Question

If cos A = 12/13, then find sin A and tan A.​

Answer 1

Step-by-step explanation:

cos A = 12/13 = B/H

B=12; H=13

By Pythagoras theorem

H^2= B^2+P^2

169= 144+P^2

P^2= 169-144

P=√25

P=5

Sin A = P/H

5/13

Tan A = P/B

5/12.

Answer 2

cosA=12/13

b/h=12/13

let,b=12k And h=13k

$so \: by \: pythagorus \: theorem \: we \: have \\ p = \sqrt{ {h}^{2} - {b}^{2} } \\ p = \sqrt{ {13k}^{2} - {12k}^{2} } \\ p = \sqrt{169 {k}^{2} - 144 {k}^{2} } \\ p = \sqrt{25 {k}^{2} } \\ p = 5k$

so now

$sin \: a = \frac{p}{h } \\ sin(a) = \frac{5k}{13k} \\ sin \: a = \frac{5}{13}$

And

$\tan(a) = \frac{p}{b} \\ \tan(a) = \frac{5k}{12k} \\ \tan(a) = \frac{5}{12}$

so hereis your answer

HOPE IT IS HELPFUL ☺