# Question #c86cf

Jun 8, 2016

I'll solve $59 i i$.

You will need the following identities to solve this equation:

$\sin 2 \theta = 2 \sin \theta \cos \theta$

$\cos 2 \theta = 1 - 2 {\sin}^{2} \theta$

#### Explanation:

$\frac{2 \sin \theta \cos \theta}{2 \sin \theta} - \cos \theta \left(1 - 2 {\sin}^{2} \theta\right) = \cos \theta$

$\cos \theta - \cos \theta - \cos \theta \left(1 - 2 {\sin}^{2} \theta\right) = 0$

$- \cos \theta = 0 \mathmr{and} {\sin}^{2} \theta = \frac{1}{2}$

$- \cos \theta = 0 \mathmr{and} \sin \theta = \pm \frac{1}{\sqrt{2}}$

$\theta = {90}^{\circ} , {270}^{\circ} , {45}^{\circ} , {135}^{\circ} , {225}^{\circ} \mathmr{and} {315}^{\circ}$

Hopefully this helps!

Jun 8, 2016

Item 60. See demonstration.

#### Explanation:

Item 60

$\tan \left({30}^{\circ}\right) = \frac{h}{\overline{B C}}$
$\tan \left({45}^{\circ}\right) = \frac{h}{\overline{C A}}$
${\left(\overline{B C}\right)}^{2} + {\left(\overline{C A}\right)}^{2} = {\left(2 r\right)}^{2}$ (mind that $\angle C = {90}^{\circ}$)

then

${\left(\frac{h}{\tan \left({30}^{\circ}\right)}\right)}^{2} + {\left(\frac{h}{\tan \left({45}^{\circ}\right)}\right)}^{2} = {h}^{2} \left(\cot {\left({30}^{\circ}\right)}^{2} + \cot {\left({45}^{\circ}\right)}^{2}\right) = 4 {r}^{2}$

but

$\left(\cot {\left({30}^{\circ}\right)}^{2} + \cot {\left({45}^{\circ}\right)}^{2}\right) = 4$

then

$r = h$