Question

list the sides of \\( \triangle abc \\) in order from longest to shortest if the angles of \\( \triangle abc \\) have the indicated measures.

1. sides:

\\( m \angle a = ( 5 x + 2 ) ^ { \circ } \\), \\( m \angle b = ( 6 x - 10 ) ^ { \circ } \\), and \\( m \angle c = ( x + 20 ) ^ { \circ } \\).
lets pretend \\( x = 10 \\). draw a picture of a triangle, and plug it in
and see which is the largest!

1. sides:

\\( m \angle a = ( x + 16 ) ^ { \circ } \\), \\( m \angle b = ( x ) ^ { \circ } \\), and \\( m \angle c = ( x + 29 ) ^ { \circ } \\).
lets pretend \\( x = 10 \\). draw a picture of a triangle, and plug it in
and see which is the largest!

Explanation:

Step1: Calculate angles for problem 13

Given \(x = 10\), \(m\angle A=(5x + 2)^{\circ}\), \(m\angle B=(6x-10)^{\circ}\), \(m\angle C=(x + 20)^{\circ}\).
\(m\angle A=(5\times10 + 2)^{\circ}=52^{\circ}\), \(m\angle B=(6\times10-10)^{\circ}=50^{\circ}\), \(m\angle C=(10 + 20)^{\circ}=30^{\circ}\).
In a triangle, the larger the angle, the longer the opposite side.
Opposite sides: \(BC\) opposite \(\angle A\), \(AC\) opposite \(\angle B\), \(AB\) opposite \(\angle C\).
Since \(52^{\circ}>50^{\circ}>30^{\circ}\), so \(BC>AC>AB\).

Step2: Calculate angles for problem 14

Given \(x = 10\), \(m\angle A=(x + 16)^{\circ}\), \(m\angle B=(x)^{\circ}\), \(m\angle C=(x + 29)^{\circ}\).
\(m\angle A=(10 + 16)^{\circ}=26^{\circ}\), \(m\angle B=10^{\circ}\), \(m\angle C=(10 + 29)^{\circ}=39^{\circ}\).
Opposite sides: \(BC\) opposite \(\angle A\), \(AC\) opposite \(\angle B\), \(AB\) opposite \(\angle C\).
Since \(39^{\circ}>26^{\circ}>10^{\circ}\), so \(AB>BC>AC\).

Answer:

1. \(BC>AC>AB\)
2. \(AB>BC>AC\)