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Question Number 214351 by MATHEMATICSAM last updated on 06/Dec/24

Commented by MATHEMATICSAM last updated on 06/Dec/24

$$\mathrm{Find}a$$

Commented by mr W last updated on 06/Dec/24

$$seealsoQ214228$$

Commented by MATHEMATICSAM last updated on 07/Dec/24

$$\mathrm{Sir}\mathrm{I}\mathrm{understood}\mathrm{your}\mathrm{process}.\mathrm{can}\mathrm{you}$$$$\mathrm{only}\mathrm{give}\mathrm{me}\mathrm{the}\mathrm{answer}\mathrm{of}\mathrm{this}?$$

Answered by mr W last updated on 07/Dec/24

Commented by mr W last updated on 07/Dec/24

$$\cos \theta =\frac{3^2+{\left(\sqrt{2}\right)}^2-2^2}{2\times 3\times \sqrt{2}}=\frac{7\sqrt{2}}{12}$$$$\Rightarrow \sin \theta =\frac{\sqrt{46}}{12}$$$$a^2=3^2+1^2-2\times 3\times 1\cos (45°+\theta )$$$$=10-6(\cos 45°\cos \theta -\sin 45°\sin \theta )$$$$=10-\frac{6}{\sqrt{2}}(\frac{7\sqrt{2}}{12}-\frac{\sqrt{46}}{12})$$$$=\frac{13+\sqrt{23}}{2}$$$$\Rightarrow a=\frac{\sqrt{2(13+\sqrt{23})}}{2}\approx 2.983$$

Commented by MATHEMATICSAM last updated on 07/Dec/24

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}$$

Commented by mr W last updated on 07/Dec/24

$$forgeneralformulaseeQ130241$$

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