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Question Number 217631 by mnjuly1970 last updated on 17/Mar/25

Commented by mnjuly1970 last updated on 17/Mar/25

$$yes.{that}^{\prime }sright.thanksalotsir$$

Answered by maths2 last updated on 17/Mar/25

$$a=\frac{1}{a^{\prime }}=\frac{1}{sin\left(10\right)};b=\frac{1}{sin\left(70\right)}=\frac{1}{b^{\prime }}$$$$c=\frac{1}{sin\left(50\right)}=\frac{1}{c^{\prime }}$$$$a^2+b^2+c^2=\frac{1}{{sin}^2\left(10\right)}+\frac{1}{{sin}^2\left(70\right)}+\frac{1}{{sin}^2\left(50\right)}$$$$sin(10.3)=sin\left(30\right);sin(3.70)=sin\left(210\right)=-sin\left(30\right)$$$$sin(3.50)=sin\left(150\right)=sin\left(30\right)$$$$sin\left(10\right);sin\left(50\right);-sin\left(70\right)rootof$$$$sin\left(3x\right)=\frac{1}{2}\iff -{sin}^3\left(x\right)+3sin\left(x\right)(1-{sin}^2\left(x\right))=\frac{1}{2}$$$$\iff {sin}^3\left(x\right)-\frac{3}{4}sin\left(x\right)+\frac{1}{8}=0$$$$\frac{1}{a^{\prime 2}}+\frac{1}{b^{\prime 2}}+\frac{1}{c^{\prime 2}}=\frac{{(a^{\prime }{b}^{\prime }+b^{\prime }{c}^{\prime }+c^{\prime }{a}^{\prime })}^2-2a^{\prime }{b}^{\prime }{c}^{\prime }(a^{\prime }+b^{\prime }+c^{\prime })}{{\left(a^{\prime }{b}^{\prime }{c}^{\prime }\right)}^2}$$$$=\frac{{(-\frac{3}{4})}^2}{{(-\frac{1}{8})}^2}=36correcteanswerisC$$$$$$$$$$

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