Question 102

As observed from the top of a lighthouse, 45 m high above the sea-level, the angle of depression of a ship, sailing directly towardsit, changes from $$30^\circ$$ to $$45^\circ$$. The distance travelled by the ship during the period of observation is: (Your answer should be correct to one decimalplace.)

Solution

Let AB be the light house of height 45 m. Let CD = $$x$$ m

In right $$\triangle$$ ABC,

=> $$tan(\angle ACB)=\frac{AB}{BC}$$

=> $$tan(45^\circ)=1=\frac{45}{BC}$$

=> $$BC=45$$ m

In right $$\triangle$$ ABD,

=> $$tan(\angle ADB)=\frac{AB}{BD}$$

=> $$tan(30^\circ)=\frac{1}{\sqrt3}=\frac{45}{45+x}$$

=> $$x+45=45\sqrt3$$

=> $$x=45(\sqrt3-1)$$

=> $$x=45\times0.732=32.9$$ m

=> Ans - (A)

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