Question-115798

Question Number 115798 by aye48 last updated on 28/Sep/20

Commented by bemath last updated on 29/Sep/20

sin α.cos β = (9/(10)) ⇒2sin α.cos β = (9/5) ⇒sin (α+β)+sin (α−β) = (9/5) ⇒ sin (α+β) = (9/5) − (4/5) = 1

$$\mathrm{sin}\:\alpha.\mathrm{cos}\:\beta\:=\:\frac{\mathrm{9}}{\mathrm{10}} \\ $$$$\Rightarrow\mathrm{2sin}\:\alpha.\mathrm{cos}\:\beta\:=\:\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\alpha+\beta\right)+\mathrm{sin}\:\left(\alpha−\beta\right)\:=\:\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\left(\alpha+\beta\right)\:=\:\frac{\mathrm{9}}{\mathrm{5}}\:−\:\frac{\mathrm{4}}{\mathrm{5}}\:=\:\mathrm{1} \\ $$

Answered by TANMAY PANACEA last updated on 29/Sep/20

sinαcosβ−cosαsinβ=(4/5) cosαsinβ=(9/(10))−(4/5)=(1/(10)) sin(α−β)=(4/5)=sin53^o sin(α+β)=sinαcosβ+cosαsinβ=(9/(10))+(1/(10))=sin90^o =1 α+β=90 α−β=53 α=((90+53)/2)=((143)/2) β=((90−53)/2)=((37)/2) ((tanα)/(tanβ))=((sinαcosβ)/(sinβcosα))=(9/(10))×((10)/1)=9

$${sin}\alpha{cos}\beta−{cos}\alpha{sin}\beta=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${cos}\alpha{sin}\beta=\frac{\mathrm{9}}{\mathrm{10}}−\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{10}} \\ $$$${sin}\left(\alpha−\beta\right)=\frac{\mathrm{4}}{\mathrm{5}}={sin}\mathrm{53}^{{o}} \\ $$$${sin}\left(\alpha+\beta\right)={sin}\alpha{cos}\beta+{cos}\alpha{sin}\beta=\frac{\mathrm{9}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{10}}={sin}\mathrm{90}^{{o}} =\mathrm{1} \\ $$$$\alpha+\beta=\mathrm{90} \\ $$$$\alpha−\beta=\mathrm{53} \\ $$$$\alpha=\frac{\mathrm{90}+\mathrm{53}}{\mathrm{2}}=\frac{\mathrm{143}}{\mathrm{2}} \\ $$$$\beta=\frac{\mathrm{90}−\mathrm{53}}{\mathrm{2}}=\frac{\mathrm{37}}{\mathrm{2}} \\ $$$$\frac{{tan}\alpha}{{tan}\beta}=\frac{{sin}\alpha{cos}\beta}{{sin}\beta{cos}\alpha}=\frac{\mathrm{9}}{\mathrm{10}}×\frac{\mathrm{10}}{\mathrm{1}}=\mathrm{9} \\ $$

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Question-115798

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