Homework Help: Need help -daughter has trig exam tomorrow

1. Jun 4, 2007

crm

1. The problem statement, all variables and given/known data

A surveyor on the ground takes two readings of the angles of elevation at the top of the tower. From 150' apart, the measures are 50 and 70 degrees. Find the tower's height to the nearest foot.

2. Relevant equations

3. The attempt at a solution

2. Jun 4, 2007

ice109

law of cosines

3. Jun 4, 2007

crm

can you work through to the solution?

4. Jun 4, 2007

crm

My daughter says that to use the law of cosines you need to know two sides and an angle. In the above problem, we only know one side. Any comments?

5. Jun 4, 2007

chroot

Staff Emeritus
Have you drawn a picture? You cannot just go blindly applying rules without understanding the problem.

If you draw a picture, you'll see that there are two triangles. You know all the angles of both, and one side of one of them. From there, you can use the law of sines to find any other side you want.

- Warren

6. Jun 4, 2007

crm

Yes, we have drawn a picture, but we do not know any sides. We know that the entire length of the two triangles together is 150, but how do we determine the length of the bases? Are they halved? and if so, how do you know? Because if they are not halved then there is no way to find the side....

7. Jun 4, 2007

chroot

Staff Emeritus
Draw a small right triangle. This represents the measurement made by the surveyor when he's close to the tower. He measures the angle to be 70 degrees in this position. Since it's a right triangle, you know all the angles of that triangle. You do not know any sides.

Imagine that the surveyor walks 150 feet backward away from the tower. When he looks at the tower again, he measures an angle of 50 degrees. Draw another triangle adjacent to this one. It will not be a right triangle. The base of this triangle is 150 feet.

(The entire length of the triangles together is not 150 feet; only the base of this second triangle is 150 feet. You do not know the sum of the bases.)

Simple inspection shows that you know all of the angles of this second triangle, and you know one of its sides. You can use the law of sines to find any other side. For example, use the law of sines to find the hypotenuse of the first triangle. From there, use the law of sines again to find the height of the tower.

- Warren

Last edited: Jun 4, 2007
8. Jun 4, 2007

crm

Okay, so the first triangle is a right triangle with the angle at the top being 50 degrees and the angle at the bottom being 40 degrees. The second triangle has a 20 degree angle at the top (70 - 50) and angles at the bottom of 140 and 20 degrees. And the length at the bottom is 150 feet.

We solve for the longest side with the formula 150/ sin 20 = X/sin 140, or 282. We then solve for the height of the tower with the formula 282/sin 90 = X/sin 20, or 96.

Unfortunately, the answer key says the correct answer is 360, so we're way off.

Any further thoughts?

9. Jun 4, 2007

chroot

Staff Emeritus
Your first triangle is incorrect. The 50 degree angle is at the bottom. The question states "angles of elevation" are 50 and 70 degrees. An angle of elevation means an angle relative the horizontal.

- Warren

10. Jun 4, 2007

Sleek

Last edited by a moderator: May 2, 2017
11. Jun 4, 2007

chroot

Staff Emeritus
Thanks for the diagram, Sleek. I didn't have time to make one.

- Warren

12. Jun 5, 2007

crm

Warren and Sleek:

Thank you so much!

crm

13. Jun 5, 2007

TheoMcCloskey

crm -- just curious, what answer did you get --

I worked this out two ways using a figure very similar to Sleek's and came up with 315.7 feet - not 360 feet.

I use tangents, eg, (ref Sleek's figure)

Tan(70 deg) = h / x

Tan(50 deg) = h / (x+150)

Solve for h

$$h = \frac{T_{50} \cdot T_{70}}{T_{70} - T_{50}} \cdot 150$$

14. Jun 5, 2007

chroot

Staff Emeritus
I also got 315.7 feet, by the way (checked in several different ways).

- Warren

15. Jun 6, 2007

crm

Warren and Sleek: Thank you so much for your help on Monday night. I found the diagram and the explanation of the angles of elevation. I found them when I got up Tuesday morning and provided them to my daughter. She reviewed this prior to her exam. She later advised that she had an almost identical question in her exam and was able to handle it. Your help made it possible. I'll ask her what answer she got when she worked through the problem, and let you know. crm

16. Jun 6, 2007

chroot

Staff Emeritus
Glad to know we could help, crm.

- Warren