Question

In a triangle right angled at B if sin A=3/4.find cos A​

Answer 1

⭐HELLO MATE⭐

✌✌ANSWER✌✌

$\sin = \frac{p}{h} = \frac{3}{4} \\ so \: p = 3 \: and \: h = 4 \\ so \: \\ {b }^{2} = {h}^{2} - {p}^{2} \\ {b}^{2} = 16 - 9 \\ {b}^{2} = 7 \\ b = \sqrt{7}$

Now for cos.

$\cos = \frac{b}{h} \\ \cos = \frac{ \sqrt{7} }{4}$

#hope it help uh..

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Answer 2

Step-by-step explanation:

Sin A = 3/4

Sin A = p/h

P/h = 3/4

So, p=3 and h=4.

Using , Pythagoras theorm

$4 {}^{2} = 3 {}^{2} + x { }^{2} \\ \\ 16 + 9 = x {}^{2} \\ \\ x = \sqrt{7}$

So, b =

$\sqrt{7}$

so, cos A =

$\frac{ \sqrt{7} }{4}$

Hope it helps