what is the sin inverse of sqrt*3/2?

Answer 1

60 degree or 1/3 radians.
sin x = sqrt*3/2
sin 60 = sqrt*3/2
60= sin inverse of sqrt*3/2

Answer 2

Let x = inverse sin[sqrt(3) / 2]

Therefore, sin(x) = sqrt(3) / 2

So we want to know the angle x.

I'm hoping you would like to know how this is derived.

Draw an equilateral triangle with each side equal to 2 units
in length. We already know that each angle is 60º.

Now draw an altitude from one vertex to the centre of the
opposite side, which creates two equal triangles. The
altitude length can be calculated from Pythagoras as
being equal to sqrt(3).

Each of these smaller triangles is right-angled
with the other two angles being 60º and 30º.

The sides are 1, sqrt(3) and 2 (this latter is the hypotenuse).

The sine of an angle is (opposite / hypotenuse),
so sin(60º) = sqrt(3) / 2. Thus, x = 60º, and so
inverse sin[sqrt(3) / 2] = 60º = pi / 3 radians.

The other ratios that can be found from this triangle are :

sin(30º) = 1 / 2
cos(60º) = 1 / 2
cos(30º) = sqrt(3) / 2
tan(30º) = 1 / sqrt(3) = sqrt(3) / 3
tan(60º) = sqrt(3) / 1 = sqrt(3)

Answer 3

60 degrees

Answer 4

It does not exist because the maximum value for sin of an angle is one.