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Saturday, July 4, 2015

UPSC Maths Problems Challenge : A Problem A Day! (Day 21)

Today's problem is one of the most basic problems (as I feel it now that we have to do several tough Trigonometry problems everyday) from Trigonometry that was in our Class X syllabus as I remember it.Anyhow,if you forget strategy of any kind of problem,you are supposed to run out of time and exams conducted by UPSC are said to be one of the toughest competitive exams in the world! And you can do anything else but not loose time.

Today's Problem(02/07/2015)

Question Credit :

(1999) A man is standing on a 6 meter pole whose shadow length is 8 meters. If the length of his shadow is 2.4m, what is the height of this man?

The man will be considered to b standing perpendicular to the ground,so we have a 90° in hand.Similarly,the pole is standing perpendicular to the ground.Now,if we imagine the shadows of both and the man and the pole,then we can get two right angled triangles.
Say,we have triangle ABC and triangle BDE,then ∠ABC = ∠BDE [ 90°;As they stand perpendicular to the ground.
Also,∠ACB = ∠BED [The elevation of the Sun is same at the same time]
So, ΔABC ΔDEF [By AA similarity criteria]
Therefore, AB/BC = BD/DE,i.e AB/2.4 = 6/8 or AB/2.4 = 3/4 or AB/0.6 = 3 or AB = 1.8 m.
So,the height of the man is 1.8 m.

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