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In triangle ABC, the perimeter root is 2 + 1 and SINB + sinc = root 2sina, and the area of triangle ABC is 1 / 6 * Sina to find Sina
SINB + sinc = √ 2sina, and use a / Sina = B / SINB = C / sinc = 2R. Substitute B + C = √ 2a, a + B + C = √ 2 + 1. The ABC area of a = 1 triangle is 1 / 6 * Sina. Know BC = 1 / 3 and know B + C = √ 2a to get the values of B and C. use a ^ 2 = B ^ 2 + C ^ 2-2 * b * c * cosa to get Sina
In triangle ABC, angle a = 60 degrees, B = 1, s triangle ABC = root 3, then (a + B + C) / (Sina + SINB + sinc) = what
S=(1/2)bcsinA=√3
(1/2)*1*c*(√3/2)=√3
c=4
a ²= b ²+ c ²- 2bccosA=1+16-2*1*4*cos60°=13
a=√13
By sine theorem
（a+b+c)/(sinA+sinB+sinC)
=a/sinA
=√13/(√3/2)
=2√39/3
In triangle ABC, angle a = 60 degrees, B = 1, area of triangle ABC = root 3, then a + B + C / Sina + SINB + sinc =?
A = 60 °, B = 1, s = sqr3 = bccosa / 2, C = 4cosa = (b ^ 2 + C ^ 2 - A ^ 2) / (2BC) = 1 / 2A = sqrt13 (sqrt is the root) Sina = sqrt (3) / 2sinb = b * Sina / asinc = C * Sina / a (a + B + C) / (Sina + SINB + sinc) = (a + B + C) / [Sina * (1 + B / A + C / a)] =
In triangle ABC, if a = 60 degrees and a = root 3, how much is a + B-C / Sina + SINB sinc
a/sinA＝√3/sin60°＝√3÷(√3/2)＝2
According to the sine theorem, a / Sina = B / SINB = C / sinc = 2
∴a=2sinA b=2sinB c=2sinC
∴a+b-c/sinA+sinB-sinC
＝(2sinA＋2sinB－2sinC)／(sinA+sinB-sinC)＝2
In △ ABC, a = 60 °, B = 1, △ ABC area is 3, then a + B + C The value of sina + SINB + sinc is () A. 2 thirty-nine three B. 26 three three C. 8 three three D. 2 three
∵S△ABC=1
2bcsinA=1
two × one × c ×
three
2=
three
∴c=4
According to the cosine theorem: A2 = B2 + c2-2bccosa = 1 + 16-2 × one × four × one
2=13
So, a=
thirteen
According to sine theorem a
sinA=b
sinB=c
Sinc, then:
a+b+c
sinA+sinB+sinC=a
sinA=2
thirty-nine
three
So choose a
It is known that in △ ABC, Sina (SINB + CoSb) - sinc = 0, SINB + cos2c = 0, calculate the size of angles a, B and C
∵ since Sina (SINB + CoSb) - sinc = 0 ∵ sinasinb + sinacosb sin (a + b) = 0. ∵ sinasinb + sinacosb-sinacosb-cosasinb = 0. ∵ SINB (Sina COSA) = 0. Because B ∈ (0, π), SINB ≠ 0, so cosa = Sina. From a ∈ (0, π), we know a = π 4, so B
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