Question #c86cf

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Question #c86cf

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Answer 1 · 2016-06-08T13:54:28

I'll solve $59 i i$ $59ii$ .

You will need the following identities to solve this equation:

$sin 2 θ = 2 sin θ cos θ$ $sin2theta = 2sinthetacostheta$

$cos 2 θ = 1 - 2 {sin}^{2} θ$ $cos2theta = 1 - 2sin^2theta$

Explanation:

$\frac{2 sin θ cos θ}{2 sin θ} - cos θ (1 - 2 {sin}^{2} θ) = cos θ$ $(2sinthetacostheta)/(2sintheta) - costheta(1 - 2sin^2theta)= costheta$

$cos θ - cos θ - cos θ (1 - 2 {sin}^{2} θ) = 0$ $costheta - costheta - costheta(1 - 2sin^2theta) = 0$

$- cos θ = 0 and {sin}^{2} θ = \frac{1}{2}$ $-costheta = 0 and sin^2theta = 1/2$

$- cos θ = 0 and sin θ = \pm \frac{1}{\sqrt{2}}$ $-costheta = 0 and sintheta =+- 1/sqrt(2)$

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

Hopefully this helps!

Answer 2 · 2016-06-08T14:20:35

Item 60. See demonstration.

Explanation:

Item 60

$tan (30^{\circ}) = \frac{h}{¯ ¯¯¯¯ ¯ B C}$ $tan(30^@)=h/(bar (BC))$
$tan (45^{\circ}) = \frac{h}{¯ ¯¯¯¯ ¯ C A}$ $tan(45^@)=h/(bar(CA))$
${(¯ ¯¯¯¯ ¯ B C)}^{2} + {(¯ ¯¯¯¯ ¯ C A)}^{2} = {(2 r)}^{2}$ $(bar (BC))^2+(bar(CA))^2=(2r)^2$ (mind that $∠ C = 90^{\circ}$ $angle C = 90^@$ )

then

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

but

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

then

$r = h$ $r = h$

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Question #c86cf

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