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