Question

If cos B=3/√13 and A+B=90 find the value of sin A

Answer 1

Answer:

sin A = $\frac{\sqrt{3} }{13}$

Step-by-step explanation:

Consider a triangle ABC right angled at C

So, according to question, cos B = $\frac{\sqrt{3} }{13}$

Now, cos θ = Adj/Hyp

cos B = $\frac{BC}{AB} = \frac{\sqrt{3} }{13}$

Now, sin θ = Opp/Adj

So, sin A = BC/AB = cos B

Thus, sin A = $\frac{\sqrt{3} }{13}$