It is known that in △ ABC, a = 1 / 2 B, B = 3 C, find a, B, C

It is known that in △ ABC, a = 1 / 2 B, B = 3 C, find a, B, C

Because a = 1 / 2 b = 1 / 3 C
Therefore, there are ∠ B = 2 ∠ a, ∠ C = 3 ∠ a
And ∠ a + B + C = 180 degree
Substituting ∠ a + 2 ∠ a + 3 ∠ a = 180 °
The solution is ∠ a = 30 °, B = 60 °, C = 90 °. This is a right triangle
The shortest side is the side corresponding to 30 degree angle, so its shortest side is 8 × 30 degree = 4cm

Let's know that a, B and C are integers, and a + B = 56; C-B = 61, a-c = 21. What is the square of the sum of a, B and C?
A,10816
B,10817
C,10804
D,10776

Add the first two formulas together
A+C=117
Add to the last formula
2A=138
A=69,C=48
Add to the first equation
A+B+C=104,
Square gain
(A+B+C)^2=10816
Choose a

-The result is a.56 B. - 56 c.28 D. - 28

-Algebraic sums of 7, + 14, - 21 = - 14
-The sum of absolute values of 7, + 14, - 21 = 42
-The algebra of 7, + 14, - 21 and the sum of their absolute values give 28
C

Two two digit numbers, their difference is 56, and the last two digits of their square number are the same, then the two numbers are______ ．

∵ X-Y = 56, x2-y2 = m × 100 (M is a positive integer). By eliminating x, we get 112Y = 100m-3136, y = 25m28-28, ∵ y is a two digit number with m < 100, ∵ M = 56 or 84, ∵ y = 22 or 47. When y = 22, X = 78; when y = 47, x = 103 (rounding). So the answer is: 22, 78

In the right triangle ABC, the angle c is 90 degrees. If a, B and C are continuous integers, then a + B + C =?

(b-1)^2+b^2=(b+1)^2
B = 4 or B = 0 (rounding off)
Then a = 4-1 = 3, C = 4 + 1 = 5
a+b+c=3+4+5=12

Given point a (4,1) and point B (- 3,2), find point C on the y-axis, so that the area of triangle ABC is 12
He is my uncle

Let the analytic expression of the straight line where AB is located be y = KX + B, and substitute the coordinates of a and B into 1 = 4K + B2 = - 3K + B. the solution is k = - 1 / 7, B = 11 / 7, so the intersection of AB and Y axis is (0,11 / 7), and the coordinate of point C is (0, c), so the area of △ ABC = the area of △ BCD + △ ACD = 1 / 2 * (CD * 3) + 1 / 2 * (CD * 4) = 1 /

Given that a (a, 0), B (B, 0), point C is on the y-axis, and the square of / A + 4 / + (b-2) is 0 (1). If the area of triangle ABC is 6, find the coordinates of point C (2) to move point C to the y-axis
It is known that a (a, 0), B (B, 0), point C is on the y-axis, and the square of / A + 4 / + (b-2) is 0
(1) If the area of triangle ABC = 6, find the coordinate of point C
(2) Translate point C to the right so that OC bisects ∠ ACB, point P is a moving point on the right side of point B on the x-axis, and PQ ⊥ OC is at point Q. when ∠ ABC - ∠ BAC = 60 °, calculate the degree of ∠ Apq
(3) Under the condition of (2), translate the line AC to get the line EF through point P, make the angle bisector of ∠ ape and intersect the extension line of OC at point M. when point P moves on the x-axis, calculate the value of ∠ M-1 / 2 ∠ ABC

(1) Let C coordinate be (0, c), then 1 / 2 * (B-A) | C | = 6. Therefore, if C = ± 2C is (0,2) or (0, - 2) (2) mark each angle, we can calculate ∠ OCA + ∠ BAC = 60 ° = ∠ POC = 90 ° - ∠ Apq, so ∠ Apq = 30 ° (3) let PM and CB intersect at K, set ∠ ACO

What are the two integers adjacent to a? What is the sum of these three integers?

The two integers adjacent to a are A-1 and a + 1
The sum of these three numbers is A-1 + A + A + 1 = 3A

In △ ABC, the sum of a and B is just twice of C, and B is 30 ° larger than A. can you find B? Explain the reason
Using binary linear equations

A+B=2C,B-A=30°,A+B+C=180°
Easy to get C = 60 degree
Then a + B = 120 ° and a = 120 ° - B is substituted by B-A = 30 ° to get
B-(120°-B)=30°
So B = 75 degree

As shown in the figure, △ ab ′ C ′ is obtained by rotating the equilateral triangle ABC 30 ° counterclockwise around point A. judge the relationship between AC and B ′ C ′ and explain the reason

The triangle ABC is similar to △ ab ′ C ′ so BC = B ′ C ′ because ABC is an equilateral triangle, AC = BC then AC = B ′ C ′