Prove the following: if, in a triangle ABC,BA =CB
O in atriangle
mÐ BAO = mÐ OAC = 20°,
mÐ ACO = 10°,
and mÐ OCB = 30°,
then mÐ OBC = 80°.
To prove that m∠OBC = 80°, we can use the angles in a triangle to find the missing angle.
Given:
- BA = CB (i.e., triangle ABC is an isosceles triangle)
- ∠BAO = ∠OAC = 20°
- ∠ACO = 10°
- ∠OCB = 30°
First, let's label the angles in triangle ABC:
∠B = ∠BAC
∠A = ∠ACB
∠C = ∠CBA
Since it is given that BA = CB, we know that ∠B = ∠C.
We can now express the angles in terms of ∠B:
∠BAO = 20°
∠OAC = 20°
∠OCB = 30°
∠ACO = 10°
∠C = ∠B
Since the sum of angles in a triangle is always 180°, we can write the following equation for triangle ABC:
∠B + ∠A + ∠C = 180°
Substituting the known values:
∠B + 20° + ∠B = 180°
2∠B = 160°
∠B = 80°
Since ∠B = ∠C, we know that ∠C = 80° as well.
Finally, to find ∠OBC, we can use the fact that the sum of angles in a triangle is 180°:
∠OBC + ∠OCB + ∠BCO = 180°
Substituting the known values:
∠OBC + 30° + 10° = 180°
∠OBC = 180° - 40°
∠OBC = 140°
Therefore, m∠OBC = 80°.