C2 Trig

Can anyone help me with this:
A tower 60m high stands on the top of hill. from a point on the ground at sea level, the angles of elevation of the top and bottom are 49 and 37 degrees. find the height of the hill.

Reply 1

Angel Interceptor

In the future please post any Mathematics questions in the Maths section of Academic Help, not in here - this forum is meant for University discussions, eg. course choices, etc.

Reply 2

z_2j96

Anyone...

Reply 3

Kolya

Just use basic trig rules?

sinA/a = sinB/b = sinC/c

sinx=o/h
cosx=a/h
tanx=o/a

Reply 4

danniella

Hello!

T = top of tower.
B = bottom of tower
P = point on the ground at sea level
H = point at sea level that's directly under the tower (i.e. inside the hill)
Draw a diagram, and you end up with two triangles; one of them goes PHB where angle PHB = 90*
Another goes PBT and is non-right angled, with BT=60m.
* means "degrees" (sorry)
You want to find HB

angle TPH = 49*
angle BPH = 37*,
so angle BPT = 12* and
BPH is a right angled triangle so HBP=53*
Also PBT = 127* (makes straight line with HBP)
Angle PTB = 41*
Using the sine rule on the top triangle PBT,
(sin 12) /60 = (sin 41) / PB
So
PB = (sin 12) / (60 sin 41)

Using the lower triangle PHB,
HB/PB = sin 37
So HB = PB sin 37

and HB = [(sin 12)/(60 sin 41)] sin 37

(No calc on me, so I can't give you an answer to 3sf, but there you go)

Love
Danniella