sines to determine the hypotenuse of the upper triangle, and then get calculate the height of the hill.

 

[jmgXD6]

Where did you get 3.5° and where would the 28.5° angle go?

 

[LCKurtz]

Call the height of the hill $h$ and the distance from the base of the hill $d$. Write the equations for the tangents of your two angles. Two equations in unknowns $h$ and $d$. Eliminate the $d$ and solve.

 

[Mark44]

Where did you get 3.5°

28.5° - 25° = 3.5°

and where would the 28.5° angle go?

There are three triangles involved: the lower triangle (a right triangle with an acute angle of 25°); the upper triangle that has one angle of 3.5°; a large right triangle that contains the two previous triangles, and that has an acute angle of 28.5°.

You have made a drawing, right?

 

[Mark44]

Call the height of the hill $h$ and the distance from the base of the hill $d$. Write the equations for the tangents of your two angles. Two equations in unknowns $h$ and $d$. Eliminate the $d$ and solve.

I like your strategy better than mine...

 

[jmgXD6]

I know where to put the lower triangle, but where would the top angle of evaluation, 28.5 degrees, go?

Here's what I drew so far.
http://s288.photobucket.com/user/AC...lusquestion85_zpse8fde88e.jpg.html?sort=3&o=0

 

[Mark44]

I know where to put the lower triangle, but where would the top angle of evaluation, 28.5 degrees, go?

Here's what I drew so far.
http://s288.photobucket.com/user/AC...lusquestion85_zpse8fde88e.jpg.html?sort=3&o=0

That's angle of elevation, not evaluation. You had that in your title, and I fixed it.

Your picture is wrong. You have the angle of elevation to the top of the antenna as 25°. It should be 28.5°. The angle of elevation to the bottom of the antenna is 25°.

 

[jmgXD6]

So where I had 25 degrees it should be 28.5, and 65 should be 61.5 degrees, correct?

 

[Mark44]

Yes.

 

[jmgXD6]

So where would the bottom angle of elevation go then?

 

[LCKurtz]

So where would the bottom angle of elevation go then?

Read your original post. It tells you.