Need help with questions (1 Viewer)

[star*eyed]

I have 2 questions:

Question 1: Solve the equation sin2x = cosx, answering as a general solution


Question 2: Two circles intersect at A and B. the tangent to the 2nd circle at A cuts the 1st circle at C. The tanget to the 1st circle at B cuts the 2nd circle at D. Prove AD is parallel to BC.

 

[SoulSearcher]

1) sin2x = cosx
sin2x - cosx = 0
2sinxcosx - cosx = 0
cosx(2sinx-1) = 0
.'. cosx = 0 = cos (pi/2)
.'. x = 2n*pi + pi/2
OR
2sinx - 1 = 0
2sinx = 1
sinx = 1/2 = sin (pi/6)
.'. x = n*pi + (-1)ⁿ * pi/6
Therefore x = n*pi + (-1)ⁿ * pi/6 or x = 2n*pi + pi/2

 

[SoulSearcher]

2) Prove: AD || BC
How do we do that: Prove that @DAB = @ABC
Now,
@DAE = @ABD (alternate segment)
@ABD = @BCA (alternate segment)
Also,
@FBC = @BAC (alternate segment)
@BAC = @ADB (alternate segment)
Therefore triangle ABC ||| triangle ABD, because @BAC = @ADB and @ABD = @BCA
Therefore @DAB = @ABC (corresponding angles in similar triangles)
Therefore AD || BC

Note: @ Stands for angle, < screws up for some reason