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a triangle with angles labeled as (5x - 27)°, (7x - 26)°, and an exteri…

Question

a triangle with angles labeled as (5x - 27)°, (7x - 26)°, and an exterior - angle (10x - 23)°. m∠a =

Explanation:

Step1: Use exterior - angle property

The exterior - angle of a triangle is equal to the sum of the two non - adjacent interior angles. In \(\triangle ABC\), the exterior angle at \(C\) is \((10x - 23)^{\circ}\), and the two non - adjacent interior angles are \((5x - 27)^{\circ}\) and \((7x - 26)^{\circ}\). So, \((5x - 27)+(7x - 26)=10x - 23\).

Step2: Simplify the left - hand side of the equation

Combine like terms: \((5x+7x)+(- 27-26)=12x - 53\). So the equation becomes \(12x - 53 = 10x - 23\).

Step3: Solve for \(x\)

Subtract \(10x\) from both sides: \(12x-10x - 53=10x-10x - 23\), which simplifies to \(2x-53=-23\). Then add 53 to both sides: \(2x-53 + 53=-23 + 53\), so \(2x = 30\). Divide both sides by 2: \(x=\frac{30}{2}=15\).

Step4: Find \(m\angle A\)

Since \(m\angle A=(5x - 27)^{\circ}\), substitute \(x = 15\) into the expression. \(m\angle A=5\times15-27=75 - 27=48^{\circ}\).

Answer:

\(48^{\circ}\)