Find angles in acute triangle ABC?

It should be clear that A=50 since <BOC=100. Also, B = 70, C=60. To prove this, I'll show a construction that gets to the stated conditions. Then I'll assert that the construction is unique.

Consider arbitrary circle O with radius r, and chord CB such that <BOC=100. Extend CO through O to E so that CE=CB. Let OE=e so that

(1) r + e = 2r cos(40) or

(2) e = r(2r cos(40) - 1). Also, recall that in general

(3) cos(3x) = 4cos³(x) - 3cos(x)

Now extend BE to intersect circle O at A (hence <A=50), and extend BO to intersect AC at D (so that <DOE=100). Since ΔCEB is isosceles, <CBA=70 and <ACB=60. So, ΔDCO is 80-20-80 so that DC=r. Find G on AC so that EG || BD, so that ΔGCE is also 80-20-80 with DG=e and <ADB=<AGE=100. Since ΔGCE is isosceles with base angle 80:

EG = 2(r+e) cos(80)

= 4r cos(40) cos(80), from (1)

= 4r cos(40) (2cos²(40) - 1)

= (2cos(120) + 2cos(40)) r, from (3)

= (2 cos(40) - 1) r

= e, from (2)

Hence, ΔDGE is 50-80-50 isosceles. Therefore, <ODE=50 which implies <DEO=30.

The above construction to produce ΔDOE as 50-100-30 for a fixed O, r is unique. Proof: Make the above construction, only extend CO to E' such that CE' ≠ CE. Except for the letters O, C, B, denote all the letters in the revised constuction with an additional apostrophe such as D' for D. Note that <D'OE' stays fixed at 100.

If CE'>CE then OD'<OD and OE'>e, hence <D'E'O < 30.

If CE'<CD then OD'>OD and OE'<e, hence <D'E'O > 30.

QED.

PS. Merry XMas to my Yahoo Answers friends.

Answer:

∠A = 50°, ∠B = 60° and ∠C = 70° as solved below. Part of the solution is obtained graphically and verified to be correct.

From the given condition, ∠ BOC = 100° => ∠ A = 50°.

Let angle DCO = x

BOC is isosceles triangle in which ∠ OBC = ∠ OCB = 40°

∠ COD = ∠ BOE = 80°

Applying sine rule to Δ ODE, Δ ODC and Δ OEB,

OD/OE = sin30°/sin50°

OD/R = sinx/sin(100°-x) and

OE/R = sin(50°-x)/sin(50°+x)

=> OD/OE = sin30°/sin50° = [sinx * sin(50°+x)] / [sin(100°-x) * sin(50°-x)]

Solving this equation for x will give ∠ B and ∠ C of Δ ABC,

as ∠ B = 40°+ x and ∠ C = 90°- x

The value of x was graphically found to be nearly 20° and verified as under to be true by plugging in the RHS of the above equation.

RHS

= [sinx * sin(50°+x)] / [sin(100°-x) * sin(50°-x)]

= sin20° sin70° / sin80° sin30°

= 2sin20°sin70° / 2sin40° cos40° [taking sin30° = 1/2]

= (cos50° - cos90°) / 2cos50°sin50°

= (1/2)cos50° / cos50° sin50°

= sin30° / sin50°

= LHS.

=> B = 40° + x = 40° + 20° = 60°

and C = 90° - x = 90° - 20° = 70°.

Found my compass. The triangle is an isoceles triangle.

B= 80 (Edit: this is not so)

A= 50 <=== this one stays, it's quite OK

C = 50 (Edit: this is not so)

..........................B

…….....….__----‾‾/|\‾‾‾----__

..…….._-‾‾......../..|...\.......‾‾-_

…...…/......…./.....|…..\....….‾-_

….…/….E../…..…|.....…\.....…\

……|……/•\……...|..…..…\…....|

……\…/……...\....•O.….……\.../

.…A.\/_________\|_________\/C

...……‾-_…………D……....._-‾

…...……‾--__……...……_--‾

.....….……….‾‾‾‾---•--‾‾‾‾

...........................M

∆ABC is an isoceles triangle (Edit: this is not an Isoceles

..................triangle, the rest stays. They are.)

∆BOC.......isoceles

∆BOA.......isoceles

∆MOC...(when connected, is an isoceles triangle)

∆MOA...(when connected, is an isoceles triangle)

∆AOC...(when connected, is an isoceles triangle)

These are isoceles triangles of varying base lengths, only the paired sides are equal to R, radius of the circle.

( I think, Quadrillerator and Mielo istetz has got it figured. I only got up to the graphics and thought it was it.)

==> Anyone of you guys who would like my diagram as reference for your notation, you may Copy and Paste it in your solution now. I will delete mine afterwards because I am accumulating more TUs although mine is an incomplete answer. It would not look good.

IF D and E are in the exact centers of the line segments they intersect, THEN...

A=80

B=50

C=50

I THINK...