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Thread: sine triangle

  1. #1
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    sine triangle

    Given a triangle PQR with sides p ,q and r and opposite angles c , b and a respectively . Show that the area of PQR , A is given by

    A= ( p^2 sin b sin a ) /2 sin ( b+a )
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  2. #2
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    Please reread what you posted.
    Does that maks any sense to you?
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  3. #3
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    Hello, mathaddict!

    Your labels are confusing . . . I'll revise them.


    Given a triangle $\displaystyle ABC$ with opposite sides $\displaystyle a,b,c$ respectively,

    show that the area of $\displaystyle ABC$ is given by: .$\displaystyle \text{Area} \;=\;\frac{c^2\sin A\sin B}{2\sin(A+B)} $
    I will assume that we know: .$\displaystyle \text{Area} \;=\;\tfrac{1}{2}\,bc\sin A$ .[1]
    . .
    (the area is one-half the product of two sides and the sine of the included angle)


    Multiply [1] by $\displaystyle \frac{\sin B}{\sin B}\!:\quad \text{Area} \;=\;\tfrac{1}{2}\,bc\,\sin A\cdot\frac{\sin B}{\sin B} \;=\;\tfrac{1}{2}\,c\,\sin A\sin B\cdot\frac{b}{\sin B}$ .[2]


    From the Law of Sines: .$\displaystyle \frac{b}{\sin B} \:=\:\frac{c}{\sin C}$

    Substitute into [2]: .$\displaystyle \text{Area} \;=\;\tfrac{1}{2}\,c\,\sin A\sin B\cdot\frac{c}{\sin C} \;=\;\frac{c^2\sin A\sin B}{2\sin C}$ .[3]


    Since $\displaystyle A + B + C \:=\:180^o$, then: .$\displaystyle C \:=\:180-(A+B)$

    . . and: .$\displaystyle \sin C \:=\:\sin[180 -(A+B)] \;=\;\sin(A+B) $


    Substitute into [3]: . $\displaystyle \text{Area} \;=\;\frac{c^2\sin A\sin B}{\sin(A+B)} \quad\hdots$ ta-DAA!

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  4. #4
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    Thanks . However , sorry for that , i think it is the textbook's mistake and i am not good in Latex ...
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