In isosceles triangle ABC, ∠A = 80°. The bisectors of the base angles, ∠B and ∠C, intersect at point P.
∠BPC =
I made my sketch so that angle B = angle C = 50°
So the bisected angles are 25° each
and angle BPC = 180 - 25 - 25 = 130°
this looks like some hard triganomitry
To find the measure of angle BPC, we can start by analyzing the properties of an isosceles triangle. In an isosceles triangle, the base angles (the angles opposite the equal sides) are congruent. So, in triangle ABC, angles B and C are congruent.
Given that angle A is 80°, we know that angles B and C must also be congruent. So, each of these angles has a measure of (180° - 80°) / 2 = 50°.
Now, let's look at the bisectors of angles B and C. The bisector of an angle divides the angle into two equal parts. Since angles B and C are each 50°, the bisectors will divide them into two equal parts with a measure of 25° each.
The bisectors of angles B and C intersect at point P. Since the bisectors divide each of the angles B and C into two equal parts, the angles formed at point P (angle BPC) will also have a measure of 25° each.
Therefore, the measure of angle BPC is 25°.
Not trigonometry involved here.
It is elementary geometry, based on the definition of an isosceles triangle, and the fact that the angles in any triangle add up to 180°