I need help solving this problem
(I had a diagram with this picture but i can not post the pic of the diagram so i described it below)
The Hypotenuse is 9
The Adjacant is 7
The Opposite Side (I do not have the lenght)
x is the angle between the hypotenuse and opposite right above the right angle
Solve for x
3 Answers

You could just use the sine rule for this:
(a / Sin A) = (b / Sin B) = (c / Sin C)
Let little a = the hypotenuse. Then capital A is the angle between the adjacent and opposite (90 degrees). So:
a / Sin A = 9 / Sin 90
= 9 / 1
= 9
Therefore b / Sin B = 9. Let little b = the adjacent. Then capital B is the angle between the opposite and hypotenuse (i.e. the angle we want to find). So:
b / Sin B = 9
Sin B = b / 9
= 7/9
B = arcsin (7/9)
= 51.06 degrees
= 51 degrees (to nearest degree)

You’re contradicting yourself ! the “Opposite ” is the side opposite the angle , you appear to want to use Adjacent to mean Opposite. To clarify what I mean, sketch a rightanbled triangle, with the right angle on the right end of the base. If x is the angle “above” the right angle, then the Opposite is the base, and the Adjacent is the upright side – the word Adjacent means “beside – it is the sidethat along with the hypotenuse forms the angle. From your description, what you have called the Adjacent is actually the Opposite ! The Opposite is actually (in this case) the base of the triangle, so in that case, use sin x = 7/9 =07777. …. so x = sin^1(0.77777…) or arcsin(0.7777….) – whichever you normally use, giving x = 51.05755… degrees – round off as you would normally do.
Source(s): Retired Maths Teacher 
sin(x) = 7/9 = 0.77778
x = 51.05 degrees
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