A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. After covering a distance of 50 m, the angle of depression of the car becomes 60°. Find the height of the tower. (Use 13 = 1.73).
Answers
Answer:
Step-by-step explanation:
A straight highway leads to the foot of a tower.
Lat the man and tower heught be AD
And let B be the point where car was seen 1st and C be the point where car was seen after covering the distance of 50m, and let the point on the imaginery line be the point P
It is given that the man observing the car 1st at an angle of depression of 30°
So,∠PAB=30°
And after covering 50m of distance by the car the man observes the car at an angle of depressin of 60°
So,∠PAC=60°
Now, we know that the tower is vertical
So,∠ADB=90°
and BC is 50m
Also, the line PA and BD are parallel
And AB is the transversal
∠ABD=∠PAB. (alternate angles)
So, ∠ABD=30°
Simillarly,
Line PA and BD are parallel
And AC is the transversal
∠ACD=∠PAC. (alternate angles)
So,∠ACD=60°
Now,in right angle triangle ACD
tanC=side opposite to angle C/side adjacent to angle C
tan60°=AD/CD
√3=AD/CD
√3CD=AD ...(1)
In right angle triangle ABD
tanB=side opposite to angle B/side adjacent to angle B
tan30°=AD/BD
1/√3=AD/BD
BD/√3=AD
AD=BD/√3 ...(2)
Now by putting the value from
FROM(1)and(2) (both L.H.S.are equal)
√3CD=BD/√3
√3(√3CD)=BD
3CD=BD
3CD=BC+CD
3CD-CD=BC
2CD=BC
CD=BC/2
As we know the value of BC is 50m
CD=50/2
CD=25m
•After covering 50m
tan60°=AD/(50-CD)
√3=AD/(50-CD)
As we know the value of CD is 25m.
√3=AD/(50-25)
√3=AD/25
AD=25√3
So,the height of the tower is 25√3m.