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Shortcut 85:
Using tanƟ
Ɵ                                tan

0                                    0

30                            \(1/\sqrt3 \)

45                                  1

60                             \( \sqrt3\)

90                              \(\infty \)
Question:
A man on a boat saw a tower of height 100m at an angle of 60o. What is the distance between the man and the tower?
Answer:
2b1
tanƟ = opp/adj
tan 60 = 100/x
√3 = 100/x
x= 100/√3

Shortcut 86:
Using Pythagoras theorem
\( hyp^2=opp^2+adj^2\)
Question:
A man looks at the top of a tower which is 400m height. The minimum distance between him and top of the tower is 500m. What is the distance between him and the base of the tower?
Answer:
2c1
hyp2 = opp2 + adj2
5002 = 4002 + x2
x2 = 90000
x = 300 m

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