Question

9sinA+40cosA=41 then prove that 41cosA=40
Please Answer Quick..​

Answer 1

Consider a triangle with leg lengths 40,9 and 41.

Then 40^2+9^2 = 41^2 is satisfied Pythagoras theorem.

So the triangle have a 90 deg angle.

So we can say either cosA = 40/41 or cosA = 9/40

But we know;

9sinA+40CosA= 41-----(1)

if cosA = 9/41 then sinA=40/41

9sinA+40CosA = (40*9+40*9)/41 = 17.56 ;not satisfied (1)

if cosA = 40/41 then sinA = 9/41

9sinA+40CosA = (9*9/41+40*40/41)/41 = 41 ;satisfied (1)

Therefore cosA = 41/40

cosA = 40/41

41cosA = 40

Answer 2

Answer:

upper answer is right

Step-by-step explanation:

it's answer is 41cos=40