A car is moving at uniform speed towards a tower. It takes 15 minutes for the angle of depression from the top of tower to the car to change from 30* to 60*. What time after this, the car will reach the base of the tower?
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Answered by
1
Step-by-step explanation:
Answer
Let speed of can be vm/min
⇒BC=v×12
In ΔAOC,tan30=
OC
AO
⇒
3
1
=
OC
AO
In ΔAOB,tan45=
OB
AO
⇒1=
OB
AO
⇒OC=
3
AO=
3
OB
⇒OB+BC=
3
OB
⇒OB(
3
−1)⇒BC⇒
BC
OB
=
3
−1
1
Now, t=
v
OB
=
v(
3
−1)
BC
=
3
−1
12
×
3
+1
3
+1
t=6(
3
+1)min
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Step-by-step explanation:
Given A car is moving at uniform speed towards a tower. It takes 15 minutes for the angle of depression from the top of tower to the car to change from 30* to 60*. What time after this, the car will reach the base of the tower?
- So there is a tower AB with triangle ABD with height h.
- So in triangle ABD which is 30 degree
- So tan30 = AB / BD
- 1/√3 = h / BD
- So BD = h√3 ----------1
- Now draw a line AC where c is the midpoint of BD.
- So in triangle ABC,
- tan 60 = AB / BC
- √3 = h / BC
- So BC = h /√3 -----------2
- According to the question it takes 15 min to travel a distance CD
- So CD = BD - BC
- = h√3 – h/√3
- CD = 2h / √3
- Now speed = distance / time
- time t = 15 min
- distance CD = 2h/√3
- Speed v = 2h / √3 / 15
- v = 2h / 15√3-----------------1
- Now distance BC = h/√3, time = t
- v = h / t√3 ------------------2
- From 1 and 2 we get
- h/t√3 = 2h / 15√3
- Or t = 15/2
- t = 7.5 min
Reference link will be
https://brainly.in/question/4203745
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