The length of the sides of a triangle are 4.2, 5.3 and 7.6. Calculate the size of the largest angle of the triangle to 1 dp. How would I know this without calculating the area using 1/2 a b sin(c)????
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Originally Posted by Mukilab The length of the sides of a triangle are 4.2, 5.3 and 7.6. Calculate the size of the largest angle of the triangle to 1 dp. How would I know this without calculating the area using 1/2 a b sin(c)???? use the law of cosines ... $\displaystyle a^2 = b^2+c^2 - 2bc\cos{A} $ rearrange ... $\displaystyle \cos{A} = \frac{b^2+c^2-a^2}{2bc} $
Originally Posted by Mukilab The length of the sides of a triangle are 4.2, 5.3 and 7.6. Calculate the size of the largest angle of the triangle to 1 dp. How would I know this without calculating the area using 1/2 a b sin(c)???? Hi Mukilab, Use the law of cosines. Let C be the largest angle (opposite 7.6) $\displaystyle c^2=a^2+b^2-2ab \cos C$
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