In the triangle ABC, ab = AC, angle BCA = 80 degrees, O is a point in the triangle ABC, and the angle OBC = 10 degrees, angle oca-20 degrees

In the triangle ABC, ab = AC, angle BCA = 80 degrees, O is a point in the triangle ABC, and the angle OBC = 10 degrees, angle oca-20 degrees

In the triangle ABC, ab = AC, angle a is 80 degrees, there is a point P in angle ABC, known angle PBC = 10 degrees, angle PCB = 30 degrees, find the degree of angle PAC? Make equilateral triangle abd, so that ∠ DAC is an acute angle, connect CD. Then: ab = BD = ad, ∠ abd = ∠ bad = 60 °. ∵ AB = AC, ∠ BAC = 80 °, ∵ ad =

In the triangle ABC, ab = AC, angle BAC = 80 degrees, point O is the inner point, and angle OBC = 10 degrees, angle OCA = 20 degrees, find: angle Bao =? Degrees

70 degrees

AB = AC, BAC = 80 degrees, O is a point in the triangle ABC, OBC = 10 degrees, OCA = 20 degrees, and Bao =? 0 points

This question is similar
In the triangle ABC, ab = AC, the angle a is 80 degrees, there is a point P in the angle ABC, the known angle PBC = 10 degrees, the angle PCB = 30 degrees, find the degree of the angle PAC?
Make an equilateral triangle abd, so that ∠ DAC is an acute angle, connecting CD
Then: ab = BD = ad, ∠ abd = ∠ bad = 60 °
∵AB=AC,∠BAC=80°,
∴AD=AC,∠DAC=80°-60°=20°,
∠ABC=50°=∠ACB
∴∠ACD=∠ADC=1/2×（180°-20°）=80°
∵∠PBC=10°∠PCB=30°
∴∠CBD=60°-50°=10°=∠PBC
∠ABP=50°-10°=40°
∠BCD=80°-50°=30°=∠PCB
∴△PBC≌△DBC,∴PB=BD=AB
∴∠BAP=∠BPA=1/2×（180°-40°）=70°,
∴∠PAC=80°-70°=10°
A: the degree of PAC is 10 degrees

In the triangle ABC, ab = AC, the angle BAC = 80 °, O is a point in the triangle ABC, and the angle OBC = 10 °, the angle OCA

If O is any point, there is no answer to this question. Do you think there are less conditions!