Question

151
15. A tower in a city is 150 m high and a multistoreyed
hotel at the city centre is 20 m high. The angle of
elevation of the top of the tower at the top of the
hotel is 59. A building, h metres high, is situated
on the straight road connecting the tower with
the city centre at a distance of 1.2 km from the
tower. Find the value of h if the top of the hotel,
the top of the building and the top of the tower
are in a straight line. Also, find the distance of
the tower from the city centre.
(Use tan 5° = 0.0875, tan 85º = 11.43).​

Answer 1

Answer:

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Answer 2

Answer:

Let the tower be AB ,hotel be EF and building be CD .

AB= 150m ,EF= 20m ,CD = h , BD = 1200m

Let DF = Xm

PE= BF= BD + DF = (1200+X)m

In triangle APE

tan angle AEP =AP /PE

tan 5 °= AP/PE

AB -PB / BF = tan 5°

AB-EF /BD + DF = tan 5°

150-20/1200+ X = tan 5°

130/ 1200+X = 0.0875

130=( 1200+X )(0.0875)

130 = 150 + 0.0875 x

X = 25/0.0875

X= 286 (approx)

Now in triangle CQE , CQ= CD-DQ= h - EF = h- 20

Also , CQ / QE = tan 5°

h- 20 =286 × 0.0875

h-20=25

h = 45m

Distance of the tower = BD + DF

BD + QE

1200 + 286

= 1486 m ( approx )

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