The angles of elevation of the top of a lighthouse from 3 boats A, B a

1.	The angles of elevation of the top of a lighthouse from 3 boats A, B and C in a straightlineof same side of the light house are a, 2a, 3arespectively. Ifthe distance between the boatsA and Band the boats B and C arex and y respectively find the height of the light house?
Answer» here, <Q = 90, PQ = h , AB = xSo, IN ∆ BPQ, tan2a = PQ/QB QB = h/tan2a----------( 1 ) now, IN ∆PQC, tan3a = PQ/QC QC = h/tan3a-----------( 2 ) similarly, IN ∆APQtana = PQ/(QA) tana = h/(QC + BC + AB) [ as QA = QC + BC + AB] (QC + BC + x) = h/tana [ AB = x]----------( 3 ) We may write [BC = QB - QC] we get, [ QC + QB - QC + X] = h/tana From-------( 1 ) , & -------( 3 ). [h/tan 2a + x] = h/tana[(h + xtan2a)/tan2a] = h/tana tana[h + xtan2a] = htan2a htana + xtana.tan2a = htan2a h(tan2a - tana) = xtana.tan2a

The angles of elevation of the top of a lighthouse from 3 boats A, B and C in a straightlineof same side of the light house are a, 2a, 3arespectively. Ifthe distance between the boatsA and Band the boats B and C arex and y respectively find the height of the light house?

Answer»

here, <Q = 90, PQ = h , AB = xSo, IN ∆ BPQ,

tan2a = PQ/QB

QB = h/tan2a----------( 1 )

now, IN ∆PQC,

tan3a = PQ/QC

QC = h/tan3a-----------( 2 )

similarly, IN ∆APQtana = PQ/(QA)

tana = h/(QC + BC + AB) [ as QA = QC + BC + AB]

(QC + BC + x) = h/tana [ AB = x]----------( 3 )

We may write [BC = QB - QC] we get,

[ QC + QB - QC + X] = h/tana

From-------( 1 ) , & -------( 3 ).

[h/tan 2a + x] = h/tana[(h + xtan2a)/tan2a] = h/tana

tana[h + xtan2a] = htan2a

htana + xtana.tan2a = htan2a

h(tan2a - tana) = xtana.tan2a