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Find-the-type-of-triangle-such-that-the-following-relationship-holds-between-its-angles-tan-B-tan-C-tan-2-B-C-2-

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Question Number 215349 by mnjuly1970 last updated on 03/Jan/25
Find the type of triangle such that the following relationship holds between its angles. tan (B)tan(C)= tan^2 (((B+C)/2) ) ■
$$ \\ $$$$\:\:\:\:{Find}\:{the}\:{type}\:{of}\:{triangle} \\ $$$$\:\:\:\:{such}\:{that}\:{the}\:{following} \\ $$$$\:\:\:\:\:{relationship}\:{holds}\:{between}\: \\ $$$${its}\:{angles}. \\ $$$$\:\:\:\:\:\: \\ $$$$\:{tan}\:\left({B}\right){tan}\left({C}\right)=\:{tan}^{\mathrm{2}} \left(\frac{{B}+{C}}{\mathrm{2}}\:\right)\:\:\:\:\blacksquare \\ $$$$ \\ $$
Answered by MrGaster last updated on 04/Jan/25
tan(B)tan(C)=tan^2 (((B+C)/2)) =tan^2 (((π−A)/2)) =cot^2 ((A/2)) tan(B)tan(C)=(((cos((A/2)))/(sin((A/2)))))^2 =(((1+cos(A))/(1−cos(A)))) tan(B)tan(C)=(((cos((A/2)))/(sin((A/2))^2 )))^2 =(((1+cos(A))/(1−cos(A)))) tan(B)tan(C)=(((1+cos(B+C))/(1−cos(B+C)))) =(((1+(cos(B)cos(C)−sin(B)sin(C))/(1−(cos(B)cos(C)−sin(B)sin(C))))) =(((cos(B)cos(C)+sin(B)sin(C)+sin(B)sin(C)))/((cos(B)cos(A)+sin(B)sin(C)−sin(B)sin(C)))) = (((cos(B−C)+sin(B)sin(C))/(cos(B−C)))) =1+tan(B)tan(C) tan(B)tan(C)=1+tan(B)tan(C) tan(B)tan(C)=tan^2 (((B+C)/2)) =tan^2 (((π−A)/2)) =cot^2 ((A/2)) tan(B)tan(C)=1 tan(B)tan(C)=tan((π/4))tan((π/4)) B=C=(π/4) A=π−(B+C)=(π/2) ∴△ABC is a right isosecles triangle
$$\mathrm{tan}\left({B}\right)\mathrm{tan}\left({C}\right)=\mathrm{tan}^{\mathrm{2}} \left(\frac{{B}+{C}}{\mathrm{2}}\right) \\ $$$$=\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi−{A}}{\mathrm{2}}\right) \\ $$$$=\mathrm{cot}^{\mathrm{2}} \left(\frac{{A}}{\mathrm{2}}\right) \\ $$$$\mathrm{tan}\left({B}\right)\mathrm{tan}\left({C}\right)=\left(\frac{\mathrm{cos}\left(\frac{{A}}{\mathrm{2}}\right)}{\mathrm{sin}\left(\frac{{A}}{\mathrm{2}}\right)}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{1}+\mathrm{cos}\left({A}\right)}{\mathrm{1}−\mathrm{cos}\left({A}\right)}\right) \\ $$$$\mathrm{tan}\left({B}\right)\mathrm{tan}\left({C}\right)=\left(\frac{\mathrm{cos}\left(\frac{{A}}{\mathrm{2}}\right)}{\mathrm{sin}\left(\frac{{A}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{1}+\mathrm{cos}\left({A}\right)}{\mathrm{1}−\mathrm{cos}\left({A}\right)}\right) \\ $$$$\mathrm{tan}\left({B}\right)\mathrm{tan}\left({C}\right)=\left(\frac{\mathrm{1}+\mathrm{cos}\left({B}+{C}\right)}{\mathrm{1}−\mathrm{cos}\left({B}+{C}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{1}+\left(\mathrm{cos}\left({B}\right)\mathrm{cos}\left({C}\right)−\mathrm{sin}\left({B}\right)\mathrm{sin}\left({C}\right)\right.}{\mathrm{1}−\left(\mathrm{cos}\left({B}\right)\mathrm{cos}\left({C}\right)−\mathrm{sin}\left({B}\right)\mathrm{sin}\left({C}\right)\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{\left.\mathrm{cos}\left({B}\right)\mathrm{cos}\left({C}\right)+\mathrm{sin}\left({B}\right)\mathrm{sin}\left({C}\right)+\mathrm{sin}\left({B}\right)\mathrm{sin}\left({C}\right)\right)}{\left(\mathrm{cos}\left({B}\right)\mathrm{cos}\left({A}\right)+\mathrm{sin}\left({B}\right)\mathrm{sin}\left({C}\right)−\mathrm{sin}\left({B}\right)\mathrm{sin}\left({C}\right)\right.}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\frac{\mathrm{cos}\left({B}−{C}\right)+\mathrm{sin}\left({B}\right)\mathrm{sin}\left({C}\right)}{\mathrm{cos}\left({B}−{C}\right)}\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\mathrm{tan}\left({B}\right)\mathrm{tan}\left({C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{tan}\left({B}\right)\mathrm{tan}\left({C}\right)=\mathrm{1}+\mathrm{tan}\left({B}\right)\mathrm{tan}\left({C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{tan}\left({B}\right)\mathrm{tan}\left({C}\right)=\mathrm{tan}^{\mathrm{2}} \left(\frac{{B}+{C}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi−{A}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{cot}^{\mathrm{2}} \left(\frac{{A}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{tan}\left({B}\right)\mathrm{tan}\left({C}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{tan}\left({B}\right)\mathrm{tan}\left({C}\right)=\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}\right)\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{B}={C}=\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{A}=\pi−\left({B}+{C}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\therefore\bigtriangleup{ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{right}\:\mathrm{isosecles}\:\mathrm{triangle} \\ $$

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