Austin's angle

Geometry Level 3

Isosceles triangle A B C ABC has A B = B C AB=BC and a vertex angle of 3 2 32^\circ . Γ \Gamma is a circle that circumscribes A B C ABC , and O O is the center of Γ \Gamma . What is the measure (in degrees) of B O C \angle BOC ?

This problem is posed by Austin C.


The answer is 148.

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3 solutions

Billy Sitompul
Aug 14, 2013

If you draw it correctly,

you can notice that B O BO cuts A B C ∠ABC right in the center.

So now, there will be new two angles that measure the same,

A B O = C B O = 1 6 ∠ABO = ∠CBO = 16^∘ .

Since B O BO and C O CO are R R ,

we can respectively say that B O = C O BO = CO ,

so we get new isosceles triangle, B C O BCO .

We now that base angles of isosceles measure the same.

That means C B O = B C O = 1 6 ∠CBO = ∠BCO = 16^∘ .

Sum of all angles of triangle will always 18 0 180^∘ ,

then we put some of our math:

B C O + C B O + B O C = 18 0 ∠BCO + ∠CBO + ∠BOC = 180^∘

1 6 + 1 6 + B O C = 18 0 16^∘ + 16^∘ + ∠BOC = 180^∘

3 2 + B O C = 18 0 32^∘ + ∠BOC = 180^∘

B O C = 18 0 3 2 ∠BOC = 180^∘ - 32^∘

B O C = 14 8 ∠BOC = 148^∘

Fouad SAadi
Aug 13, 2013

the mesure of the angle <BOC is the double of the angle <BAC . Since the some of mesures of angle of triangle is 180° and ABC isosceles and <ABC is 32° we can deduce that mesure of <BAC is 74° then mesure of <BOC is 148°

Moderator note:

From circle properties, we know that the angle at the center is twice the angle at the circumference.

TIKI TAKA ...

Konstantinos Angelis - 7 years, 10 months ago

I can't believe I had the answer but I decided to put something else on...... -.-

Jannat Khusbo - 7 years, 10 months ago
Kenneth Gravamen
Sep 23, 2015

One of the base angles of the triangle B A C \angle{BAC} is the inscribe angle of O \circ{O} . Thus, B O C \angle{BOC} = 2 B A C \angle{BAC} . B O C \angle{BOC} = 148 \text{148} .

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