Question

in $\triangle abc$, $m\angle a=(10x)^{\circ}$, $m\angle b=(x^{2})^{\circ}$, and $m\angle c=(21x)^{\circ}$. which of the following orders the sides from longest to shortest? a ac, bc, ab b ab, ac, bc c bc, ac, ab d ab, bc, ac

Explanation:

Step1: Use the triangle - angle - sum theorem

The sum of the interior angles of a triangle is \(180^{\circ}\). So, \(m\angle A+m\angle B + m\angle C=180^{\circ}\). Substitute \(m\angle A=(10x)^{\circ}\), \(m\angle B=(x^{2})^{\circ}\), and \(m\angle C=(21x)^{\circ}\) into the equation:
\(x^{2}+10x + 21x=180\)
\(x^{2}+31x - 180=0\)
Factor the quadratic equation: \((x + 36)(x-5)=0\)
Solve for \(x\): \(x=-36\) or \(x = 5\). Since \(x>0\) (angle measures are non - negative), \(x = 5\).

Step2: Calculate the angle measures

\(m\angle A=(10x)^{\circ}\), when \(x = 5\), \(m\angle A=10\times5=50^{\circ}\)
\(m\angle B=(x^{2})^{\circ}\), when \(x = 5\), \(m\angle B=5^{2}=25^{\circ}\)
\(m\angle C=(21x)^{\circ}\), when \(x = 5\), \(m\angle C=21\times5 = 105^{\circ}\)

Step3: Use the side - angle relationship

In a triangle, the larger the angle, the longer the side opposite it.
The side opposite \(\angle A\) is \(BC\), the side opposite \(\angle B\) is \(AC\), and the side opposite \(\angle C\) is \(AB\)
Since \(m\angle C>m\angle A>m\angle B\), then \(AB>BC>AC\) (by the side - angle relationship: if \(\angle X>\angle Y>\angle Z\) in \(\triangle XYZ\), then \(YZ>XZ>XY\))

Answer:

D. \(AB, BC, AC\)