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The edges corresponding to ABC are the edges corresponding to angle ABC, acosb + bcosa= The process is detailed
Cosine theorem CoSb = (A & sup2; + C & sup2; - B & sup2;) / 2Ac
cosA=(b²+c²-a²)/2bc
∴acosB+bcosA=c/2
In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, and acosb = bcosa + 3 / 5C
(1) Find the value of Tana / tanb
(2) Finding the maximum of Tan (a-b)
(3) If the perimeter of △ ABC is 5 + 3 root sign 5, when Tan (a-b) reaches the maximum, calculate the area of △ ABC
1. A / Sina = B / SINB = C / sinc, so C = asinc / Sina = asin (a + b) / Sina = a (sinacosb + cosasinb) / Sina. Substituting this formula into the known formula, we get that asinacosb acosasinb = 2bsinacosa, because a / Sina = B / SINB, so sinacosb = 3sinbcosa, that is, Tana / tanb = 3
2、tan(A-B)=(tanA-tanB)/(1+tanAtanB)=(tanA/tanB-1)/(ctanB+tanA)=2/(ctanB+3tanB)
When ctanb = 3tanb, the maximum value of Tan (a-b) is 3
In this case, tanb = root 3, Tana = root 3, so the shape of triangle ABC is a right triangle of 60 degrees, 30 degrees and 90 degrees
If acosb + bcosa = csinc, B ^ 2 + C ^ 2-A ^ 2 = √ 3 BC, then ∠ B =?
If acosb + bcosa = csinc
From the sine theorem a = 2rsina, B = 2rsinb, C = 2rsinc
Substituting: sinacosb + sinbcosa = sincsinc
sin（A+B）=sinCsinC
sinC=sinCsinC
Sinc = 1, or sinc = 0
So C = 90
b^2+c^2-a^2=√3 bc
According to the cosine theorem, cosa = √ 3 / 2, a = 30, so B = 60,
In △ ABC, a, B and C are the opposite sides of angles a, B and C respectively, and S is the area of △ ABC. If acosb + bcosa = csinc and s △ ABC = 14 (B2 + c2-a2), then angle B is equal to ()
A. 30°B. 45°C. 60°D. 90°
According to the sine theorem, acosb + bcosa = 2rsinacosb + 2rsinbcosa = 2rsin (a + b) = 2rsinc = 2rsinc · sinc sinc = 1, C = 90 °, s = 12ab = 14 (B2 + c2-a2), the solution is a = B, so ∠ B = 45 °. So B is chosen
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