Hi all,

could y’all me solve this proof? i’ve shown some of my work, although i’m pretty sure it’s (pic is attached below)

i’ll mark !

Arc BC = arc DC because their subtended chords BC and DC are equal

Step-by-step explanation:

∵ CE is a diameter of circle A(chord passing through the center of the circle)

∵ CE ⊥ to the chord BD and intersect it at F

∴ BF = FD ⇒ fact in the circle

In the 2 Δ CFB and CFD

∵ BF = DF

∵ m∠BFC = m∠DFC = 90°

∵ CF is a common side

∴ The two triangles are congruent

∴ BC = DC

In the circle if the two chords are equal in lengths then their subtended arcs are equal in lengths and measures

∴ Arc BC = arc DC

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92