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9. what is the base angle of isosceles if the vertex is 132 degrees

Question

  1. what is the base angle of isosceles if the vertex is 132 degrees

Explanation:

Step1: Recall the property of isosceles triangle

In an isosceles triangle, the sum of interior angles is \(180^{\circ}\), and the base angles are equal. Let the base angles be \(x\) each and the vertex angle \(y = 132^{\circ}\).

Step2: Use the angle - sum formula

We know that \(x + x+y=180^{\circ}\) (sum of angles in a triangle). Substitute \(y = 132^{\circ}\) into the formula: \(2x+132^{\circ}=180^{\circ}\).

Step3: Solve for \(x\)

First, subtract \(132^{\circ}\) from both sides: \(2x=180^{\circ}- 132^{\circ}=48^{\circ}\). Then divide both sides by \(2\): \(x=\frac{48^{\circ}}{2}=24^{\circ}\).

Answer:

\(24^{\circ}\)