PLEASE FIND THE ANGLE I CANT GET IT
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The sine rule is a/sinA=b/sinB=c/sinC. The area of the triangle is (bcsinA)/2, or (absinC)/2, or (acsinB)/2.
We also know that C=180-(A+B)) so sinC=-sin(A+B).
sinB=(bsinA)/a from the sine rule. sin(A+B)=sinAcosB+cosAsinB.
a/sinA=c/sinC=-c/(sinAcosB+cosAsinB).
Cross-multiply: asinAcosB+acosAsinB=-csinA.
Substitute sinB=(bsinA)/a and cosB=sqrt(a^2-b^2sin^2A)/a, sinA.sqrt(a^2-b^2sin^2A)+bcosAsinA=-csinA.
Divide out sinA: sqrt(a^2-b^2sin^2A)=-bcosA-c. Square both sides: a^2-b^2sin^2A=b^2cos^2A+c^2+2bccosA. a^2-b^2sin^2A=b^2(1-sin^2A)+c^2+2bccosA/a.
a^2-b^2-c^2=2bc.cosA, so cosA=(a^2-b^2-c^2)/2bc
Substitute values for a, b and c: (256-(189+441))/714=121.59 degrees.
SinA=0.8518, so area=bcsinA/2=21*17*0.8518/2=152.05.
by Top Rated User (815k points)

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