Join ExamsbookAnswer : 3. "66.6 mts"
The top and bottom of a tower were seen to be at angles of depression 30° and 60° from the top of a hill of height 100 m. Find the height of the tower ?5
Q: The top and bottom of a tower were seen to be at angles of depression 30° and 60° from the top of a hill of height 100 m. Find the height of the tower ?
- 142.2 mtsfalse
- 233.45 mtsfalse
- 366.6 mtstrue
- 458.78 mtsfalse
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Answer : 3. "66.6 mts"
Explanation :
Answer: C) 66.6 mts Explanation: From above diagramAC represents the hill and DE represents the tower Given that AC = 100 m angleXAD = angleADB = 30° (∵ AX || BD ) angleXAE = angleAEC = 60° (∵ AX || CE) Let DE = h Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1) tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)∵ BD = CE and Substitute the value of CE from equation 1 100/√3 = 100−h(√3) => h = 66.66 mts The height of the tower = 66.66 mts.