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Did #70 on today's test have a typo, or can you actually solve with only the given values?

why is everybody talking about the test! I thought you're not supposed to be discuss any part of it!

You technically can, as today was the only day where you can take the test. I didn't take it, so I don't know what people are talking about.

So we can...Let it Rip!

How do you do number 1-70

How is number 70 done along with 69?

I get you but I dont really get how to work those.

I could help, if I knew the question(s)... -_-

Oh well...

Sorry its cause they r geometry questions

#69
The shaded area is equal to the total circle area minus the little, unshaded circle. If we denote the little circle radius as r, then the area of the shaded section is

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Solve for r,

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Your total radius then is r, the little circle radius, plus the rest of the radius.
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#70
No typo, notice that the hypotenuse of the smaller right triangle and the bottom leg are both of length r, the radius of the circle. Denote the angle you're looking for by theta and note that the other acute angle is 90 minus theta. It's superfluous to use the other acute angle, or cotagent for that matter, but it helped me organize my thoughts better for some reason. Now you have the two trigonometric ratios

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Taking advantage of trig identities, we can solve the second equation for r and then plug that into the first equation.

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Canceling out the sines on both sides, and isolating the cosine, we are left with

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