How Can I Calculate the Height of a Building Using Right Angle Trigonometry?

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To calculate the height of a smaller building using right angle trigonometry, the observer measures angles of depression of 12° and 34° from a height of 150 ft. The observer initially attempts to use the tangent function with an angle of 56° but encounters errors. The correct approach involves creating two right triangles using the angles of depression and the height difference to find the distance to the smaller building and its height. Ultimately, the height of the smaller building is determined to be 103 ft after correcting the calculations.
ciubba
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Problem was initially posted in a technical math section, so is missing the homework template.
From a position 150 ft above the ground, an observer in a building measures angles of depression of 12° and 34° to the top and bottom, respectively, of a smaller building, as in the picture on the right. Use this to find the height h of the smaller building.

I have found that the angle between the observer and the ground is 56°; however, when I use the tangent function to get the distance between the smaller building and the observer [150 tan(56)] I get the wrong answer and I do not understand why. The only solutions to this problem that I have found involve the law of sines, which has not been covered yet. If I knew the distance between the observer and the smaller building (let's call it x), then I would be able to construct a right triangle from the observing to above the smaller building, where the triangle would have acute angles 78 and 12 and a leg of length x; however, I would not know how to proceed from there.

My questions are: why can I not use [150 tan(56)] to calculate x, how do I calculate x, and how do I find h from the triangle with leg x and angles 78 and 12.
 
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ciubba said:
From a position 150 ft above the ground, an observer in a building measures angles of depression of 12° and 34° to the top and bottom, respectively, of a smaller building, as in the picture on the right. Use this to find the height h of the smaller building.

I have found that the angle between the observer and the ground is 56°; however, when I use the tangent function to get the distance between the smaller building and the observer [150 tan(56)] I get the wrong answer and I do not understand why. The only solutions to this problem that I have found involve the law of sines, which has not been covered yet. If I knew the distance between the observer and the smaller building (let's call it x), then I would be able to construct a right triangle from the observing to above the smaller building, where the triangle would have acute angles 78 and 12 and a leg of length x; however, I would not know how to proceed from there.

My questions are: why can I not use [150 tan(56)] to calculate x, how do I calculate x, and how do I find h from the triangle with leg x and angles 78 and 12.
A drawing would be very helpful.

One thing I need to ask - is your calculator in degree mode?
 
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I did my best to draw it, and yes, my calculator is in degree mode. I used variables to represent any quantity that wasn't explicitly given to me by the problem. By the complement rule, I think theta is 56°; however, I get the wrong answer for x when I do [150 tan(56)]. It is apparently solvable without the law of sines as the book has not covered that idea yet.

I believe that A is 90° and that B is 78°. I want to find the variable "h," which is the height of the small building.
 

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Your equation of x = 150 tan(56°) is fine. I did it another way, but the two ways are equivalent. All you need is another equation that involves x and h.

In your drawing, extend the line segment of length h all the way to the top so that you have a rectangle that is 150' by x'. Using the 34° you can use the dimensions of the triangle whose acute angle is 34° - that's your second equation. You don't need the Law of Sines or the Law of Cosines - just some ordinary right triangle trig.

BTW, this is a homework problem, so I am moving it to the Homework & Coursework sections, which is where problems of this sort should be posted.
 
Mark44 said:
Your equation of x = 150 tan(56°) is fine. I did it another way, but the two ways are equivalent. All you need is another equation that involves x and h.

In your drawing, extend the line segment of length h all the way to the top so that you have a rectangle that is 150' by x'. Using the 34° you can use the dimensions of the triangle whose acute angle is 34° - that's your second equation. You don't need the Law of Sines or the Law of Cosines - just some ordinary right triangle trig.

BTW, this is a homework problem, so I am moving it to the Homework & Coursework sections, which is where problems of this sort should be posted.

Oh, it looks like I forgot to subtract h from 150! The answer is then 103. Thank you for the advice!
 
ciubba said:
Problem was initially posted in a technical math section, so is missing the homework template.
From a position 150 ft above the ground, an observer in a building measures angles of depression of 12° and 34° to the top and bottom, respectively, of a smaller building, as in the picture on the right. Use this to find the height h of the smaller building.

I have found that the angle between the observer and the ground is 56°; however, when I use the tangent function to get the distance between the smaller building and the observer [150 tan(56)] I get the wrong answer and I do not understand why. The only solutions to this problem that I have found involve the law of sines, which has not been covered yet. If I knew the distance between the observer and the smaller building (let's call it x), then I would be able to construct a right triangle from the observing to above the smaller building, where the triangle would have acute angles 78 and 12 and a leg of length x; however, I would not know how to proceed from there.

My questions are: why can I not use [150 tan(56)] to calculate x, how do I calculate x, and how do I find h from the triangle with leg x and angles 78 and 12.
wherw did you get 56 degree?
 
Cess said:
wherw did you get 56 degree?
Fill in the unknown angles based on the known angles.
 
ciubba said:
Oh, it looks like I forgot to subtract h from 150! The answer is then 103. Thank you for the advice!
Good. In the future, it would be good to show the intermediate results of your calculations. Then it may be possible for you and others to spot a mistake quickly.
 
FactChecker said:
Good. In the future, it would be good to show the intermediate results of your calculations. Then it may be possible for you and others to spot a mistake quickly.
What? Post #6 comes six years after the initial few posts. Maybe the original poster member has moved on already.
 
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symbolipoint said:
What? Post #6 comes six years after the initial few posts. Maybe the original poster member has moved on already.
Just trying to help. I didn't notice the dates of the posts.
 
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FactChecker said:
Just trying to help. I didn't notice the dates of the posts.
I have done the same thing at times.
 
  • #12
symbolipoint said:
What? Post #6 comes six years after the initial few posts. Maybe the original poster member has moved on already.
Yep, the OP was last seen about four years ago. I thought about deleting the new post, as he would not likely get a reply from the orig. poster, but decided against it, and posted a reply instead.
 
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