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ParthKohli

  • one year ago

@dan815 @Kainui @ganeshie8 @Astrophysics Help me regain my creativity!

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  1. anonymous
    • one year ago
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    Try sleeping?

  2. anonymous
    • one year ago
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    try practicing art like painting or drawing

  3. ParthKohli
    • one year ago
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    Reciting a question from my memory: \(ABC\) is a triangle and it's given that \(b>c\). \(AD\) is an altitude and is given to be equal to \(\dfrac{abc}{b^2 - c^2}\). If \(C = 23^{\circ} \) then find angle \(B\).

  4. ParthKohli
    • one year ago
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    |dw:1438169365378:dw|

  5. ganeshie8
    • one year ago
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    |dw:1438169558232:dw|

  6. ParthKohli
    • one year ago
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    \[c \sin B = \frac{abc}{b^2 - c^2}\]\[\sin B = \frac{bc}{b^2 - c^2}\]

  7. ganeshie8
    • one year ago
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    Oh you re looking for numerical value of angle B is it

  8. ParthKohli
    • one year ago
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    Yes.

  9. ganeshie8
    • one year ago
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    |dw:1438169798797:dw|

  10. ganeshie8
    • one year ago
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    thats just sine law anyways

  11. Astrophysics
    • one year ago
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    it's that b>c

  12. ganeshie8
    • one year ago
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    b > c implies B > 23

  13. Astrophysics
    • one year ago
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    Yeah Ik, did you get the numerical value?

  14. ganeshie8
    • one year ago
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    no, still clueless..

  15. ganeshie8
    • one year ago
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    how this \[\frac{abc}{b^2-c^2}=b\sin{c}\] gives this \[\sin{A}=\sin^2{B}-\sin^2{C}\] ?

  16. ganeshie8
    • one year ago
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    I mean how this \[\frac{abc}{b^2-c^2}=b\sin{\color{red}{C}}\] gives this \[\sin{A}=\sin^2{B}-\sin^2{C}\] ?

  17. ganeshie8
    • one year ago
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    @amilapsn

  18. amilapsn
    • one year ago
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    \[\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}=k\]

  19. amilapsn
    • one year ago
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    \[\therefore a=k\sin{A},b=k\sin{B},c=k\sin{C}\]

  20. amilapsn
    • one year ago
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    \[\therefore \frac{abc}{b^2-c^2}=b\sin{C}\Rightarrow \sin{A}=\sin^2{B}-\sin^2{C}\]

  21. ganeshie8
    • one year ago
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    With you so far !

  22. amilapsn
    • one year ago
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    \[\Rightarrow \sin{(B-C)}=1\Rightarrow B=113^0\]

  23. ganeshie8
    • one year ago
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    how \(\sin A = \sin^2B-\sin^2C\) becomes \(\sin(B-C) = 1\) ?

  24. amilapsn
    • one year ago
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    \[\sin{A}=\sin^2{B}-\sin^2{C}=(\frac{1-\cos{2B}}{2})-(\frac{1-cos2C}{2})\]

  25. ganeshie8
    • one year ago
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    \[\begin{align}\sin{A}=\sin^2{B}-\sin^2{C}&=(\frac{1-\cos{2B}}{2})-(\frac{1-cos2C}{2})\\~\\ &=\frac{1}{2}(\cos 2C - \cos 2B)\\~\\ &= -\sin(C+B)\sin(C-B)\\~\\ &=-\sin(A)\sin(C-B)\\~\\ \end{align}\] \(\implies \sin(B-C)=1=\sin(90) \implies B = C+90 = 113\) Awesome!

  26. ganeshie8
    • one year ago
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    I did nothing, pls medal amila :)

  27. Astrophysics
    • one year ago
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    Sure thing!

  28. amilapsn
    • one year ago
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    \[\Huge\unicode{x1f63a}\]

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is replying to Can someone tell me what button the professor is hitting...

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