Quantitative Aptitude
HEIGHT AND DISTANCE MCQs
Let AB be the lighthouse and C and D be the positions of the ships.
Then,\(AB=100m, \angle ACB=30^{0} and \angle ADB=45^{0}
\)
\(\frac{AB}{AC}=\tan30^{0}=\frac{1}{3}\Rightarrow AC=AB\times3=1003m.\)
\(\frac{AB}{AD}=\tan45^{0}=1 \Rightarrow AD=AB=100m.\)
Therefore CD=\(\left(AC+AD\right)=\left(1003+100\right)m\)
= 100(3 + 1)
= (100 x 2.73) m
= 273 m
A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the mans eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?
Let angle of elevation of right angle triangle ABC of angle C = θ
and height AB = shadow BC = x
tan θ = \(\frac{AB}{BC}=\frac{x}{x}=1=45^{0}\)
Let in right angle triangle ABC of angle C = θ
height AB = x and shadow BC = √3 x
tan θ = \(\frac{AB}{BC}=\frac{x}{\sqrt{3}x}=\frac{1}{\sqrt{3}}=30^{0}\)
lat pole AB = 14 m
pole CD = 20 m
CD – AB = DE
20 m -14 m = 6 m
sin \(30^{0}=\frac{DE}{BD}=\frac{6}{BD} \)
\(\frac{1}{2}=\frac{6}{BD}\)
\(BD=6\times2= 12\)m