0
$\begingroup$

A radio antenna that is 40 feet high stands on top of the Wentworth Building. From a point in front of Bailey's Drugstore, the angle of elevation of the top of the pole is 54°54'and the angle of elevation of the bottom of the pole is 47°30'.How tall is the Wentforth Building.

I think the Wentforth Building is 40-X

I have a diagram I don't think I'm right but here enter image description here

Thats how angle of elevation is supposed to be right, like straight and then diagonal? I don't really know what to do after this, How am I supposed to solve this?

BD is Bailey's Drugstore and WB is Wentworth Building

  • 3
    Well, I don't know, but I'd assume that the angle of elevation to the base of the tower was a sub-angle of the angle of elevation to the top. That is to say, sighting at $47^\circ\,30'$ spots the base and sighting at $54^\circ\,54'$ spots the top. Assuming this is correct, let $X$ be the height of WB, so $X+40$ is the height of WB+tower, let $Y$ be the distance to WB and compute the two tangents. – 2017-01-04
  • 1
    [Here's a better diagram.](https://i.imgur.com/dGZXU4T.jpg) The two angles of elevation overlap. – 2017-01-04
  • 0
    but aren't we just finding the height of WB isn't that X-40 or are we counting the tower as well, so you're saying 54°54′ would be like near the radio tower? – 2017-01-04
  • 0
    @AkivaWeinberger Oh I don't really get it can you explain? so is the height of the tower 40-x or is it 40+x? – 2017-01-04
  • 0
    Using @lulu's variables, $X$ is the height of the tower (not including the antenna). $X+40$ is the height of the tower including the antenna. $54^\circ54'$ is the angle of elevation to the top of the antenna. – 2017-01-04
  • 0
    @AkivaWeinberger There's two angles, which one will I use to find X? – 2017-01-04
  • 0
    Both. Call the distance between the buildings (the length of the horizontal line on the bottom) $Y$. Using the definition of tangent (opposite over adjacent), what's the tangent of $54^\circ54'$? What's the tangent of $47^\circ30'$? – 2017-01-04
  • 0
    @AkivaWeinberger so $$Tan(54°54′)=40+x/y$$ but there's two unknowns right? – 2017-01-04
  • 0
    @AkivaWeinberger How would I solve this? – 2017-01-04
  • 0
    What's the tangent of the other angle? – 2017-01-04
  • 0
    the same I think $$Tan(47°30')=x+40/y $$ – 2017-01-04
  • 0
    Should I equal them like find x in one and then plug that value in the other? – 2017-01-04

1 Answers 1

2

Adapting the earlier diagram, consider the triangle ABC.

We know that $AB=40$, that $\angle ACB=7^{\circ}24'$ (difference between two given angles) and that $\angle CAB=35^{\circ}06'$.

Use the Sine Rule to find $BC$.

Then the height of the building is the angle opposite a $47^{\circ}30'$ angle in a right-angled triangle with hypotenuse $BC$. Just use straightforward trigonometry for that.

enter image description here