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.
canyou explain with a diagram and explain better
∆PQT=65-25=40°. ∆PST=90+65=115.
PT=? ST=20m
PTsin40°=20sin115°
PT=28.2m
i dont understand break it down that i can understand.
Height of the flagpole
Sin65°=X/28.2
X=28.2sin65°=25.6m
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
I don't understand at all is like u don't no what u are doing
Pls I don't understand ,pls explain with diagram
Correct
Plese explain better by using normal arithmetic signs
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
I understand 😊
He knows what he is doing, you just don't understand!
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