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As shown in Figure 1, in triangle ABC, OB and OC are bisectors of angle ABC and angle ACB respectively. If angle a = x degrees, what is the degree of angle BOC
Tip: the sum of inner angles of triangle is 180 degrees!
(1) In triangular BOC:
Angle BOC + (1 / 2 angle ABC + 1 / 2 angle ACB) = 180
Angle BOC = 180 - (1 / 2 angle ABC + 1 / 2 angle ACB)
Angle BOC = 180-1 / 2 (angle ABC + angle ACB)
(2) In triangle ABC:
Angle a + angle ABC + angle ACB = 180
Angle ABC + angle ACB = 180-x
(3) It can be seen from (1) and (2)
Angle BOC = 180-1 / 2 (180-x)
The angle BOC = 90 + X / 2)
It is known that in △ ABC, the bisector of ∠ ABC and ∠ ACB intersects at point o It is proved that: ∠ BOC = 90 ° + 1 2∠A．
It is proved that: the bisector of ABC and ∠ ACB intersects at the point O, OBC = 12 ∠ ABC, OCB = 12 ∠ ACB, ? OBC + ∠ OCB = 12 (∠ ABC + ∠ ACB). In △ OBC, ∠ BOC = 180 ° - (∠ OBC + ∠ OCB) = 180 ° - 12 (180 ° - ∠ a) = 90 ° + 12 ∠ a
(1) As shown in the figure, in triangle ABC, the bisector of angle ABC and angle ACB intersect at point O. if angle a = 42 degrees, calculate the degree of angle BOC (2) Remove the condition that angle a = 42 degrees in (1) and try to find out the quantitative relationship between angle BOC and angle A
Because angle a is equal to 42 degrees, angle ABC and angle ACB are equal to 138 degrees. Because Bo bisects angle ABC and co bisects angle ACB, so angle OBC angle OCB equals 69 degrees, so angle BOC equals 111 degrees
It is known that in △ ABC, the bisector of ∠ ABC and ∠ ACB intersects at point o It is proved that: ∠ BOC = 90 ° + 1 2∠A．
It is proved that the bisector of ABC and ACB intersects at point o,
∴∠OBC=1
2∠ABC，∠OCB=1
2∠ACB，
∴∠OBC+∠OCB=1
2（∠ABC+∠ACB），
In △ OBC, ∠ BOC = 180 ° - (∠ OBC + ∠ OCB)
=180°-1
2（∠ABC+∠ACB）
=180°-1
2（180°-∠A）
=90°+1
2∠A，
Namely: ∠ BOC = 90 ° + 1
2∠A．
As shown in the figure, O is the outer center of the triangle ABC. (1) if the angle BOC = 130 °, find the degree of angle BAC; (2) if the angle a = n, find the degree of angle b0c We only studied the vertical diameter theorem
Halo, what about graphs? We only learned the vertical diameter theorem, but didn't learn what is the outer center of a triangle?
(1) There are two situations in this question
① Because o is the outer center of the triangle ABC
So when o is in the triangle, 2 ∠ BAC = ∠ BOC = 130
So ∠ BAC = 65
② Because o is the outer center of the triangle ABC
Therefore, when o is outside the triangle, BAC + BOC = 180
So ∠ BAC = 50
(2) There are two situations in this question
① When o is in the triangle ABC, BOC = 2n
② When o is outside the triangle ABC
∠BOC＝180－n
As shown in the figure, if ⊙ o is the inscribed circle of ⊙ ABC, and ∠ BAC = 50 °, then ∠ BOC is______ Degree
∵ OB and OC are the angular bisectors of ᙽ ABC and ᙽ ACB,
∴∠OBC+∠OCB=1
2（∠ABC+∠ACB）=1
2（180°-50°）=65°，
∴∠BOC=180°-65°=115°．
In the triangle ABC, the angle c is 90 ° and the circle O is its inscribed circle. The angle BOC is 105 ° and ab is 3 / 8. Find the length of BC
Through O, OM is perpendicular to BC and o
The angle BCO is 45 degrees
The angle MOC is 45 degrees
So the angle mob is 60 degrees
So the angle MBO is 30 degrees
So the angle ABC is 60 degrees
BC is one half of AB, three of sixteen
(Note: the center of the inscribed circle of a triangle is the intersection of the bisectors of the three corners of the triangle)
In the triangle ABC, the angle a = 80 degrees, and the bisector of angle B and angle c intersect at point O. (1) calculate the degree of angle BOC; (2) connect OA to find the degree of angle OAC
(1) ∵ ob, OC is the angular bisector
Qi
As shown in the figure, the point O is the center of the inscribed circle of △ ABC, ∠ BAC = 80 ° and the degree of ∠ BOC is obtained
∵∠BAC=80°，
∴∠ABC+∠ACB=180°-80°=100°，
∵ point O is the center of the inscribed circle of ∵ ABC,
ν Bo and Co are the angular bisectors of ∠ ABC and ∠ BCA respectively,
∴∠OBC+∠OCB=50°，
∴∠BOC=130°．
Circle O is the inscribed circle of right triangle ABC, angle c = 90 degrees, if angle BOC = 105 degrees, ab = 4cm, calculate the degree of angle BOC
Angle AOB = angle OAC + angle OCA + angle OBC + angle OCB = angle BAC / 2 + angle c + angle ABC / 2 = (180 degrees + angle c) / 2 = 135 degrees
Angle AOC = 360 degrees - angle AOB - angle BOC = 120 degrees
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