Question

A flagpole 12.5m high stands at the top of a bldg. From a pt. P on the street the angle of elevation of the top of the pole is 54 degrees and the angle of elevation of the foot of the pole is 46 degrees.
How high is the bldg.?

Answer 1

Did you make your sketch?
On mine, I labeled the bottom of the building Q,
the top of the flagpole B and its bottom A
I have a right-angled triangle APQ and a scalene triangle ABP.
Angles can be easily found
In triangle ABP, I have
angle BPA = 8°
angle PBA = 36°
angle PAB = 136° and AB = 12.5
So we can use the Sine Law to find AP
AP/sin36 = 12.5/sin8
AP = 12.5sin36/sin8

in triangle APQ, which is right-angled
AQ/AP = sin46°
AQ = APsin46
= (12.5sin36/sin8)sin46
= 12.5sin36sin46/sin8
= ...
only now would I touch my calculator

I used to do this type of question using the cotangent ratios in two right-angled triangles. This seems to be the common way of doing this.
I started doing it the above way when I noticed that too many errors were made in the cotangent method, since it involves factoring and the fact that reciprocal calculations were required on your calculator to obtain the cotangents.

Answer 2

how do you get the angle of triangle ABP?

Answer 3

what about the 54 degrees of the top of the pole?
srry i have too many question

Answer 4

The angle of "elevation" means it was measured from ground level up
So angle QPB = 54
angle QPA = 46
leaving you with 54-46 or 8° or angle APB = 8°

In the right-angled triangle APQ, angle P = 54°
so angle B = 90-54 = 36°

Answer 5

now i get it i.. :) im so happy thank you reiny :) thank you

Answer 6

Angle ABP = 54 - 46 = 8

Angle PBQ = 90- 54 = 36 (note used 54)
so
Angle PAB = 180 - 8 - 36 = 136

Answer 7

thank you damon.. i'll try to help too