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Question Number 198913 by mr W last updated on 25/Oct/23

Commented by mr W last updated on 25/Oct/23

$$find\theta intermsofa,b,c.$$

Answered by ajfour last updated on 25/Oct/23

$$\sin 2\theta =\frac{a^2+b^2+c^2}{2Rabc(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})}$$

Commented by mr W last updated on 26/Oct/23

$$igot\tan \theta =\frac{4\mathrm{\Delta }}{a^2+b^2+c^2}=\frac{abc}{R(a^2+b^2+c^2)}$$$$dotheymatch?$$

Answered by mr W last updated on 26/Oct/23

Commented by mr W last updated on 26/Oct/23

$$\mathrm{\Delta }=areaof\mathrm{\Delta }ABC$$$$\frac{pc\sin \theta }{2}+\frac{qa\sin \theta }{2}+\frac{rb\sin \theta }{2}=\mathrm{\Delta }$$$$\Rightarrow (pc+qa+rb)\sin \theta =2\mathrm{\Delta }...\left(i\right)$$$$q^2=c^2+p^2-2pc\cos \theta$$$$r^2=a^2+q^2-2qa\cos \theta$$$$p^2=b^2+r^2-2rb\cos \theta$$$$\mathrm{\Sigma }:q^2+r^2+p^2=c^2+p^2+a^2+q^2+b^2+r^2-2(pc+qa+rb)\cos \theta$$$$\Rightarrow (pc+qa+rb)\cos \theta =\frac{a^2+b^2+c^2}{2}...\left(ii\right)$$$$\left(i\right)/\left(ii\right):$$$$\Rightarrow \tan \theta =\frac{4\mathrm{\Delta }}{a^2+b^2+c^2}$$$$or$$$$\Rightarrow \tan \theta =\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{a^2+b^2+c^2}$$$$◼$$

Answered by ajfour last updated on 26/Oct/23

$$\frac{\sin (\theta +\alpha )}{a}=\frac{\sin (\theta +\beta )}{b}=\frac{\sin (\theta +\gamma )}{c}=\frac{1}{2R}$$$$\frac{\sin (\theta +\alpha )}{b}=\frac{\sin \theta }{p}=\frac{\sin \alpha }{r}from△ACO$$$$\Rightarrow \frac{1}{b}\left(\frac{a}{2R}\right)=\frac{\sin \theta }{p}$$$$\frac{1}{c}\left(\frac{b}{2R}\right)=\frac{\sin \theta }{q}$$$$\frac{1}{a}\left(\frac{c}{2R}\right)=\frac{\sin \theta }{r}......set\left(I\right)$$$$$$$$2aq\cos \theta =a^2+q^2-r^2$$$$2br\cos \theta =b^2+r^2-p^2$$$$2cp\cos \theta =c^2+p^2-q^2$$$$2\cos \theta (aq+br+cp)=a^2+b^2+c^2...\left(II\right)$$$$from..set\left(I\right)$$$$p=\left(2R\sin \theta \right)\frac{b}{a}$$$$q=\left(2R\sin \theta \right)\frac{c}{b}$$$$r=\left(2R\sin \theta \right)\frac{a}{c}$$$$Nowfrom\left(II\right)$$$$4R^2\left(\sin \theta \cos \theta \right)(\frac{ca}{b}+\frac{ab}{c}+\frac{bc}{a})=a^2+b^2+c^2$$$$\left(2abcR\sin 2\theta \right)(\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{a^2})=a^2+b^2+c^2$$$$\sin 2\theta =\frac{a^2+b^2+c^2}{2abcR(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})}$$

Commented by mr W last updated on 26/Oct/23

$$nicesolutionsir!$$

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