Trigonometry

Question

Can someone provide a solution to this problem?
The angle of elevation from the roof of a 10 m high building to the top of another building is 41º . The two buildings are 18 m apart at the base.

- How tall, to the nearest metre, is the taller building?

Answer 1

Draw a diagram - This results in a right triangle with an adjacent side of 18m.  Let "x" be the opposite side for this triangle.

 

tan 41 = opposite/adjacent = x/18m.  Therefore, x = 18tan41.  This "x" is the additional height above the 10m that the other building goes up.

 

For the total height, add 10 to x.  Thus, the height of the other building is 10 + 18tan41.

 

Answer 2

Okay, you need to follow along on paper with this one. Grab some paper and a pencil. I'll wait...

...Okay, ready? Great. first, draw a right triangle. The vertical side represents the height of the taller building compared to the smaller building (we'll deal with their actual heights later). The horizontal side (the base) is the 18m. Finally, the lower right angle is 41°. Relative to this angle, the two sides we know are the opposite and the adjacent. Put them together and you get the tangent function:

Tan (θ) = Opp / Adj

Plug in everything we know:

Tan (41°) = 18 / X

.869 = 18 / X

Put the left side over 1...

.869 / 1 = 18/ X

...and you can now cross-multiply:

.869X = 18

Now solve:

.869X / .869 = 18 / .869

X = 20.707                    This tells us the taller building exceeds the shorter building by 20.707m

                                    Remember that the smaller building was already 10M high. That means you total:

20.707 + 10 = 30.707m ≈ 31m to the nearest meter