Asked by Moses

A flagpole and a building stands on the same horizontal level.From the point P at the bottom of the building.the angle of elevation of the top T of the flagpole is 65 degrees from the top Q of the building the angle of elevation of the point T is 25 degrees.If the building is 20m high,Calculate
(a)Distance PT
(b)Height of the flagpole
(c)Distance QT.

Answers

Answered by Steve
Draw a horizontal line from Q to the flagpole. Label the intersection S.
The height of the pole is thus 20+ST

since
SQ/ST = cot25
SQ/(ST+20) = cot65

equate and solve for ST
Now you can get SQ
PT^2 = SQ^2 + (ST+20)^2
QT^2 = SQ^2 + ST^2

Answered by mathew
I don't understand
Answered by mathew guru
∆PQT=65-25=40°. ∆PST=90+65=115.
PT=? ST=20m
PTsin40°=20sin115°
PT=28.2m
Answered by mathew guru
Height of the flagpole
Sin65°=X/28.2
X=28.2sin65°=25.6m
Answered by mathew guru
Distance QT
Cos65°=25.6/X
X=25.6/cos65°=11.9m
To find distance QT
Cos25°=11.9/X
X=11.9/cos25°=13.1m
Answered by sultan
Correct
Answered by glory
i dont understand break it down that i can understand.
Answered by Margaret
Plese explain better by using normal arithmetic signs
Answered by lopez
canyou explain with a diagram and explain better
Answered by Anonymous
I don't understand at all is like u don't no what u are doing
Answered by Finest
Ndi ala una dey mad u don't even no wat u re doing
Answered by Apotieri
Pls I don't understand ,pls explain with diagram

Answered by Emmanuella
for distance QT its not cos 65, you wouldn't get 11.9 with that, it is Tan 65, then u would get 11.9
Answered by Seun
I understand 😊
Answered by David
He knows what he is doing, you just don't understand!
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