1.

From a point on the ground40 m away from the foot of a tower, the angle of elevation of the top of thetower is 30^(@). The angle of elevation of

Answer»


Solution :BC be the tower and CD be thewater tank.
A be the towerpoint of observation. Then, `angleBAC = 30^(@), ANGLEBAD = 45^(@)`and `AB = 40 m`.
From right `DeltaABD `, we have
`(BD)/(AB) = tan 45^(@) = 1 RARR (BD)/(40 m) = 1 rArr BD = 40 m`.
From right `DeltaABC`,we have
`(BC)/(AB) = tan 30^(@) = 1/(sqrt(3)) rArr (BC)/(40 m)=1/(sqrt(3))`.
`rArr BC = (40 m)/(sqrt(3)) rArr BC = (40 m)/(sqrt(3)) xx (sqrt(3))/(sqrt(3))= (40sqrt(3))/(3) m`.
(i) HEIGHT of the tower `= BC = (40sqrt(3))/(3) m = 23.1 m`
(ii) Deapth of tank`= CD = (BD - BC) = (40-23.1) m = 16.9 m`.


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