Question

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.

Answer 1

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

Answer 2

I don't understand

Answer 3

∆PQT=65-25=40°. ∆PST=90+65=115.
PT=? ST=20m
PTsin40°=20sin115°
PT=28.2m

Answer 4

Height of the flagpole
Sin65°=X/28.2
X=28.2sin65°=25.6m

Answer 5

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

Answer 6

Correct

Answer 7

i dont understand break it down that i can understand.

Answer 8

Plese explain better by using normal arithmetic signs

Answer 9

canyou explain with a diagram and explain better

Answer 10

I don't understand at all is like u don't no what u are doing

Answer 11

Ndi ala una dey mad u don't even no wat u re doing

Answer 12

Pls I don't understand ,pls explain with diagram

Answer 13

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

Answer 14

I understand 😊

Answer 15

He knows what he is doing, you just don't understand!