NIMCET 2015 Question Paper with solutions

Question : 29 of 120

Marks: +1, -0

A harbor lies in a direction 60° south of west from a fort and at a distance 30 km from it, a ship sets out from the harbor at noon and sails due east at 10 km an hour. The time at which the ship will be 70 km from the fort is;

7 PM
8 PM
5 PM
10 PM

Validate

Solution:

In ΔOAC,

\sin30°={AC}/{OA}

⇒1/2={AC}/30

⇒AC=15

km
Also,

\cos30°={OC}/{OA}

√3/2={OC}/30

⇒OC=15√3

In ΔOBC,

\cosθ={OC}/{OB}={15√3}/70

Also, In ΔOBC,

\sinθ={BC}/{OB}

⇒\sinθ={BC}/70

⇒1-\cos^2θ={BC^2}/4900

⇒1-(15√3)^2/4900={BC^2}/4900

⇒BC^2=4225

⇒BC=65

∴ Total distance travelled

=AC+BC=15+65=80

The time at which ship will cover 70 km from the fort

=80/10=8h

[∵ Speed of ship = 10 km/h]
∴ The ship will reach at 8 pm.

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