1) A tower in a city is 150m high and a multi storeyed hotel at the ci

1.	1) A tower in a city is 150m high and a multi storeyed hotel at the city centre is 20m high. The angle ofelevation of the top of the tower at the top of hotel is 30". A building h m high, s situated on the straightroad connecting the tower withs the city cernre at a distance of 200m from the tower. Find the value of h, ifthe top of the hotel and top of the building and top of the tower are in straight line. Also, find the distanceof the tower from the city centre.
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 )

1) A tower in a city is 150m high and a multi storeyed hotel at the city centre is 20m high. The angle ofelevation of the top of the tower at the top of hotel is 30". A building h m high, s situated on the straightroad connecting the tower withs the city cernre at a distance of 200m from the tower. Find the value of h, ifthe top of the hotel and top of the building and top of the tower are in straight line. Also, find the distanceof the tower from the city centre.

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 )