In the triangle ABC, the degree of angle a is 100 degrees. If the bisector of angle B and angle c intersects point O, then the angle BOC and so on are equal to? choice question A. 100 degrees B. 140 degrees C. 60 degrees D. None of the above answers is correct

In the triangle ABC, the degree of angle a is 100 degrees. If the bisector of angle B and angle c intersects point O, then the angle BOC and so on are equal to? choice question A. 100 degrees B. 140 degrees C. 60 degrees D. None of the above answers is correct

Because the bisectors of angles B and C intersect at o
So angle OBC = 1 / 2 angle ABC
Angle OCB = 1 / 2 angle ACB
Because angle a + angle ABC + angle ACB = 180 degrees
Angle a = 100 degrees
So angle ABC + angle ACB = 80 degrees
So angle OBC + angle OCB = 40 degrees
Because the angle OBC + angle OCB + angle BOC = 180 degrees
So BOC = 140 degrees
So choose B

In the triangle ABC, the bisector of angle B and angle c intersects point O. if angle a = 60 degrees, try to find the degree of angle BOC?

∠BOC=90+1/2∠A=90+30=120°

It is known that in the isosceles triangle ABC, ∠ BAC = 90 °, AF is the height on the side of BC, and the arc AB with a as the center and AF as the radius intersects at point D,
If BC = 10, AF = 5, find the perimeter of the shadow

3.14*AF/4+2*AF=3.15*5/4+2*5=13.9

In the triangle ABC, AC is perpendicular to BC, AC = BC, D is a point on AB, AF is perpendicular to CD, the extension line of intersecting CD is perpendicular to F, be is perpendicular to CD and E

In RT triangle ABC, angle ACF = 90 degrees minus BCE, in RT triangle ACF, angle CAF = 90 degrees minus BCE, so angle ACF is equal to angle caf. In triangle ACF and triangle CBE, AC = BC, angle AFC = angle BEC = 90 degrees, angle ACF is equal to angle CAF, so triangle ACF and triangle CBE are congruent, so CE = AF