# Thread: How high and how far problem?

1. ## 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

2. 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

3. 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

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

$\displaystyle h = x \tan(20.2)$

$\displaystyle h = (x+24)\tan(18.8)$

solve the system for $\displaystyle x$ and $\displaystyle h$

5. 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.

. . $\displaystyle \begin{array}{cccccccc}A&=&159.8^o && a &=& 339.19 \\ B &=& 18.8^o && b &=& 316.57 \\ C &=& 1.4^o && c &=& 24 \end{array}$

$\displaystyle h \:=\: 339.19\sin18.8^o \:=\: 109.3$

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

. . $\displaystyle \begin{array}{ccccccccc}A &=& 20.2^o && a &=& 109.3 \\ B &=& 90^o && b &=& 316.37 \\ C &=& 69.8^o && c &=& 297.07 \end{array}$

Thus, she is 297.07ft away.

How is my logic?

Once I figured out your labeling scheme, it all became clear.