Asked by Precious
a woman looking out from a window of a building at a height of 30meters observed that the angle of depression of a flag pole was 44°. if the foot of the pole is 25meters from the foot of the building from the same horizontal, find (a) correct to the nearest whole number the angle of depression of the foot of the pole from the woman (b) the height of the flag pole.
Answers
Answered by
Reiny
The key is a good diagram.
On mine I labeled the window position as A and the base of the building as B, so AB = 30
I labeled the top of the flag pole as P, and its base as Q
So BQ = 25
We can find PQ by
PQ^2 = 30^2 + 25^2
PQ = ....
No look at triangle AQP,
angle APQ = 90+44 = 134°
and AQ we found above.
Back to triangle ABQ,
tan(angle AQB = 30/25
so we can find angle AQB = .....
and angle PQA = 90° - angle BQA = ....
allowing us to find angle QAP
now we can use the sine law in triangle AQP
PQ/ sin (QAP) = AQ/sin 134°
take over with the calculations
On mine I labeled the window position as A and the base of the building as B, so AB = 30
I labeled the top of the flag pole as P, and its base as Q
So BQ = 25
We can find PQ by
PQ^2 = 30^2 + 25^2
PQ = ....
No look at triangle AQP,
angle APQ = 90+44 = 134°
and AQ we found above.
Back to triangle ABQ,
tan(angle AQB = 30/25
so we can find angle AQB = .....
and angle PQA = 90° - angle BQA = ....
allowing us to find angle QAP
now we can use the sine law in triangle AQP
PQ/ sin (QAP) = AQ/sin 134°
take over with the calculations
Answered by
Steve
Draw a diagram. It should be clear that
(a) tanθ = 30/25
(b) (30-h)/25 = tan44°
(a) tanθ = 30/25
(b) (30-h)/25 = tan44°
Answered by
Steve
seems to me that
AQ^2 = 30^2 + 25^2
AQ^2 = 30^2 + 25^2
Answered by
Reiny
Steve, thanks for the catch.
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