How to apply the cosine rule here?











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ABC is an equilateral triangle. D is a point on BC and AD is produced to E such that $angle EAC= angle EBC.$ Find the length of AE given that BE=5 and CE= 12










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    Its equilateral!
    – QuIcKmAtHs
    Nov 14 at 14:38










  • Hint. Observe that $E$ must lie on the circumcircle of $Delta ABC$, which would mean $angle BEC = frac{2pi}{3}$.
    – Faustus
    Nov 14 at 14:50










  • Yes, you are right but I have no idea about how to use this
    – Shivansh J
    Nov 14 at 14:52










  • @ShivanshJ I got the answer i-e the length of AE=17. But i used triangle calculator.Cosine rule is $a=sqrt{b^2+c^2-2*b*c*cos{A}}$
    – Dhamnekar Winod
    Nov 14 at 17:21












  • @ShivanshJ,you must use sine rule as well.
    – Dhamnekar Winod
    Nov 15 at 4:01

















up vote
-4
down vote

favorite












ABC is an equilateral triangle. D is a point on BC and AD is produced to E such that $angle EAC= angle EBC.$ Find the length of AE given that BE=5 and CE= 12










share|cite|improve this question




















  • 1




    Its equilateral!
    – QuIcKmAtHs
    Nov 14 at 14:38










  • Hint. Observe that $E$ must lie on the circumcircle of $Delta ABC$, which would mean $angle BEC = frac{2pi}{3}$.
    – Faustus
    Nov 14 at 14:50










  • Yes, you are right but I have no idea about how to use this
    – Shivansh J
    Nov 14 at 14:52










  • @ShivanshJ I got the answer i-e the length of AE=17. But i used triangle calculator.Cosine rule is $a=sqrt{b^2+c^2-2*b*c*cos{A}}$
    – Dhamnekar Winod
    Nov 14 at 17:21












  • @ShivanshJ,you must use sine rule as well.
    – Dhamnekar Winod
    Nov 15 at 4:01















up vote
-4
down vote

favorite









up vote
-4
down vote

favorite











ABC is an equilateral triangle. D is a point on BC and AD is produced to E such that $angle EAC= angle EBC.$ Find the length of AE given that BE=5 and CE= 12










share|cite|improve this question















ABC is an equilateral triangle. D is a point on BC and AD is produced to E such that $angle EAC= angle EBC.$ Find the length of AE given that BE=5 and CE= 12







trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 14 at 15:10









Dhamnekar Winod

358414




358414










asked Nov 14 at 14:37









Shivansh J

126




126








  • 1




    Its equilateral!
    – QuIcKmAtHs
    Nov 14 at 14:38










  • Hint. Observe that $E$ must lie on the circumcircle of $Delta ABC$, which would mean $angle BEC = frac{2pi}{3}$.
    – Faustus
    Nov 14 at 14:50










  • Yes, you are right but I have no idea about how to use this
    – Shivansh J
    Nov 14 at 14:52










  • @ShivanshJ I got the answer i-e the length of AE=17. But i used triangle calculator.Cosine rule is $a=sqrt{b^2+c^2-2*b*c*cos{A}}$
    – Dhamnekar Winod
    Nov 14 at 17:21












  • @ShivanshJ,you must use sine rule as well.
    – Dhamnekar Winod
    Nov 15 at 4:01
















  • 1




    Its equilateral!
    – QuIcKmAtHs
    Nov 14 at 14:38










  • Hint. Observe that $E$ must lie on the circumcircle of $Delta ABC$, which would mean $angle BEC = frac{2pi}{3}$.
    – Faustus
    Nov 14 at 14:50










  • Yes, you are right but I have no idea about how to use this
    – Shivansh J
    Nov 14 at 14:52










  • @ShivanshJ I got the answer i-e the length of AE=17. But i used triangle calculator.Cosine rule is $a=sqrt{b^2+c^2-2*b*c*cos{A}}$
    – Dhamnekar Winod
    Nov 14 at 17:21












  • @ShivanshJ,you must use sine rule as well.
    – Dhamnekar Winod
    Nov 15 at 4:01










1




1




Its equilateral!
– QuIcKmAtHs
Nov 14 at 14:38




Its equilateral!
– QuIcKmAtHs
Nov 14 at 14:38












Hint. Observe that $E$ must lie on the circumcircle of $Delta ABC$, which would mean $angle BEC = frac{2pi}{3}$.
– Faustus
Nov 14 at 14:50




Hint. Observe that $E$ must lie on the circumcircle of $Delta ABC$, which would mean $angle BEC = frac{2pi}{3}$.
– Faustus
Nov 14 at 14:50












Yes, you are right but I have no idea about how to use this
– Shivansh J
Nov 14 at 14:52




Yes, you are right but I have no idea about how to use this
– Shivansh J
Nov 14 at 14:52












@ShivanshJ I got the answer i-e the length of AE=17. But i used triangle calculator.Cosine rule is $a=sqrt{b^2+c^2-2*b*c*cos{A}}$
– Dhamnekar Winod
Nov 14 at 17:21






@ShivanshJ I got the answer i-e the length of AE=17. But i used triangle calculator.Cosine rule is $a=sqrt{b^2+c^2-2*b*c*cos{A}}$
– Dhamnekar Winod
Nov 14 at 17:21














@ShivanshJ,you must use sine rule as well.
– Dhamnekar Winod
Nov 15 at 4:01






@ShivanshJ,you must use sine rule as well.
– Dhamnekar Winod
Nov 15 at 4:01

















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