Saved Bookmarks
| 1. |
Op05 |
|
Answer» Let AB be the buildingAnd CD be the lamp post. While DE is the horizontal line parallel to the ground from the top of the lamp post to the building.So in Triangle ABC,AB=60mθ=60°tanθ= perpendicular /basetan 60°= AB / BC√3=60 / BCBC = 60 /√3On rationalising denominator, we getBC=60 * √3 /3 =[20 * √3]m Distance between building and lamp post =20 √3 cm =20*1.732(√3=1.732) =34.64m(Ans) EBCD is a rectangle , hence BC =EDIn Triangle AEDθ=30°tan 30° = AE / ED1 /√3 = AE / 20√3AE * √3 = 20√3AE = 20mAB = AE + EBEB =60 - 20=40mSince EB = CD (EBCD being a rectangle)CD or Height of lamp post = 40m (Ans) |
|