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Question Number 172202 by mr W last updated on 24/Jun/22

Commented by infinityaction last updated on 24/Jun/22

$$?=\sqrt{\frac{14\times 16}{14}}=4$$

Answered by Rasheed.Sindhi last updated on 24/Jun/22

$${(?)}^2=3^2+2^2-2\left(3\right)\left(2\right)\cos \left(\theta \right)=4^2+2^2-2\left(4\right)\left(2\right)\cos (\pi -\theta )$$$$9+4-12\cos \theta =16+4+16\cos \theta$$$$28\cos \theta =-7$$$$\cos \theta =-\frac{1}{4}$$$$?=\sqrt{3^2+2^2-2\left(3\right)\left(2\right)\cos \theta }$$$$=\sqrt{9+4-12(-\frac{1}{4})}$$$$=\sqrt{13+3}=4$$$$\bullet \mathrm{Cyclic}\mathrm{quadrilateral}\mathrm{has}\mathrm{its}\mathrm{opposite}$$$$\mathrm{angles}\mathrm{supplementary}.$$$$\bullet \cos (\pi -\theta )=-\cos \theta$$$$\bullet a^2=b^2+c^2-2bc\cos \theta$$

Answered by infinityaction last updated on 24/Jun/22

Commented by infinityaction last updated on 24/Jun/22

$$usethisformula$$

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