This question was uploaded on 02/11/21 on social media accounts.
Solution:
From first two relations:
\[\alpha+\beta+\gamma=180\]
\[\alpha+\gamma=2\beta\]
\[\Rightarrow\beta=60\]
From 3rd relation:
\[\sin(\alpha)+\sin(60)+sin(\gamma)=\frac32+\frac{\sqrt3}2\]
\[\Rightarrow\sin(\alpha)+sin(\gamma)=\frac32\]
Clerly,
\[(\alpha,\gamma)=(30,90)\]
But we have to find it:
By sine rule:
\[\frac{\sqrt3}2=\frac{\sin(\alpha)}a=\frac{\sin(\gamma)}b\]
\[\sin(\alpha)=\frac{a\sqrt3}2,⠀\sin(\gamma)=\frac{b\sqrt3}2\]
Now our previous equation:
\[\frac{a\sqrt3}2+\frac{b\sqrt3}2=\frac32\]
\[\Rightarrow a+b=\sqrt3\]
\[\Rightarrow a^2+b^2=3-2ab\]
By cosine rule:
\[\cos(60)=\frac{a^2+b^2-1}{2ab}\]
\[\Rightarrow ab=\frac23\]
Now we have to find:
\[\sin(\alpha)\cdot\sin(\beta)\cdot\sin(\gamma)\]
\[=\frac{a\sqrt3}2\cdot\frac{\sqrt3}2\cdot\frac{b\sqrt3}2\]
\[=ab\cdot\frac{3\sqrt3}8\]
\[=\frac23\cdot\frac{3\sqrt3}8\]
\[=\frac{\sqrt3}4\]
