Thread: How high and how far problem?

Hi all... could someone check me here. I have the right process I think, but I'm not 100% sure.

Here is the problem...

A student stands a certain distance from a building and measures the angle to the top of the building as 18.8°. The student then walks 24 ft straight towards the building and measures the angle to the top of the building as 20.2°.
A What is the height of the building, and
b) How close was the student to the bottom of the building after she had walked the 24 ft?

This is what I came up with to solve the height of the building.

A = 159.8
B = 18.8
C = 1.4
a = 339.19
b = 316.57
c = 24

h = 339.19sin(18.8) = 109.3

This is what I have for the distance from the building.

A = 20.2
B = 90
C = 69.8
a = 109.3 (the height from the previous answer)
b = 316.57
c = 297.07

thus...She is 297.07ft away.

How is my logic?

By the way, I am so glad to be a part of this forum. I am taking Math 122 online and need all the help I can get. I hope I can also contribute to this forum for others as well.

Neil

Originally Posted by Smokinoakum

Hi all... could someone check me here. I have the right process I think, but I'm not 100% sure.

Here is the problem...

A student stands a certain distance from a building and measures the angle to the top of the building as 18.8°. The student then walks 24 ft straight towards the building and measures the angle to the top of the building as 20.2°.
A What is the height of the building, and
b) How close was the student to the bottom of the building after she had walked the 24 ft?

This is what I came up with to solve the height of the building.

A = 159.8
B = 18.8
C = 1.4
a = 339.19
b = 316.57
c = 24

h = 339.19sin(18.8) = 109.3

This is what I have for the distance from the building.

A = 20.2
B = 90
C = 69.8
a = 109.3 (the height from the previous answer)
b = 316.57
c = 297.07

thus...She is 297.07ft away.

How is my logic?

By the way, I am so glad to be a part of this forum. I am taking Math 122 online and need all the help I can get. I hope I can also contribute to this forum for others as well.

Neil

You are going to need to provide us with a diagram or some explanation of what A, B, C, a, b, and c are.

-Dan

Originally Posted by Smokinoakum

Hi all... could someone check me here. I have the right process I think, but I'm not 100% sure.

Here is the problem...

A student stands a certain distance from a building and measures the angle to the top of the building as 18.8°. The student then walks 24 ft straight towards the building and measures the angle to the top of the building as 20.2°.
A What is the height of the building, and
b) How close was the student to the bottom of the building after she had walked the 24 ft?

This is what I came up with to solve the height of the building.

A = 159.8
B = 18.8
C = 1.4
a = 339.19
b = 316.57
c = 24

h = 339.19sin(18.8) = 109.3

This is what I have for the distance from the building.

A = 20.2
B = 90
C = 69.8
a = 109.3 (the height from the previous answer)
b = 316.57
c = 297.07

thus...She is 297.07ft away.

How is my logic?

By the way, I am so glad to be a part of this forum. I am taking Math 122 online and need all the help I can get. I hope I can also contribute to this forum for others as well.

Neil

The way I got the first measurement for the height was as follows.
The first angle was given to be 18.8°. I labeled that angle B. Then we're given the base as 24ft. I labeled that c. Then we're told the next angle is 20.2°. I subtracted that from 180 to get the opposite angle, which is 159.8 and I labeled that A. I solved for angle C by adding 159.8 + 18.8 = 1.4°. I then solved for side a and b with the given angles.

Now for the distance problem I simply took the second angle given as 20.2° and the 90° angle that is next to the building and subtracted that from 180 as well given me 180 - 20.0 - 90 = 69.8. I also used the height of the building of 109.3 as side a. Then I solved for side b and c.

I hope this clarifies this a bit. I would like to show a diagram but I'm not sure how to do that here. Maybe you could give me a tip for that as well.

Thanks for you help!

Neil

using "paint" to sketch a diagram ... save as a jpg file and upload as an attachment.





solve the system for and

Hello, Smokinoakum!

A student stands a certain distance from a building
and measures the angle to the top of the building as 18.8°.
The student then walks 24 ft straight towards the building
and measures the angle to the top of the building as 20.2°.

(a) What is the height of the building?

(b) How close was the student to the bottom of the building
after she had walked the 24 ft?


This is what I came up with to solve the height of the building.

. .




This is what I have for the distance from the building.

. .

Thus, she is 297.07ft away.

How is my logic?

Once I figured out your labeling scheme, it all became clear.
Your logic and your work is excellent!


skeeter's approach is the best way.

Thanks guys...It's nice to know I'm kinda getting this stuff.

Thanks for the tip about using paint to label my diagram out.

Neil