One of a plus one of B plus one of C equals four fifths, a + B + C =?

One of a plus one of B plus one of C equals four fifths, a + B + C =?

A+B+C=17

In △ ABC, ∠ B = 45 ° and ∠ C = 72 °, then an external angle adjacent to ∠ A is equal to______ °．

As shown in the figure: ∵ ABC, in ∵ ABC, ∵ B = 45 °, ∵ C = 72 °, ∵ 1 = ∵ B + ⊙ C = 45 ° + 72 ° = 117 °. So the answer is: 117

In the triangle ABC, BD.CD Angle D is equal to 90 degrees minus 1 / 2 angle A
This is the first day of junior high school, so please try to be simple

Because the angle of the triangle is 180 degrees
So angle B + C = 180 - angle A
Because the complement sum is 180
So the sum of the outer angles of the two base angles is (180 angle b) + (180 angle a) = 360 - (angle B + C)
=180+A
Because BD and DC divide the outer corner equally
So the angle AB * + AC @ = 1 / 2 the sum of the outer angles of the two base angles = 90 + 1 / 2 A
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Because the outer angle of a triangle is equal to the sum of the two inner angles that are not adjacent to it
So angle d = angle AB * + AC @ - angle a = 90 + 1 / 2 angle a - angle a = 90 ° - 1 / 2 angle a
A little bit on the extension line*
A little bit on the DC extension line@

One third of a equals one fourth of B equals one fifth of C. calculate the degree of a + B

Without the condition "in the triangle ABC"
From 1 / 3 ∠ a = 1 / 4 ∠ B = 1 / 5 ∠ C
∠A=3/5∠C,∠B=4/5∠C,
∵∠A+∠B+∠C=180°
∴3/5∠C+4/5∠C+∠C=180°
The solution is ∠ C = 75 degree
∴∠A+∠B=180°-75°=105°