Question #c86cf

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Answer 1 · 2016-06-08T13:54:28

I'll solve $59ii$ .

You will need the following identities to solve this equation:

$sin2theta = 2sinthetacostheta$

$cos2theta = 1 - 2sin^2theta$

$(2sinthetacostheta)/(2sintheta) - costheta(1 - 2sin^2theta)= costheta$

$costheta - costheta - costheta(1 - 2sin^2theta) = 0$

$-costheta = 0 and sin^2theta = 1/2$

$-costheta = 0 and sintheta =+- 1/sqrt(2)$

$theta = 90^@, 270^@, 45^@, 135^@, 225^@ and 315^@$

Hopefully this helps!

Answer 2 · 2016-06-08T14:20:35

Item 60. See demonstration.

Item 60

$tan(30^@)=h/(bar (BC))$
$tan(45^@)=h/(bar(CA))$
$(bar (BC))^2+(bar(CA))^2=(2r)^2$ (mind that $angle C = 90^@$ )

then

$(h/(tan(30^@)))^2+(h/(tan(45^@)))^2 = h^2(cot(30^@)^2+cot(45^@)^2)= 4 r^2$

but

$(cot(30^@)^2+cot(45^@)^2) = 4$

then

$r = h$