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

Asked by Abhisek | 1 year ago |  87

##### 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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