2d motion range , find angle (Really simple, not sure what I'm missing)

[Chas3down]

Homework Statement

Firing something at 40m/s at angle X, it lands 65meters away, give 2 angles it could be.. (Assume g=10m/s^2)(Fired at y=0)(No air resistance)

Homework Equations

D = v^2/g sin(2x)
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The Attempt at a Solution

d= v^2/g sin(2x)

65 = 40^2/10 sin(2x)

650/1600 = sin(2x)

13/32 = sin(2x)

arcsin(13/32)/2 = x
23.97/2 = x

x = 11.985
and
x = 90-11.985

But it's wrong?

Heres another form it could be in, but don't know where to go from here..
13/32 = 2sinxcosx
13/64 = sinxcosx

 

[Orodruin]

Your approach looks reasonable. What is the given answer?

 

[Chas3down]

Your approach looks reasonable. What is the given answer?

There isn't one, it just says if it is correct or not (anwser must be exact)

With further research, it seems one angle is
"arcsin(13/32)/2" , which is what I had but in decimal form, however can't find the second angle.. , i tried 90 - arcsin(13/32)/2, 180 - arcsin(13/32)/2, 360 - arcsin(13/32)/2, - arcsin(13/32)/2, and arcsin(-13/32)/2, and arccos(13/32)/2

"90 - arcsin(13/32)/2" and "arccos(13/32)/2" work, but I am just not sure what form they want it in, probably a manipulation of 13/32 so the form would be arcsin(??)/2 for second angle

 

[Orodruin]

But it's wrong?

This question seemed to indicate that you had compared it with a solutions manual or equivalent which had stated something else.

But I can't seem to get the second angle, i tried 180 - that and 90 - that

You can argue for this either completely through mathematics - or by physical reasoning: If the angle would be 180 - 12 degrees, would the projectile land in front of the firing point?

The mathematical approach is to note that sin(2x) = a implies that 2x = asin(a) or 2x = 180^o - asin(a).