Question

Two men are on the opposite side of a tower.They measure the elevation of the top of the tower as 30° and 60° respectively. If the height of the tower is 40m, find the distance between the ships

Answer 1

The distance between the two men is 92.37 m.

Step-by-step explanation:

Given :

∠ABD = tan 30°

∠ACD = tan 60°

AD = 40 m

To Find ,

Distance between the men = BC = BD+DC

In ΔADC,

tan 60° = AD/DC

$\sqrt{3}$ = $\frac{40}{DC}$ m

DC = $\frac{40}{\sqrt{3} }$

     = $\frac{40*\sqrt{3} }{\sqrt{3}*\sqrt{3} }$

     = $\frac{40\sqrt{3} }{3}$

   DC  = $\frac{40 * 1.732}{3}$ = 23.09 m

In ΔADB,

tan 30° = AD/ BD

$\frac{1}{\sqrt{3} }$ = $\frac{40}{BD}$ m

BD = 40 $\sqrt{3}$

BD = 69.28 m

∴Distance between the two men = BC = BD+DC

                                                = 69.28 + 23.09

                                               = 92.37 m

To Learn More....

1. The angle of depression from the top of a tower of a point A on the ground is 30 degrees.On moving distance of 20m from the point A towards the foot of the tower to a point B ,the angle of elevation of the top of the tower from the point B is 60 degrees.Find the height of the tower and its distance from point A

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2. The angle of elevation of the top of a tower is observed to be 60 degree at a point 30 M vertically above the first point of observation derivation is found to be 45 degree find the height of the tower and its horizontal distance from the point of observation

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