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As shown in the figure, in △ ABC, ∠ B = 90 ° o is a point on AB, and a circle with o as its center and ob as its radius intersects with point e at point E
Write all the inscriptions and pass on the picture
In the triangle ABC, a, B and C form an arithmetic sequence with the circumcircle radius of 1, and Sina sinc + √ 2 / 2cos (A-C) = √ 2 / 2 Find the size of (1) a (2) Triangle area
Because a, B, C form the arithmetic sequence
So a + C = 2B
A + B + C = π
So B = π / 3, a + C = 2 π / 3
sinA-sinC
=2cos(A+C)/2cos(A-C)/2
=-cos(A-C)/2
So Sina sinc + √ 2 / 2cos (A-C)
=-cos(A-C)/2+√2/2[2cos²(A-C)/2-1]=√2/2
That is, 2cos (A-C) / 2 - √ 2cos (A-C) / 2-2 = 0
The solution is cos (A-C) / 2 = - √ 2 / 2
Also - 2 π / 3
It is known that the three inner angles of the triangle ABC form an arithmetic sequence, the circumcircle radius is 1, and there is Sina COSC + 2 ^ (- 1 / 2) cos (A-C) = 2 ^ (- 1 / 2) to find the size of a, B, C
A=B-dC=B+dA+B+C=3B=180B=602^(1/2)*(sinA-cosC)+cos(A-C)=12^(1/2)*[sin(B-d)-cos(B+d)]+cos(2d)=12^(1/2)*[(sinBcosd-cosBsind)-(cosBcosd-sinBsind)]+cos(2d)=12^(1/2)*[sinBcosd-cosBcosd-cosBsind+sinBsind)]+c...
In △ ABC, the sizes of three inner angles a, B and C are arithmetic sequence. The value range of sina + sinc is calculated
∵ the size of three internal angles a, B, C is the arithmetic sequence
If a + B + C = π, a + C = 2B, then 3B = π, that is, B = π
Three
∴A+C=2π
Three
∴sinA+sinC=sinA+sin（2π
3-A）=3
2sinA+
Three
2cosA=
3sin（A+π
6）
∵0＜A＜2π
Three
∴π
6＜A+π
6＜5π
Six
∴1
2＜sin（A+π
6）≤1
The value range of sina + sinc is（
Three
2，
3]
As shown in the figure, ∠ ACB = 90 °, AC = BC, D is the point outside △ ABC, and the extension line of ad = BD, de ⊥ AC crossing CA is at point E
Proof: connect CD,
∵AC=BC，AD=BD，
ν C is on the vertical bisector of AB, D is on the vertical bisector of ab,
ν CD is the vertical bisector of ab,
∵∠ACB=90°，
∴∠ACD=1
2∠ACB=45°，
∵DE⊥AC，
∴∠CDE=∠ACD=45°，
∴CE=DE，
∴DE=AE+AC=AE+BC．
As shown in the figure, ∠ ACB = 90 °, AC = BC, D is the point outside △ ABC, and the extension line of ad = BD, de ⊥ AC crossing CA is at point E
Proof: connect CD,
∵AC=BC，AD=BD，
ν C is on the vertical bisector of AB, D is on the vertical bisector of ab,
ν CD is the vertical bisector of ab,
∵∠ACB=90°，
∴∠ACD=1
2∠ACB=45°，
∵DE⊥AC，
∴∠CDE=∠ACD=45°，
∴CE=DE，
∴DE=AE+AC=AE+BC．
In the triangle ABC, ∠ ACB = 90 degrees, AC = AB, D is the outer point of △ ABC and ad = AB, de ⊥ AC intersect the extension line of Ca at e. it is proved that de = AE + BC In the triangle ABC, ∠ ACB = 90 degrees, AC = BC, D is the outer point of △ ABC and ad = AB, de ⊥ AC intersect the extension line of Ca at e. it is proved that de = AE + BC
If we change ad = AB to ad = BD, it is right to prove: connecting CD, ∵ AC = BC, ad = BD, CD = CD ? ACD ≌ △ BCD (SSS) ACD = ≌△ BCD (SSS) ACD = ≌≌△ BCD (SSS) ? ACD = ≌≌△ BCD (SSS) ACD = ≌≌△ CED is isosceles right triangle de = ce
As shown in the figure, in the triangle ABC, the angle ABC = 50 degrees, the angle ACB = 80 degrees, extend CB to D, make BD = Ba, extend BC = Ca, connect ad, AE to find the angle D, angle e, angle DAE To process
In the title, extend BC to point e so that CE = ca
The problem should be to find the degree of each angle
We can see that the triangle ABC is an isosceles triangle
Because BD = Ba, angle d = angle DAB,
Because the angle ABC = angle D + angle DAB = 2 angle D
The angle D is 25 degrees
Similarly, the angle e is 40 degrees,
The angle DAE = 115 degrees
It is known that in △ ABC, ab = AC, BC = BD, ad = de = EB, then the degree of ∠ A is () A. 30° B. 36° C. 45° D. 50°
In this paper, we have the idea that ᙽ EBD = x
It is known that in △ ABC, ab = AC, BC = BD, ad = de = EB, then the degree of ∠ A is () A. 30° B. 36° C. 45° D. 50°
Let ∠ EBD = x °,
∵BE=DE，
∴∠EDB=∠EBD=x°，
∴∠AED=∠EBD+∠EDB=2x°，
∵AD=DE，
∴∠A=∠AED=2x°，
∴∠BDC=∠A+∠ABD=3x°，
∵BD=BC，
∴∠C=∠BDC=3x°，
∵AB=AC，
∴∠ABC=∠C=3x°，
∵∠A+∠ABC+∠C=180°，
∴2x+3x+3x=180，
The solution is: x = 22.5,
∴∠A=2x°=45°．
Therefore, C
RELATED INFORMATIONS
- 1. It is known that in △ ABC, ab = AC, BC = BD, ad = de = EB, then the degree of ∠ A is () A. 30° B. 36° C. 45° D. 50°
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