the angle of the tower to the top of the tower is 60° but the height of the tower in 100 m from the tower word____
Answers
Answered by
0
Answer:
Let AB be the tower and CD be the hill. Then, ∠ACB=30
o
,∠CAD=60
o
and AB=50 m.
Let CD=x m
In right △BAC, we have,
cot30
o
=
AB
AC
3
=
50
AC
AC=50
3
m
In right △ACD, we have,
tan60
o
=
AC
CD
3
=
50
3
x
x=50×3=150 m
Therefore, the height of the hill is 150 m.
Answered by
0
Answer:
Given height of tower CD=50m
Let the height of the building, AB=h
In right angled △BDC,
⇒ tan60
o
=
BD
CD
⇒
3 =
BD
50
⇒ BD=
3
50metres
In right angled △ABD,
⇒ tan30
=
BD
AB
⇒
31
=
350
h
∴
31
=
503
∴ h=
3
50
=16.66m
∴ The height of the building is 16.66m.
s
Step-by-step explanation:
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