Question

A tower in a city $b=150m$ high and a multistoreyed hotel at the city center is $20m$ high. The angle of elevation of the top of tower from the top of the hotel is $5^0$. A building, $h$ m high is situated on the straight road connecting the tower with the city center at a distance of $1.2km$ from the tower. If the top of the hotel, the top of the building and the top of the tower are in a straight line, then find the height of building $h$ and distance of tower from the city center. $(\tan 5^0=0.0875,\tan {85}^0=11.43)$

Answer 1

Let $AB$ is a tower of height $AB=150m$. $CD$ is a building of height $h$ m. and $EF$ is a hotel of height $EF=20m$. Distance of building from tower $BD=1.2km=1200m$.

Let $DF=x$ m

Given, $\angle AEP=5^0$

From figure $PE=BF=(1200+x)m$

From right angled $\Delta APE$

$\tan 5^0=\frac{AP}{PE}$

$\tan 5^0=\frac{AB-PB}{PE}$

$0.0875=\frac{AB-20}{1200+x}$

$=\frac{130}{1200+x}$ $[\because AB=150m,PB=EF=20m]$

$(1200+x)\left(0.0875\right)=130$

$1200\times 0.0875+0.0875x=130$

$105+0.0875x=130$

$0.0875x=130$

$0.0875x=130–105=25$

$x=\frac{25}{0.0875}$

$=286m$ (approx)

$x=DF=QE$

Again from right angled $\Delta CQE$

$\tan 5^0=\frac{CQ}{QE}$

$CQ=QE\tan 5^0$

$=286\times 0.0875$

$=25.025m$

Hence, height of building $h=CD$

$=QD+QC$

$=20+25.025$

$=45m$ (approx.)

Distance of tower $BF=BD+DE$

$=BD+QE[\because DF=QE]$

$=1200+286$

$=1486m$

Hence, distance of tower from the city center $=1486m$.