if the side of triangle are 16,20 and 27 respectivly,what is the lenght of the bisector of the largest angle?
Using cosine rule;
272=162+202−2×16×20 cos 2θ27^2=16^2+20^2-2×16×20\>cos \>2\theta272=162+202−2×16×20cos2θ
2θ=96.552\theta=96.552θ=96.55
θ=48.28\theta=48.28θ=48.28
202=272+162−2×16×27 Cos β20^2=27^2+16^2-2×16×27\>Cos\>\beta202=272+162−2×16×27Cosβ
β=47.38\beta=47.38β=47.38
16Sin 84.34=xSin 47.38\frac{16}{Sin\>84.34}=\frac{x}{Sin\>47.38}Sin84.3416=Sin47.38x
x=11.83x=11.83x=11.83
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