The lower window of a house is at a height of 2m above the ground and its upper window is 4m vertically above the tower window. At certain instant the angles of elevation of a balloon from these windows are observed to be `60^(@)` and `30^(@)`, respectively. Find the height of the balloon above the ground.

The lower window of a house is at a height of 2m above the ground and

1.	The lower window of a house is at a height of 2m above the ground and its upper window is 4m vertically above the tower window. At certain instant the angles of elevation of a balloon from these windows are observed to be `60^(@)` and `30^(@)`, respectively. Find the height of the balloon above the ground.
Answer» Let the height of the balloon from above the grond is H. A and OP = `w_(2)R=w_(1)Q=x` Given that, height of lower window from the above the ground `=w_(2)P=2m=OR` Height of upper window from above in the lower window `=w_(1)w_(2)=4m=QR` `therefore BQ=OB=(QR+RO)` `=H-(4+2)` `=H-6` and `angleBw_(1)Q=30^(@)` `rArr angleBw_(2)R=60^(@)` Now, in `DeltaBw_(2)R` `tan60^(@)= (BR)/(w_(2)R) = (BQ+QR)/(x)` `rArr sqrt(3) = ((H-6)+4)/(x)` `rArr x=(H-2)/(sqrt(3))`................(i) and in `DeltaBw_(1)Q`, `tan30^(@)= (BQ)/(w_(1)Q)` `tan30^(@)= (H-6)/x = 1/sqrt(3)` `rArr x=sqrt(3)(H-6)`...........(ii) From Eqs. (i) and (ii), ltbgt `sqrt(3)(H-6) = ((H-2))/(sqrt(3))` `3(H-6) = H-2 = 3H-18 = H-2` `rArr 2H=16 rArr H=8` So, the required height is 8m. Hence, the required height of the ballon from above ground is 8 m.