Please enter the following details
A person standing on the ground at point A saw an object at point B on the ground at a distance of 600 meters. The object started flying towards him at an angle of 30° with the ground. The person saw the object for the second time at point C flying at 30° angle with him. At point C, the object changed direction and continued flying upwards. The person saw the object for the third time when the object was directly above him. The object was flying at a constant speed of 10 kmph.
Find the angle at which the object was flying after the person saw it for the second time. You may use additional statement(s) if required.
Statement I: After changing direction the object took 3 more minutes than it had taken before.
Statement II: After changing direction the object travelled an additional 200√3 meters.
Which of the following is the correct option?
If the object does not change direction at point C then it will be at D above the person. But since it is given that object changed direction and continued flying upwards, thus object would reach point E.
$$\triangle$$ ABC is isosceles triangle ($$\because \angle CAG = \angle ABC$$)=> CG is perpendicular bisector to AB => BG = 300 m
In $$\triangle$$ BCG
=> $$cos 30 = \frac{BG}{BC}$$
=> $$\frac{\sqrt{3}}{2} = \frac{300}{BC}$$
=> $$BC = \frac{600}{\sqrt{3}} = 200 \sqrt{3}$$
Now, according to statement I, time can increase only when the angle with ground level will increase.
But, according to statement II, if CD = $$200 \sqrt{3}$$ (= BC), then at constant speed, it will not take any additional time and thus there should not be any increase in angle.
But in first statement direction has been changed whereas in second statement direction has not been changed.
So both the statements are inconsistent with each other.
Create a FREE account and get:
- All Quant Formulas and shortcuts PDF
- 15 XAT previous papers with solutions PDF
- XAT Trial Classes for FREE
