Just can't figure this angles of elevation and depression problem out? Involves sun and building's shadow?

Question

When the sun is 62 degrees above the horizon, a building casts a shadow 18 m long. How tall is the building?

If you cant at least tell me how to set it up, it would be great.
THANK YOU!

Answer 1

The 62 degrees is the angle of elevation from the tip of the shadow to the top of the building. So draw the building straight up, its shadow on the ground and connect with the hypotenuse (which actually isn't needed but draw it anyway). So the acute angle on the ground is 62, the building (b) is opposite leg, and the shadow (18) is adjacent, making the equation be tan 62 = b/18

And b = 18 tan 62,

Answer 2

Think trigonometry
Sin = o/h ; Cos = a/h ; Tan = o/a

We have 18 m base adjacent(a) to the angle.
We need to find height which is opposite(o) to the given angle.
The is angle is 62 degrees.
So the Trig function to select is 'Tan' because its ratio is (o) height over (a) base.

Tan (62) = o/18
o = 18 x Tan62

Find the value of Tan 62 first (calculator or Trig Tables).

o = 18 x 1.880726465
o= 21.40558295 m
o = 21. 405 m (3 dp) This level of decimal points takes the height to 21 m 40 cm 5 mm. Any further decimals is trivial against a building height.