if we have the opposite and hypotenuse what do we need to use to work out the angle?
+354 Remeber SOH CAH TOA.
You have the oppoisite and hypotenuse whic are to O and H, so we use sine or sin().
Because we don't know the angle, we need to rearrange the equation to get the angle on its own. In this case, the angle will be x and O and H are known numbers.
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{x}}\right)} = {\frac{{\mathtt{O}}}{{\mathtt{H}}}}$$
$${\mathtt{x}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{O}}}{{\mathtt{H}}}}\right)}$$
Note I didn't use sin(O/H), i used sin^-1(O/H), which is the inverse of sine.
+354 Remeber SOH CAH TOA.
You have the oppoisite and hypotenuse whic are to O and H, so we use sine or sin().
Because we don't know the angle, we need to rearrange the equation to get the angle on its own. In this case, the angle will be x and O and H are known numbers.
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{x}}\right)} = {\frac{{\mathtt{O}}}{{\mathtt{H}}}}$$
$${\mathtt{x}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{O}}}{{\mathtt{H}}}}\right)}$$
Note I didn't use sin(O/H), i used sin^-1(O/H), which is the inverse of sine.
