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Question Number 84441 by Power last updated on 13/Mar/20

Commented by Power last updated on 13/Mar/20

sir please solution

$$\mathrm{sir}\:\mathrm{please}\:\mathrm{solution} \\ $$

Answered by mr W last updated on 13/Mar/20

Commented by mr W last updated on 13/Mar/20

only for rhombus α=110°. otherwise α is not unique.

$${only}\:{for}\:{rhombus}\:\alpha=\mathrm{110}°.\:{otherwise} \\ $$$$\alpha\:{is}\:{not}\:{unique}. \\ $$

Commented by mr W last updated on 13/Mar/20

((sin (40−β))/(2a))=((sin β)/a)=((sin 40)/(DE)) ((sin 40)/(tan β))−cos 40=2 ⇒tan β=((sin 40)/(2+cos 40)) DE=((a sin 40)/(sin β)) DF=DE−FE=((a sin 40)/(sin β))−2a cos β ((DF)/(sin (γ−β)))=((DC)/(sin γ)) ((sin (γ−β))/(sin γ))=((DF)/(DC))=((sin 40)/(sin β))−2 cos β cos β−((sin β)/(tan γ))=((sin 40)/(sin β))−2 cos β 3 cos β−((sin 40)/(sin β))=((sin β)/(tan γ)) (1/(tan γ))=(3/(tan β))−sin 40(1+(1/(tan^2 β))) (1/(tan γ))=((1−cos 40)/(sin 40)) tan γ=((sin 40)/(1−cos 40)) ⇒γ=70° ⇒α=180−γ=110°

$$\frac{\mathrm{sin}\:\left(\mathrm{40}−\beta\right)}{\mathrm{2}{a}}=\frac{\mathrm{sin}\:\beta}{{a}}=\frac{\mathrm{sin}\:\mathrm{40}}{{DE}} \\ $$$$\frac{\mathrm{sin}\:\mathrm{40}}{\mathrm{tan}\:\beta}−\mathrm{cos}\:\mathrm{40}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{tan}\:\beta=\frac{\mathrm{sin}\:\mathrm{40}}{\mathrm{2}+\mathrm{cos}\:\mathrm{40}} \\ $$$${DE}=\frac{{a}\:\mathrm{sin}\:\mathrm{40}}{\mathrm{sin}\:\beta} \\ $$$${DF}={DE}−{FE}=\frac{{a}\:\mathrm{sin}\:\mathrm{40}}{\mathrm{sin}\:\beta}−\mathrm{2}{a}\:\mathrm{cos}\:\beta \\ $$$$\frac{{DF}}{\mathrm{sin}\:\left(\gamma−\beta\right)}=\frac{{DC}}{\mathrm{sin}\:\gamma} \\ $$$$\frac{\mathrm{sin}\:\left(\gamma−\beta\right)}{\mathrm{sin}\:\gamma}=\frac{{DF}}{{DC}}=\frac{\mathrm{sin}\:\mathrm{40}}{\mathrm{sin}\:\beta}−\mathrm{2}\:\mathrm{cos}\:\beta \\ $$$$\mathrm{cos}\:\beta−\frac{\mathrm{sin}\:\beta}{\mathrm{tan}\:\gamma}=\frac{\mathrm{sin}\:\mathrm{40}}{\mathrm{sin}\:\beta}−\mathrm{2}\:\mathrm{cos}\:\beta \\ $$$$\mathrm{3}\:\mathrm{cos}\:\beta−\frac{\mathrm{sin}\:\mathrm{40}}{\mathrm{sin}\:\beta}=\frac{\mathrm{sin}\:\beta}{\mathrm{tan}\:\gamma} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\gamma}=\frac{\mathrm{3}}{\mathrm{tan}\:\beta}−\mathrm{sin}\:\mathrm{40}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\beta}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\gamma}=\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{40}}{\mathrm{sin}\:\mathrm{40}} \\ $$$$\mathrm{tan}\:\gamma=\frac{\mathrm{sin}\:\mathrm{40}}{\mathrm{1}−\mathrm{cos}\:\mathrm{40}} \\ $$$$\Rightarrow\gamma=\mathrm{70}° \\ $$$$\Rightarrow\alpha=\mathrm{180}−\gamma=\mathrm{110}° \\ $$

Commented by Power last updated on 13/Mar/20

thank you sir

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

Commented by Power last updated on 13/Mar/20

Commented by Power last updated on 13/Mar/20

romb ?

$$\mathrm{romb}\:? \\ $$

Commented by mr W last updated on 13/Mar/20

what′s your concrete question?

$${what}'{s}\:{your}\:{concrete}\:{question}? \\ $$

Commented by Power last updated on 13/Mar/20

for rhombus or parallelogram

$$\mathrm{for}\:\mathrm{rhombus}\:\mathrm{or}\:\mathrm{parallelogram} \\ $$

Commented by Power last updated on 13/Mar/20

is it true that in a parallelogram, then angle will be changed.

$$\mathrm{is}\:\mathrm{it}\:\mathrm{true}\:\mathrm{that}\:\mathrm{in}\:\:\mathrm{a}\:\:\mathrm{parallelogram},\:\mathrm{then}\:\mathrm{angle}\:\mathrm{will}\:\mathrm{be}\:\mathrm{changed}. \\ $$

Commented by Power last updated on 13/Mar/20

thanks

$$\mathrm{thanks} \\ $$

Commented by Power last updated on 13/Mar/20

work with the alpha of 40°

$$\mathrm{work}\:\mathrm{with}\:\mathrm{the}\:\mathrm{alpha}\:\mathrm{of}\:\mathrm{40}° \\ $$

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