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[squexy]

View attachment 2941

Answer: K

View attachment 2942

Answer: BView attachment 2943

Answer: C

 

[mathmari]

https://www.physicsforums.com/attachments/2941

Answer: K

38. $$\sin M=\frac{\text{ opposite }}{\text{ hypotenuse }}=\frac{KL}{MK}$$

From the Pythogorean Theorem: $$(KL)^2+(ML)^2=(MK)^2 \Rightarrow (KL)^2=144-100=44 \Rightarrow KL=\sqrt{44}$$

Therefore, $$\sin M=\frac{\sqrt{44}}{12}$$

 

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What have you tried so far?

 

[squexy]

What have you tried so far?

37
By estimative I can find the answer, since radius is 5 coordinate units I decrease 5 from Y and add 5 to X having (7,-2) 39
a = b = c3x = 180
x = 60

 

[Prove It]

37
By estimative I can find the answer, since radius is 5 coordinate units I decrease 5 from Y and add 5 to X having (7,-2) 39
a = b = c3x = 180
x = 60

For 37 I would find the equation of the line passing through your two known points. Then try to find the equation of the normal to this line through the centre of the circle (keep in mind that the gradients of perpendicular lines multiply to give -1). Once you have the equation of this normal you can find where it intersects with the circle, solving your problem.

For 39 I have no idea what you are talking about. What are a, b, c? Are you using these letters to represent the sides of the triangle. If so, how do you know the sides of the triangle are all equal in length?