Given:
In triangle ABC, m∠A=(8x-2)°, m∠B=(2x-8)° and m∠C=(94-4x)°.
To find:
The sides of the triangle ABC in order from shortest to longest.
Solution:
In triangle ABC,
[tex]m\angle A+m\angle B+m\angle C=180^\circ[/tex] (Angle sum property)
[tex](8x-2)^\circ+(2x-8)^\circ+(94-4x)^\circ=180^\circ[/tex]
[tex](6x+84)^\circ=180^\circ[/tex]
[tex]6x+84=180[/tex]
[tex]6x=180-84[/tex]
[tex]6x=96[/tex]
Divide both sides by 6.
[tex]x=\dfrac{96}{6}[/tex]
[tex]x=16[/tex]
Now,
[tex]m\angle A=(8(16)-2)^\circ[/tex]
[tex]m\angle A=(128-2)^\circ[/tex]
[tex]m\angle A=126^\circ[/tex]
Similarly,
[tex]m\angle B=(2(16)-8)^\circ[/tex]
[tex]m\angle B=(32-8)^\circ[/tex]
[tex]m\angle B=24^\circ[/tex]
And,
[tex]m\angle C=(94-4(16))^\circ[/tex]
[tex]m\angle C=(94-64)^\circ[/tex]
[tex]m\angle C=30^\circ[/tex]
In a triangle the smaller angle has shorter opposite side and larger angle has longer opposite side.
[tex]24^\circ<30^\circ<126^\circ[/tex]
[tex]m\angle B<m\angle C<m\angle A[/tex]
[tex]AC<AB<BC[/tex]
List the sides of triangle ABC in order from shortest to longest is AC:AB:BC.
Therefore, the correct option is A.
