Can somebody help me with this please ?!
What is the value of sin x°?
i think it's 13 ÷ 5 ??
\[\huge\ Sinx=P/H\]
sinx=5/13,x=\[\sin^{-1} (5/13)=22.6\]
Remember SOH CAH TOA to remind you of the sin, cos and tan for teh acute angles of a right triangle. SOH stands for (S)ine equals (O)pp over (H)ypotenuse CAH stands for (C)osine equals (A)djacent over (H)ypotenuse TOA stands for (T)angent equals (O)pposite over (A)djacent When you look at an acute angle of a right triangle, one leg is next to that angle and forms the angle with the hypotenuse. That's the (A)djacent side. The other leg of the triangle is opposite the angle and is the (O)pposite side, and the (H)ypotenuse is always opposite the right angle. In your case, sin x = opp/hyp = 5/13
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