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Question Number 164339 by mnjuly1970 last updated on 16/Jan/22

In AB^Δ C : cos^( 2) (A )+ cos^( 2) (B )+ cos^( 2) ( C )=1 . Prove that AB^Δ C is right angled. −−−−−−−−

$$ \\ $$$$\:\:\:\:{In}\:\:{A}\overset{\Delta} {{B}C}\:\:\::\:\:\:{cos}^{\:\mathrm{2}} \left({A}\:\right)+\:{cos}^{\:\mathrm{2}} \left({B}\:\right)+\:{cos}^{\:\mathrm{2}} \left(\:{C}\:\right)=\mathrm{1}\:\:. \\ $$$$\:\:\:\:\:\:\:\:{Prove}\:{that}\:\:{A}\overset{\Delta} {{B}C}\:\:\:{is}\:\:\:{right}\:{angled}. \\ $$$$\:\:\:\:\:\:−−−−−−−− \\ $$$$\:\:\:\:\:\: \\ $$

Answered by mindispower last updated on 17/Jan/22

cos^2 (c)−1=−sin^2 (A+B) =−sin^2 (A)cos^2 (B)−sin^2 (B)cos^2 (A)−((sin(2A)sin(2B))/2) cos^2 (A)+cos^2 (B)−sin^2 (A)cos^2 (B)−sin^2 (B)cos^2 (A)−=0 ⇔2cos^2 (A)cos^2 (B)−2sin(A)sin(B)cos(A)cos(B)=0 cos(A)cos(B)(cos(A+B))=0 cos(A)=0⇒A=(π/2),Or cos(B)=0⇒B=(π/2) A+B=(π/2)⇒C=π−(A+B)=(π/2)

$${cos}^{\mathrm{2}} \left({c}\right)−\mathrm{1}=−{sin}^{\mathrm{2}} \left({A}+{B}\right) \\ $$$$=−{sin}^{\mathrm{2}} \left({A}\right){cos}^{\mathrm{2}} \left({B}\right)−{sin}^{\mathrm{2}} \left({B}\right){cos}^{\mathrm{2}} \left({A}\right)−\frac{{sin}\left(\mathrm{2}{A}\right){sin}\left(\mathrm{2}{B}\right)}{\mathrm{2}} \\ $$$${cos}^{\mathrm{2}} \left({A}\right)+{cos}^{\mathrm{2}} \left({B}\right)−{sin}^{\mathrm{2}} \left({A}\right){cos}^{\mathrm{2}} \left({B}\right)−{sin}^{\mathrm{2}} \left({B}\right){cos}^{\mathrm{2}} \left({A}\right)−=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{2}{cos}^{\mathrm{2}} \left({A}\right){cos}^{\mathrm{2}} \left({B}\right)−\mathrm{2}{sin}\left({A}\right){sin}\left({B}\right){cos}\left({A}\right){cos}\left({B}\right)=\mathrm{0} \\ $$$${cos}\left({A}\right){cos}\left({B}\right)\left({cos}\left({A}+{B}\right)\right)=\mathrm{0} \\ $$$${cos}\left({A}\right)=\mathrm{0}\Rightarrow{A}=\frac{\pi}{\mathrm{2}},{Or}\:{cos}\left({B}\right)=\mathrm{0}\Rightarrow{B}=\frac{\pi}{\mathrm{2}} \\ $$$${A}+{B}=\frac{\pi}{\mathrm{2}}\Rightarrow{C}=\pi−\left({A}+{B}\right)=\frac{\pi}{\mathrm{2}} \\ $$

Commented by mnjuly1970 last updated on 18/Jan/22

In AB^Δ C : cos^( 2) (A )+ cos^( 2) (B )+ cos^( 2) ( C )=1 . Prove that AB^Δ C is right angled. −−−−−−−− thank you very much sir... grateful

$$ \\ $$$$\:\:\:\:{In}\:\:{A}\overset{\Delta} {{B}C}\:\:\::\:\:\:{cos}^{\:\mathrm{2}} \left({A}\:\right)+\:{cos}^{\:\mathrm{2}} \left({B}\:\right)+\:{cos}^{\:\mathrm{2}} \left(\:{C}\:\right)=\mathrm{1}\:\:. \\ $$$$\:\:\:\:\:\:\:\:{Prove}\:{that}\:\:{A}\overset{\Delta} {{B}C}\:\:\:{is}\:\:\:{right}\:{angled}. \\ $$$$\:\:\:\:\:\:−−−−−−−− \\ $$$$\:{thank}\:{you}\:{very}\:{much}\:{sir}... \\ $$$$\:\:{grateful}\:\:\:\:\:\:\:\:\:\: \\ $$

Commented by mindispower last updated on 19/Jan/22

wihe Pleasur have a nice day

$${wihe}\:{Pleasur}\:{have}\:{a}\:{nice}\:{day} \\ $$

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