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

Commented by MATHEMATICSAM last updated on 06/Dec/24

Find a

$$\mathrm{Find}\:{a} \\ $$

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

see also Q214228

$${see}\:{also}\:{Q}\mathrm{214228} \\ $$

Commented by MATHEMATICSAM last updated on 07/Dec/24

Sir I understood your process. can you only give me the answer of this?

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

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

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

cos θ=((3^2 +((√2))^2 −2^2 )/(2×3×(√2)))=((7(√2))/(12)) ⇒sin θ=((√(46))/(12)) a^2 =3^2 +1^2 −2×3×1 cos (45°+θ) =10−6 (cos 45° cos θ−sin 45° sin θ) =10−(6/( (√2))) (((7(√2))/(12))−((√(46))/(12))) =((13+(√(23)))/2) ⇒a=((√(2(13+(√(23)))))/2)≈2.983

$$\mathrm{cos}\:\theta=\frac{\mathrm{3}^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}×\sqrt{\mathrm{2}}}=\frac{\mathrm{7}\sqrt{\mathrm{2}}}{\mathrm{12}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{46}}}{\mathrm{12}} \\ $$$${a}^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} −\mathrm{2}×\mathrm{3}×\mathrm{1}\:\mathrm{cos}\:\left(\mathrm{45}°+\theta\right) \\ $$$$\:\:\:\:=\mathrm{10}−\mathrm{6}\:\left(\mathrm{cos}\:\mathrm{45}°\:\mathrm{cos}\:\theta−\mathrm{sin}\:\mathrm{45}°\:\mathrm{sin}\:\theta\right) \\ $$$$\:\:\:\:=\mathrm{10}−\frac{\mathrm{6}}{\:\sqrt{\mathrm{2}}}\:\left(\frac{\mathrm{7}\sqrt{\mathrm{2}}}{\mathrm{12}}−\frac{\sqrt{\mathrm{46}}}{\mathrm{12}}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{13}+\sqrt{\mathrm{23}}}{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{\sqrt{\mathrm{2}\left(\mathrm{13}+\sqrt{\mathrm{23}}\right)}}{\mathrm{2}}\approx\mathrm{2}.\mathrm{983} \\ $$

Commented by MATHEMATICSAM last updated on 07/Dec/24

Thank you sir

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

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

for general formula see Q130241

$${for}\:{general}\:{formula}\:{see}\:{Q}\mathrm{130241} \\ $$

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