Help Solving.

[aroberts5124]

Hello, I have been trying to solve this all night. I think I have successfully solved the 30° angle. But I can not seem to get the 40° angle. We are only given the two angles. These are angles of elevation and depression. We also know the angles are from eye level and that is 6' above the ground. I am needing to figure out the total height. Also, we can not assume that there is a 90° angle even though it looks to be one, it could have a smaller angle. It is not to sale.

 

[Otis]

I think I have successfully solved the 30° angle. But I can not seem to get the 40° angle. We are only given the two angles.

Hi aroberts. I'm not sure what you mean, when talking about solving/getting the given angles.

… we can not assume that there is a 90° angle even though it looks to be one …

If the line segments having lengths x and 6 are not perpendicular to the line-of-sight, then we don't have enough information to find x. Let's assume the angles that look like right angles are 90º.

L = base length of right triangle containing 40º angle

Then, base length of right triangle containing 30º angle = L+12

Using the right-triangle identity for tangent, we can write two equations and solve for x (using substitution).

?

 

[aroberts5124]

Hi aroberts. I'm not sure what you mean, when talking about solving/getting the given angles.


If the line segments having lengths x and 6 are not perpendicular to the line-of-sight, then we don't have enough information to find x. Let's assume the angles that look like right angles are 90º.

L = base length of right triangle containing 40º angle

Then, base length of right triangle containing 30º angle = L+12

Using the right-triangle identity for tangent, we can write two equations and solve for x (using substitution).

?

Okay, that is really the conclusion that I came to as well. It is an assignment where we have to measure something outside that is too tall to measure. Mine happens to be a power pole. Theoretically, it should be straight, however it could be leaning. I think it has to be a 90° to make it work. The angles 40° and 30° are measurements I took. We have to come up with x and all we get is the 2 angles (40°,30°) the distance between (12') and the hight from the ground (6') I think we have to assume the angle is 90° to be able to solve.

 

[Cubist]

Okay, that is really the conclusion that I came to as well. It is an assignment where we have to measure something outside that is too tall to measure. Mine happens to be a power pole. Theoretically, it should be straight, however it could be leaning. I think it has to be a 90° to make it work. The angles 40° and 30° are measurements I took. We have to come up with x and all we get is the 2 angles (40°,30°) the distance between (12') and the hight from the ground (6') I think we have to assume the angle is 90° to be able to solve.

Does this diagram help you to make any different conclusions?



EDIT: by the way the green monstrosity is my attempt at drawing a non-vertical electric pylon!

 

[Deleted member 4993]

This the principle surveyors use to estimate height of a mountain peak - without actually climbing it.

Note:

tan(40^o) = x/DB

tan(30^o) = x/(DB+12)

 

[Otis]

… we have to measure something outside that is too tall to measure. Mine happens to be a power pole. Theoretically, it should be straight, however it could be leaning …

I'm thinking that you've been asked to measure an object's height as it currently stands (i.e., the vertical distance from your object's highest point to the ground). If so, then it doesn't matter whether the current height would change, by tilting the pole one way or another. (That would be a question for a different exercise.)

We could certainly investigate how the height would change, were the pole to deviate from perpendicular by various measures of degrees. For example, assuming your diagram shows angle measures for a perpendicular pole, how would a tilt of 1º from perpendicular alter the height? More than half an inch?