In the triangle ABC, the angle a = 80 degrees, e, F, P are a point on the edge of AB, AC, BC respectively, be = BP, CP = CF to calculate the angle EPF

In the triangle ABC, the angle a = 80 degrees, e, F, P are a point on the edge of AB, AC, BC respectively, be = BP, CP = CF to calculate the angle EPF


50°
I don't know, right



In the triangle ABC, the intersection of the bisector of the angle ABC and the bisector of the angle ACD is at E. if the angle a is equal to 78, the angle e is obtained


∵ be and CE divide ∠ ABC and ∠ ACB equally,
∴∠EBC+∠ECB=1/2∠ABC+1/2∠ACB
=1/2(∠ABC+∠ACB)
=1/2(180°-∠A)
=90°-1/2∠A,
∴∠E=180°-(∠EBC+∠ECB)
=90 ° + 1 / 2 ∠ a - it can be used as a formula in filling in the blanks and multiple choice questions,
=129°.



If point O is a point in △ ABC, and the distance from point O to three sides is equal, ∠ BOC = 6150 °, then what is the degree of ∠ BAC


The distance between ∵ O and △ ABC is equal ∵ ob bisection ∵ ABC, OC bisection ∵ ACB ∵ CBO ∵ ABC / 2, ∵ BCO ∵ ACB / 2 ∵ BOC = 180 - (∵ CBO + ∵ BCO) = 180 - (∵ ABC + ∵ ACB) / 2 = 180 - (180 - ? BAC) / 2 = 90 + ? BAC / 2 = 150 ∵ BAC = 120 ° mathematics auxiliary



As shown in the figure, point O is the outer center of △ ABC, angle a = 72 ° to find the degree of angle BOC


Degree of angle BOC = 2, angle a = 144 degree