Q1.A man finds the angle of elevation of a tower to be 60°. When he walks away at a distance of 50m he finds the angle to be 30°, then the height of the tower is:
Answers
Explanation:
✬ Height = 25√3 ✬
Step-by-step explanation:
Given:
Angle of elevation of tower is 60°.
After walking 50 m away from tower it changes to 30°.
To Find:
What is height of tower ?
Solution: Let the height of tower AB be h m and distance between the man initial position and foot of tower be x m. Therefore,
AB (perpendicular) = h
DB (base) = x
∠ABD = 90°
∠ADB (angle of elevation) = 60°
[ Now he walked 50 m away from point D. Let he is at now point C. ]
DC = 50 m
BC = DB + DC
∠ACB (angle of elevation) = 30°
In ∆ABD , using tanθ
★ tanθ = Perpendicular/Base ★
➯ tan60° = AB/DB
➯ √3 = h/x
➯ √3x = hㅤㅤㅤㅤㅤ(eqⁿ i)
Now in ∆ABC , again by tanθ
➯ tan30° = AB/BC
➯ 1/√3 = h/x + 50
➯ x + 50 = √3h
➯ x + 50 = √3 × √3x
➯ 50 = 3x – x
➯ 50/2 = x
➯ 25 m = x
Putting the value of x in eqⁿ (i)
⟹ √3x = h
⟹ √3 × 25 = h
⟹ 25√3 = h
Hence, the height of tower is 25√3 m.