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 )
Hello, mathaddict!
Your labels are confusing . . . I'll revise them.
I will assume that we know: .$\displaystyle \text{Area} \;=\;\tfrac{1}{2}\,bc\sin A$ .[1]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)} $
. . (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!