In a triangle ABC,m angle A, m angle B, m ang | vedclass

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(D) Given that $\angle A, \angle B, \angle C$ are in $A$.$P$.,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,which im

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$5-\sqrt{6}$

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(D) Given that $\angle A, \angle B, \angle C$ are in $A$.$P$.,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,which implies $B = 60^{\circ}$. Let the sides opposite to $A, B, C$ be $a, b, c$ respectively. Given $a = 10$ and $b = 9$ (the two larger sides). Using the Law of Cosines: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$. Substituting the values: $\cos 60^{\circ} = \frac{10^2 + c^2 - 9^2}{2(10)c}$. $\frac{1}{2} = \frac{100 + c^2 - 81}{20c} \Rightarrow 10c = c^2 + 19$. $c^2 - 10c + 19 = 0$. Using the quadratic formula: $c = \frac{10 \pm \sqrt{100 - 76}}{2} = \frac{10 \pm \sqrt{24}}{2} = 5 \pm \sqrt{6}$. Since $b=9$ is the second largest side,$c$ must be the smallest side. Thus,$c = 5 - \sqrt{6}$.

In a triangle $ABC$,$m \angle A, m \angle B, m \angle C$ are in $A$.$P$. and the lengths of the two larger sides are $10$ units and $9$ units respectively. Then the length (in units) of the third side is:

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