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In the triangle ABC, if a + C = 2B, C = 2, a = radical 3 + 1, find the value of B as in the title
A + C + B = 3B = 180, B = 60 ° B = under root sign (a ^ 2 + C ^ 2-2accosb) = under root sign (3 + 1 + 2 √ 3 + 4-2x2x2x (√ 3 + 1) / 2) = root sign (8 + 2 √ 3-2 √ 3-2) = √ 6
In the triangle ABC, the sides of angles a, B and C are a, B, C respectively. It is known that B = root sign 3a. When C = 1 and the area of triangle ABC is root 3 / 4, find a In the triangle ABC, the sides of angles a, B and C are a, B, C respectively. It is known that B = (root 3) A. when C = 1 and the area of triangle ABC is (root 3) / 4, the value of a is calculated, When COSC = root 3 / 3, find the value of COS (B-A)
B = √ 3A (1) cosine theorem C? = a? + B? - 2abcosc1 = 4A? - 2 √ 3A? Cosccosc = (4a? - 1) / (2 √ 3A?) (2) s Δ ABC = (1 / 2) absinc = √ 3 / 4C = 1, B = √ 3a a? Sinc = 1 / 2, sinc = 1 / (2a?) sin? C + cos
In △ ABC, the edges to which ∠ a, B and C are a, B and C respectively. If (root 3b-c) cosa = a COSC, then cosa=
Using the sine theorem a / Sina = B / SINB = C / sinc ∵ (√ 3b-c) cosa = acosc ? (√ 3sinb-sinc) cosa = sinacosc ? 3sinbcosa = sinacosc + sinccosa ? 3sinbcosa = sin (a + C) ? a + C = π - B, ? M (√ 3b-c) cosa = Sina COSC √ 3sinbcosa = 1 cosa = √ 3 / 3
In the triangle ABC, the radical 3B = 2asinb, and cosa = COSC, try to judge the shape of the triangle
Radical 3B = 2asinb,
therefore
b/sinB=a/√3/2=a/sinA
That is, Sina = √ 3 / 2,
also
cosA=cosC
therefore
A. C is an acute angle, so
A = 60 degrees = C
thus
Triangle is equilateral triangle!
The lengths of the opposite sides of the three inner angles a, B and C of △ ABC are a, B, C respectively, if a = Five 2B, a = 2B, then CoSb = () A. Five Three B. Five Four C. Five Five D. Five Six
∵ △ in ABC
a＝
Five
2B
A＝2B
According to the sine theorem
sinA＝
Five
2sinB
sinA＝sin2B＝2sinBcosB
∴cosB＝
Five
Four
Therefore, B is selected;
In △ ABC, the three inner angles a, B and C are respectively a, B and C. We know that 2B = a + C, a + radical 2B = 2c, and find the value of sinc Senior one compulsory 5 mathematics learning method p24 side 5 question
A + B + C = 180 degrees B = 60 degrees according to the sine theorem, a / Sina = B / SINB = C / sinc has Sina + √ 2sinb = 2sincb = 60 degrees, a = 120 ° - C is substituted, expanded and simplified √ 3 / 2 * sinc-1 / 2 * COSC = √ 2 / 2, that is sin (c-30) = √ 2 / 2 / / c-30 ° = 45 ° or 135 °
The lengths of the opposite sides of the three inner angles a, B and C of △ ABC are a, B, C respectively, if a = Five 2B, a = 2B, then CoSb = () A. Five Three B. Five Four C. Five Five D. Five Six
∵ △ in ABC
a＝
Five
2B
A＝2B
According to the sine theorem
sinA＝
Five
2sinB
sinA＝sin2B＝2sinBcosB
∴cosB＝
Five
Four
Therefore, B is selected;
In △ ABC, the edges of angles a, B and C are a, B, C respectively, if a = 2，b=2，sinB+cosB= 2, then the size of angle a is () A. π Two B. π Three C. π Four D. π Six
∵sinB+cosB=
2，
Qi
2sin(B+π
4)＝
Two
∴sin(B+π
4)＝1
∵ B is the inner angle of ᙽ ABC, ᙽ B = π
Four
∵a＝
2，b=2，
Qi
Two
sinA＝2
sinπ
Four
∴sinA=1
Two
∵a＜b，∴A=π
Six
Therefore, D
In △ ABC, the edges of angles a, B and C are a, B, C respectively, if a = 2，b=2，sinB+cosB= 2, then the size of angle a is () A. π Two B. π Three C. π Four D. π Six
∵sinB+cosB=
2，
Qi
2sin(B+π
4)＝
Two
∴sin(B+π
4)＝1
∵ B is the inner angle of ᙽ ABC, ᙽ B = π
Four
∵a＝
2，b=2，
Qi
Two
sinA＝2
sinπ
Four
∴sinA=1
Two
∵a＜b，∴A=π
Six
Therefore, D
In △ ABC, ∠ ABC = 45 degrees, ad is the bisector of ∠ BAC, CE ⊥ ad is at e (1) when ∠ BAC = 60 degrees, it is proved that AE + EC = ab
It is proved that DF is perpendicular to AB and F by D
Firstly, ∠ ACD = 180 ° - 45 ° - 60 ° = 75 °∠ ADC = ∠ bad + ∠ abd = 30 ° + 45 ° = 75 °
So the triangle ACD is isosceles triangle, so ad = AC
In addition, due to ∠ fad = ∠ EAC = 30 °∠ AFE = ∠ AEC = 90 °
So triangle AFD is equal to triangle AEC
So we have AF = AE, EC = FD
Then from the right triangle BFD ∠ B = 45 ° so DF = BF, so EC = BF
So AE + EC = AF + FB = ab
AE + EC = ab
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