Question

Can someone please solve number 31 and 32? and can you show me how to do it?

Answer 1

$tan(x)=2.5\implies \measuredangle x=tan^{-1}(2.5)\implies \measuredangle x\approx \begin{cases} \stackrel{I~Quadrant}{68.2^o}\\\\ \stackrel{III~Quadrant}{248.2^o} \end{cases}$

Make sure your calculator is in Degree mode, Check the picture below on the left side.

well, the tangent recall is sine/cosine, or namely y/x, for that fraction if either of them is negative, the whole fraction is negative, if both are either positive or negative, the fraction is positive.

given such, on the 1st quadrant and 3rd quadrant that fraction y/x is positive because on the 1st quadrant both are positive and on the 3rd quadrant both are negative, yielding the fraction, namely tangent, positive.

$cos(x)=-0.7\implies \measuredangle x=cos^{-1}(-0.7)\implies \measuredangle x\approx \begin{cases} \stackrel{II~Quadrant}{134.43^o}\\\\ \stackrel{III~Quadrant}{225.57^o} \end{cases}$

now, check the picture below on the right side, cosine or namely "x", is negative on the 2nd and 3rd quadrants only, the first angle you'd get from your calculator will be 134.43°, now, this one is on the 2nd quadrant, so it has a margin from 180° of 180 - 134.43 = 45.57, that means the angle of reference is 45.57° in red in the picture.

to get the 2nd angle on the 3rd quadrant, we add to 180° the reference angle, so 180 + 45.57 = 225.57°.