#1**+2 **

Let's call the unknown scale factor "s'"..

1s + 2s + 4s + 5s = 360º

Solve for s.

12s = 360º

s = 30º

So...

a. m arc PQ = 1s = 1(30º) = 30º

b. m arc QR = 2s = 2(30º) = 60º

c. m arc RS = 4s = 4(30º) = 120º

d. m arc SP = 5s = 5(30º) = 150º

e. m∠S = (1/2) * (m arc PQ + m arc QR) = (1/2) * (30 + 60) = 45º

f. m∠Q = (1/2) * (m arc RS + m arc SP) = (1/2) * (120 + 150) = 135º

g. m∠R = (1/2) * (m arc PQ + m arc SP) = (1/2) * (30 + 150) = 90º

h. m∠P = (1/2) * (m arc RS + m arc QR) = (1/2) * (120 + 60) = 90º

*edit* added ºs :)

hectictar
Apr 28, 2017

#2**+1 **

Hi Mellie :)

I am really not convinced Hectictar,

Your angles are correct but I do not think that there is enough info to give the arc lengths.

Your s relates to angles subtended from the centre. The 360 is 360 degrees. It is not related to the length of the circumference.

\(\text{You could say that } arc {PQ}= \dfrac{\pi r} {6}\\ \text{and } \qquad \qquad \qquad arc \;{QR}= \dfrac{2\pi r} {6}\\ \text{and } \qquad \qquad \qquad arc \;{RS}= \dfrac{4\pi r} {6}\\ \text{and } \qquad \qquad \qquad arc \; {SP}= \dfrac{5\pi r} {6}\\ \)

Where r is the radius but I do not see how you can give exact values for these. :/

Melody
Apr 28, 2017