Question 75

# A person is standing at a distance of 1800 meters facing a giant clock at the top of a tower. At 5.00 p.m., he can see the tip of the minute hand of the clock at 30 degree elevation from his eye - level. Immediately, the person starts walking towards the tower. At 5.10 pm., the person noticed that the tip of the minute hand made an angle of 60 degrees with respect to his eye - level. Using three - dimensional vision, find the speed at which the person is walking. The length of the minutes hand is 200√3 meters (√3 = 1.732).

Solution

Let O be the centre of the clock. Let the person's eye be at A and the tip of minute hand at 5.00 p.m. is at P and at 5.10 p.m. at Q

AM = 1800 m and OP = OQ = $$200\sqrt{3}$$ m

In $$\triangle$$ APM

=> $$tan 30 = \frac{PM}{AM}$$

=> $$\frac{1}{\sqrt{3}} = \frac{PM}{1800}$$

=> $$PM = \frac{1800}{\sqrt{3}} = 600 \sqrt{3}$$

=> $$OM = PM - OP = 600 \sqrt{3} - 200 \sqrt{3} = 400 \sqrt{3}$$

In $$\triangle$$ OBM

=> $$tan 60 = \frac{OM}{BM}$$

=> $$\sqrt{3} = \frac{400 \sqrt{3}}{BM}$$

=> $$BM = 400$$ m

=> $$AB = AM - BM = 1800 - 400 = 1400$$ m

Time taken to reach B from A = 10 minutes = 600 sec

$$\therefore$$ Speed of the person = $$\frac{1400}{600} = \frac{7}{3}$$ m/s

= $$(\frac{7}{3} \times \frac{18}{5})$$ km/hr = $$8.4$$ km/hr

• All Quant Formulas and shortcuts PDF
• 40+ previous papers with solutions PDF

##### Ashish Kumar

2 years, 9 months ago

We should consider BQM not BOM as he is looking at tip not middle part. Kindly revert If I am wrong.

OR