Answer:
cos A= 9/41 and tan A is 40/9
Step-by-step explanation:
SOHCAHTOA.
Sine = opposite/hypotenuse
Cosine=adjacent/hypotenuse
Tangent=opposite/adjacent
Let's go back to your problem
41 sin A = 40
To get sin A by itself, divide both sides of the equation by 41. You get sin A = 40/41. As sine = opposite/hypotenuse, 40 will be the leg and 41 will be the hypotenuse. What is the value of the other leg?
Use the pythagorean theorem.
a^2 + b^2 = c^2
a^2 + 40^2 = 41^2
a^2 + 1600 = 1681
a^2 = 81
a=9
This is a 9-40-41 right triangle
Now, solve cos A
cos=adjacent/hypotenuse
cos=9/41
Finally, solve for tan A
tan=opposite/adjacent
tan=40/9
Hope this helps :)
