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# In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A. Q.1(1)

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Today i am solving ncert class 10 introduction to trigonometry  questions of exercise 8.1 question no.1(1)  but i can’t solve the question properly  ,please help me to solve the question easily . In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A.

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1. In a given triangle ABC, right angled at B = ∠B = 90°

Given: AB = 24 cm and BC = 7 cm

According to the Pythagoras Theorem,

In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

By applying Pythagoras theorem, we get

AC2=AB2+BC2

AC2 = (24)2+72

AC2 = (576+49)

AC2 = 625cm2

AC = √625 = 25

Therefore, AC = 25 cm

(i) To find Sin (A), Cos (A)

We know that sine (or) Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side. So it becomes

Sin (A) = Opposite side /Hypotenuse = BC/AC = 7/25

Cosine or Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side and it becomes,

Cos (A) = Adjacent side/Hypotenuse = AB/AC = 24/25

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