Jika α + β =π/3, alpha,beta sudut sudut lancip dan tan alpha=1/6 tan beta, maka sin alpha+ sin beta = a. 1/7 √ 7+1/5√5 b. 1/10+1/5 √ 5 c. 1/4+1/5akar 5

Mapel : Matematika
Kelas : X SMA
Bab : Trigonometri

Pembahasan :
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Sin A + Sin B
= (√3)/(2√21) + (2√3)/(√21)
= (√3 + 4√3)/(2√21)
= (5√3)/(2√21)
= (15√7)/(42)
= 15/24 √7
= 5/8 √7

[tex]\displaystyle \tan a=\frac16\tan b\\\tan(\frac13\pi-b)=\frac16\tan b\\\frac{\tan\frac13\pi-\tan b}{1+\tan\frac13\pi\tan b}=\frac16\tan b\\\frac{\sqrt3-\tan b}{1+\sqrt3\tan b}=\frac16\tan b\\6\sqrt3-6\tan b=\tan b+\sqrt3\tan^2b\\\sqrt3\tan^2b+7\tan b-6\sqrt3=0\\\sqrt3\tan^2b+9\tan b-2\tan b-6\sqrt3=0\\\sqrt3\tan b(\tan b+3\sqrt3)-2(\tan b+3\sqrt3)=0\\(\sqrt3\tan b-2)(\tan b+3\sqrt3)=0\\\tan b=\frac2{\sqrt3}\vee\tan b=-3\sqrt3\\\\\tan a=\frac16\tan b\\\tan a=\frac16\cdot\frac2{\sqrt3}\\\tan a=\frac1{3\sqrt3}[/tex]
[tex]\displaystyle \sin a+\sin b=\sin(\frac13\pi-b)+\sin b\\\sin a+\sin b=\sin\frac13\pi\cos b-\cos\frac13\pi\sin b+\sin b\\\sin a+\sin b=\sin(\frac13\pi-b)+\sin b\\\sin a+\sin b=\frac12\sqrt3\cos b-\frac12\sin b+\sin b\\\sin a+\sin b=\frac12\sqrt3\cos b+\frac12\sin b\\\sin a+\sin b=\frac12\sqrt3\cdot\frac{\sqrt3}{\sqrt7}+\frac12\cdot\frac{2}{\sqrt7}\\\sin a+\sin b=\frac{3}{2\sqrt7}+\frac{2}{2\sqrt7}\\\sin a+\sin b=\frac{5}{2\sqrt7}\\\boxed{\boxed{\sin a+\sin b=\frac{5}{14}\sqrt7}}[/tex]