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by **redhawk2015** » Mon Apr 30, 2012 1:10 pm

Was going through my region test to see how stupid my mistakes/skips were, when I realized I didn't have an answer key. Could someone with an answer key to the regional test please verify/invalidate the following answers?

2. C
6. D
25. E
33. D
38. B
41. B
44. A
45. D
48. C
49. A
50. D
52. D
57. D
58. C
60. What is the answer?

by **FabensMath** » Mon Apr 30, 2012 4:02 pm

2.D
6.D
25.E
33.C
38.B
41.B
44.A
45.D
48.C
49.A
50.D
52.D
57.D
58.C.
60.D

I think these are right, hopefully I didn't make a typo.

by **redhawk2015** » Mon Apr 30, 2012 5:20 pm

Thanks Fabens. In that case I need some help on these two:

33. Let $\triangle ABC$ exist such that D is the midpoint of segment AC, $m\angle CAB = m\angle CBD$ , and AB = 12. Find the square of the length of BD.

Answer: 72

60. $\triangle ABC$ is an acute triangle. Point D lies on segment AB such that CD is the length of a median of $\triangle ABC$ , $m\angle ACD=30^\circ$ , $m\angle BCD=20^\circ$ , and AC = 10 cm. Find BC.

Answer: 14.6

by **FabensMath** » Mon Apr 30, 2012 5:23 pm

Salvadoramerican wrote:Anyways i figure a quick way for 60 on regions 2012, since we have amedian the area of the two split triangles are equal then we can solve by having a=median and x=wanted side $\frac{1}{2}*a*10*\sin 30=\frac{1}{2}*a*x*\sin 20$ if we solve sor x the a woud cancel out and we get 14.61
though check my work

by **007math** » Mon Apr 30, 2012 6:29 pm

redhawk2015 wrote:33. Let $\triangle ABC$ exist such that D is the midpoint of segment AC, $m\angle CAB = m\angle CBD$ , and AB = 12. Find the square of the length of BD. Answer: 72

Well, I decided to make this a right triangle. I let $m\angle \text{ABC} = 90^\text{o}$ and therefore $m\angle \text{CAB} = m\angle \text{CBD} = 45^\text{o}$ . From here, $\text{AC} = 12\sqrt2 \implies [\text{DB}]^2 = 12^2 - (6\sqrt2)^2 = 72$

by **Salvadoramerican** » Tue May 01, 2012 7:13 am

can you guys explain 30 and ,39(just humor me on this one)

by **ohernandez** » Tue May 01, 2012 10:58 am

Salvadoramerican wrote:can you guys explain 30

$\displaystyle sin(\theta)=\frac{y}{30}$
$30sin(\theta)=y$

Differentiation with respect to time

$\displaystle 30cos(\theta) \frac{d\theta}{dt} =\frac{dy}{dt}$

$\displaystyle 30( \frac{adj}{hyp}) \frac{d\theta}{dt} =\frac{dy}{dt}$

So, now it's a plug-n-chug:

$\displaystyle 30( \frac{24ft}{30ft}) \frac{d\theta}{dt} =\frac{1ft}{2sec}$

$\displaystyle \frac{d\theta}{dt} =\frac{1ft}{2sec} \cdot \frac{1}{24ft}$

E. $\displaystyle \frac{d\theta}{dt} =\boxed{\frac{1rads}{48sec}}}$

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