In △ABC, the internal bisectors of ∠B and ∠C meet at O. If ∠BAC = 72°, then the value of ∠BOC is:
126°
In triangle ABC, the internal bisectors of ∠B and ∠C meet at point O (the incenter). Using the angle bisector theorem: ∠BOC = 90° + (A/2)
Given ∠A = 72° → ∠BOC = 90° + 72°/2 = 90° + 36° = 126°
Similar Questions
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Important Questions from Geometry
In a circle with center O, an arc ABC subtends an angle of 138° at the centre of the circle. The chord AB is produced to a point P. Then, the measure of ∠CBP is:
A secant PAB is drawn from an external point P to the circle with the centre at O, intersecting it at A and B. If OP = 17 cm, PA = 12 cm and PB = 22.5 cm, then the radius of the circle is:
The side BC of △ABC is produced to a point D. If AC = BC and ∠BAC = 70°, then find the value of 2.5∠ACD – 1.5∠ABC.
If the radius and height of a right circular cylinder are 21 cm and 5 cm, respectively, then the total surface area of the cylinder is (use π = 22 / 7):
The central angle of a sector is 80° and whose length is 96π. What is the radius of the circle?
