Mathematics: The tower is 650 meters high. suppose a building is erected such that the base of the building is on the same plane as the base of the tower, the angle of elevation from the top of the building to the top of the tower is 80.38° and the angle of depression from the top of the building to the foot of the tower is 65.05°. how high would the building have to be?

Here is the point A (2, 3) graphed. You can access a graphing calculator on desmos as well....

The tower is 650 meters high. suppose a building

kiki4832

01.01.2022

$173.57\ m$

Step-by-step explanation:

Let

h -----> the height of the building

x -----> the horizontal distance between the building and the tower

we know that

$tan(80.38\°)=(650-h)/x$

solve for x

$x=(650-h)/tan(80.38\°)$ -----> equation A

$tan(65.05\°)=h/x$

solve for x

$x=h/tan(65.05\°)$ ------> equation B

equate equation A and equation B and solve for h

$h/tan(65.05\°)=(650-h)/tan(80.38\°)$

$tan(80.38\°)h=(650-h)tan(65.05\°)$

$tan(80.38\°)h=(650)tan(65.05\°)-(h)tan(65.05\°)$

$h[tan(80.38\°)+tan(65.05\°)]=(650)tan(65.05\°)$

$h=(650)tan(65.05\°)/[tan(80.38\°)+tan(65.05\°)]$

$h=173.57\ m$

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