Find the Distance and Angle to a Tall Building with Calc/Trig Homework Help!"

[anthonym44]

A surveyor stands on flat ground at an unknown distance from a tall building. She measures the angle from the horizonal ground to the top of the building; this angle is pi/3. next she paces 40ft further away from the building. the angle from the ground to the top of the building is now measured to be pi/4.

i think i got this one right, but I am new to the site so i thought i would just check myself with this problem. i got equation 40 + x = y beause the horizontal axis must equal the vertical when the angle is 45 degrees (pi/4). Then i used tan(60) = y/x to solve for x and then i subsituted eventually solving for y. my final answer is 94.8ft. part B asks for the angle if the person moves further back 20 ft. i just added that length to 94.8 and used tan to solve eventually getting the angle to be 39.5 degrees. If anyone has the time can they check the work to make sure i did it correctly? Thanks.

 

[HallsofIvy]

A surveyor stands on flat ground at an unknown distance from a tall building. She measures the angle from the horizonal ground to the top of the building; this angle is pi/3. next she paces 40ft further away from the building. the angle from the ground to the top of the building is now measured to be pi/4.

i think i got this one right, but I am new to the site so i thought i would just check myself with this problem. i got equation 40 + x = y beause the horizontal axis must equal the vertical when the angle is 45 degrees (pi/4).
Then i used tan(60) = y/x to solve for x and then i subsituted eventually solving for y. my final answer is 94.8ft. part B asks for the angle if the person moves further back 20 ft. i just added that length to 94.8 and used tan to solve eventually getting the angle to be 39.5 degrees. If anyone has the time can they check the work to make sure i did it correctly? Thanks.

Wouldn't it be better to check it yourself? You say that the building has height 94.8 feet. The first distance from building is 94.8- 40= 54.8. Is tan(pi/3)= 94.8/54.8? The last distance is 94.8+ 20= 114.8. Is tan(39.4)= 94.8/114.8?

By the way, why are you converting from radians (as given) to degrees? Certainly your answer should be given in radians.

And, of course, this is not in any sense "calculus and analysis". I am moving the thread to "general math".

 

[tacman]

Your approach and final answer seem correct! Here is a step-by-step explanation in case you or other readers need it:

First, let's label the unknown distance from the building as x and the height of the building as y. We can then use the tangent function to set up an equation for the first scenario:

tan(pi/3) = y/x

We know that tan(pi/3) = sqrt(3), so our equation becomes:

sqrt(3) = y/x

Next, we can set up a similar equation for the second scenario, where the person moves 40ft further away:

tan(pi/4) = y/(x+40)

We know that tan(pi/4) = 1, so our equation becomes:

1 = y/(x+40)

Now, we have two equations and two unknowns (x and y), so we can solve for them. We can rearrange the first equation to solve for y:

y = sqrt(3)x

Then, we can substitute this expression for y into the second equation:

1 = (sqrt(3)x)/(x+40)

We can then solve for x by cross-multiplying and simplifying:

x = 40/(sqrt(3) - 1) ≈ 94.8ft

So the distance from the building is approximately 94.8ft. To find the angle for the second scenario, where the person moves 20ft further back, we can simply add 20ft to our previous result:

x + 20 = 114.8ft

And then use the tangent function again to solve for the angle:

tan^-1(y/(x+20)) ≈ 39.5 degrees

So the angle is approximately 39.5 degrees. Your approach of adding the distance to the previous result and using the tangent function to find the angle is correct. Great job!