Question

Expert sir try to solve the following question.
A light house is 300 meter high from the sea level.from the top of the light house the angle of depression of two boats are 30 and 45 degrees .If the line joining the boats passing through the foot of the light house find the distance between the boats when they are on the same side of the light house. (my answer is 300-(100/sqrt(3)) but book answer is "300-100srt(3)" Best wishes
MARIA JAN

Answer 1

A light house is 300 meter high from the sea level. From the top of the light house the angle of depression of two boats are 30 and 45 degrees. If the line joining the boats passing through the foot of the light house find the distance between the boats when they are on the same side of the light house.

Solution:

According to condition, AC=300m, angle ACD=30°, angle ACD=45°. Our boats are D and B, so we have to find DB.

Consider the triangle CAB, angle CAB=90°. So, angle CBA= 90-45=45(°). So, AB=AC=300m.

Consider the triangle CAD, angle CAD=90°. $\tan(< ACD) = \frac{AD}{CA} = > AD = CA * \tan(< ACD) = 300 * \frac{1}{\sqrt{3}} = 100\sqrt{3}$

So, we can see, that DB=AB-AD=300-100$\sqrt{3}$ (m)

Answer: 300-100$\sqrt{3}$ (m)