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.