Question

In right triangle ABC, angle B is a right angle and sin C = x. sin A =

Answer 1

Comment
The answer to this depends on what you have been told. I don't want to go into a whole lot of talking when you don't need it. 

It also depends where C is (in what quadrant). If it is just an ordinary triangle, not drawn in a coordinate system, then the answer will be plus.

Step One
To made things simple, draw ABC with BC on the horizontal.
Label the left acute angle as C and the other end of the hypotenuse as A.
 
Step Two
Label the hypotenuse as having a length of 1 which means that x is less than 1.

Step 3
Label len BC as x. You should have AC = 1 and AB = x

Step 4
Find CB
Use the pythagorean theorem to find CB.
a^2 + b^2 = c^2
c = 1
b = AB = x
a = ??

a^2 + x^2 = 1^1 Subtract x^2 from both sides.
a^2  = 1 - x^2 Take the square root of both sides.
sqrt(a^2) = sqrt(1 - x^2)
a = sqrt(1 - x^2)

Answer
sin(A) = sqrt(1 - x^2)/1

Comment.
Any length can be given to the hypotenuse. Suppose it is y = z*x . Then AB can be reduced to  x by dividing all sides by z.

Using a much simpler example suppose you are given a right triangle of 6 8 and 10. To reduce the hypotenuse to 1 you would divide by 10. to give you 0.6 0.8 and 1. That would not change the value of sin(C) and in that case you would call x = 0.8 or x = 0.6 depending on how you drew the triangle.

I'm sorry, this got very long because I had to get it correct. 

The key step is getting sqrt(1 - x^2)

If this is giving you a royal headache, please PM me and I'll try and say another way. 

Answer 2

Answer:

sin A = $\sqrt{1-x^{2} }$

Step-by-step explanation:

∵ABC is a right angled triangle at B

∴sin C = cos A

∴sin A = cos C

∵$sin^{2}C+cos^{2}C=1$

∴$cos^{2}C=1-sin^{2}C$

∴$cosC=\sqrt{1-sin^{2}C }$

∵sin C = x

∴$sinA=\sqrt{1-x^{2} }$