Question

Someone please help me with Trig it's the simple kind

Answer 1

`\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies \measuredangle \theta=tan^(-1)\left( \cfrac{opposite}{adjacent} \right) \\\\\\ adjacent=\cfrac{opposite}{tan(\theta)}\\\\ -----------------------------\\\\ thus \\\\\\ 30)\qquad tan(63^o)=\cfrac{3√(2)}{x}\implies x=\cfrac{3√(2)}{tan(63^o)} \\\\\\ 32)\qquad tan(x)=\cfrac{43}{30}\implies \measuredangle x=tan^(-1)\left( \cfrac{43}{30} \right) \\\\\\ 34)\qquad tan(x)=\cfrac{5.94}{5.66}\implies \measuredangle x=tan^(-1)\left( \cfrac{5.94}{5.66} \right)`

just bear in mind that, if you were to put an eye on the angle given, the side you'll see on the other end, is the "opposite side"

the adjacent side, is the one, well, adjacent(next to, touching) the angle

and the hypotenuse, in a right-triangle, is always the slanted side