Question

An observer from the top of a lighthouse 100m high above sea level the angle of depression of ship sailing directly towards it changes from 30 to 60 find the distance travelled by the ship during the period of observation

Answer 1

tantheta = AD/AB
tan 60°=100/AB
√3=100/AB
√3AB=100
AB=100/√3----------(1)

tan theta = AD/AC
tan30=100/(AB+BC)
1/√3=100/(AB+BC)
(AB+BC)/√3=100-----------(2)
(1)=(2) so,
(100/√3+BC)/√3=100
(100/3)+(BC/√3)=100
BC/√3=100-100/3
=200/3
BC= 200√3/3
BC= 115.470054m

by SK mathematics

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

$\tt{\rightarrow tan\theta=\dfrac{Perpendicular}{Base}}$

$\tt{\rightarrow tan60=\dfrac{100}{QR}}$

$\tt{\rightarrow\sqrt{3}=\dfrac{100}{QR}}$

$\tt{\rightarrow QR=\dfrac{100}{\sqrt{3}}}$

QR = 57.735 metre

Now,

$\tt{\rightarrow tan30=\dfrac{100}{QS}}$

$\tt{\rightarrow\dfrac{1}{\sqrt{3}}=\dfrac{100}{QS}}$

$\tt{\rightarrow QS=\dfrac{100}{1/(\sqrt{3}}}$

QS =173.205 metre

$\text{Therefore}$

RS = QS - QR

RS = 173.205 - 57.735

RS = 115.47 metre

$\fbox{Total\;distance\;travelled\;by\;the\;ship}$

= 115.47 metre