how do you find the height of a right triangle given two elevation angles?

Question

this is the question i am having trouble with:

 

At a certain distance the angle of elevation to the top of a building is 60 deg. 

From 40 feet further back, the angle of elevation is 45 deg. what is the height of the building

Answer 1

Enrico, the first step is draw the figure as described.

 

Let x=the certain distance in which its angle of elevation to the roof top of the building is 60 degrees.

Due to this triangle being a 30-60 degree right triangle, this would indicate the height of the building is x√3.

 

Then x+40 is the distance from the building when the angle of elevation is 45 degrees.

Since the triangle is now a 45-45 degree right triangle would indicate the horizontal distance is equal to the height of the building. With this being said, now the equation can be set up to solve for x.

 

x+40=x√3 Square both sides.

(x+40)^2=3x^2

x^2+80x+1600=3x^2 Rewrite the equation so that you could use the quadratic formula to solve for x.

 

2x^2-80x-1600=0

 

x=(80±√(80^2-4(2)(-1600)))/(2*2)=(80±√(6400+12800))/4=(80±√(19200))/4

x≅(80+138.564)/4≅54.641

 

Hence, the height of the building is 54.641+40=94.641 feet.

Answer 2

Hi Enrico,

 

Let the distance from the building to the first point be x

 

so the distance from the building to the 2nd point is x + 40

 

and since the second angle is 45 deg our right triangle will be an isosceles rt triangle

 

that means the two legs will be equal

 

so the height of the building will also be x + 40

 

now go back to the first point and you have a 30 60 90 triangle

 

with one side x the other side x+40 and the hypothenuse 2x

 

and then the sin of 60 deg will equal

 

sin 60 = (x+40)/2x

 

sqrt(3)/2 = (x+40)/2x

 

1.4641x = 80

 

x = 54.641  and the building is x + 40

 

so

 

= 94.641   (this is the height of the building

 

Hope this helps