If sin A = \( \dfrac{3}{4}\), Calculate cos A and tan A.

Asked by Abhisek | 1 year ago |  87

1 Answer

Solution :-

Let us assume a right angled triangle ABC, right angled at B

Given: Sin A = \( \dfrac{3}{4}\)

We know that, Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side.

Therefore, Sin A 

= \( \dfrac{Opposite \;side}{Hypotenuse}\)\( \dfrac{3}{4}\)

Let BC be 3k and AC will be 4k

where k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to 

the sum of the squares of the other two sides of a right angle triangle and we get,

AC2=AB+ BC2

Substitute the value of AC and BC

(4k)2=AB2 + (3k)2

16k2−9k=AB2

AB2=7k2

Therefore, AB = \( \sqrt{7k}\)

Now, we have to find the value of cos A and tan A

We know that,

Cos (A) = \( \dfrac{ Adjacent \;side}{Hypotenuse}\)

Substitute the value of AB and AC and cancel the constant k in both numerator and denominator, we get

 \( \dfrac{AB}{AC}=\) \( \dfrac{\sqrt{7k}}{4}= \dfrac{\sqrt{7}}{4}\)

Therefore, cos (A) = \( \dfrac{\sqrt{7}}{4}\)

tan(A) = \( \dfrac{Opposite \;side}{Adjacent \;side}\)

Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

 \(\dfrac{BC}{AB}=\dfrac{3k}{\sqrt{7k}}=\dfrac{3}{\sqrt{7}} \)

Therefore, tan A =\( \dfrac{3}{\sqrt{7}}\)

Answered by Pragya Singh | 1 year ago

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