In the triangle ABC, ab = AC, BAC = 80 degrees, O is a point in the triangle ABC, OBC = 10 degrees, OCA = 20 degrees The picture shows

In the triangle ABC, ab = AC, BAC = 80 degrees, O is a point in the triangle ABC, OBC = 10 degrees, OCA = 20 degrees The picture shows

In the triangle ABC, ab = AC, angle a is 80 degrees, there is a point P in the 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 degree. ∵ AB = AC, ∠ BAC = 80 degree, ∵ ad = AC

In isosceles △ ABC, ab = AC, Bo and co intersect at point O, and ∠ ABO = ∠ ACO, try to judge what triangle △ OBC is and explain the reason

Δ OBC is an isosceles triangle
It is proved that ∵ AB = AC, ∵ ABC = ACB,
∵∠ABO=∠ACO,∴∠ABC-∠ABC=∠ACB-∠ACO,
That is, OBC = OCB,
That is, Δ OBC is an isosceles triangle

In the isosceles right triangle ABC, ∠ BAC = 90 degrees, ab = AC, O is a point in the triangle, ∠ ACO is 20 degrees, and ∠ OBC is 10. Find ∠ Bao

The final result obtained by trigonometric function is arctan (sin 25 ° / Tan 10 ° - 1 / Tan 20 ° + cos 25 °)
It should be 29.0573291 by calculator
It is estimated that the final result should be 30 degrees
But I didn't think of the clever way
Is this a junior high school question or a senior high school question?

Let o be a point out of the plane of the triangle ABC, OA = AB = AC, angle oba = angle OCA = 45 degrees

∵ OA = AC, ∵ AOC = OCA, and ∵ OCA = 45 °, ∵ AOC = 45 °, ∵ OA ⊥ AC
∵ OA = AB, ∵ AOB = oba, and ∵ oba = 45 °, ∵ AOB = 45 °, ∵ OA ⊥ ab
From OA ⊥ AC, OA ⊥ AB, AC ∩ AB = a, Ao ⊥ plane ABC