Question

1. m∠a = ______ in △abc, m∠a = (3x + 15)°, m∠b = 4x° and m∠c = (6x - 30)°. what type of triangle is △abc? m∠b = ____ m∠c = ____ 4. m∠a = ____ in △abc, m∠a = (3x + 4)°, m∠b = (2x - 4)° and m∠c = (4x - 18)°. what type of triangle is △abc? m∠b = ____ m∠c = ______

Explanation:

Step1: Use the triangle - angle sum theorem

The sum of the interior angles of a triangle is \(180^{\circ}\). For \(\triangle ABC\) with \(m\angle A=(3x + 15)^{\circ}\), \(m\angle B = 4x^{\circ}\), and \(m\angle C=(6x-30)^{\circ}\), we have the equation \((3x + 15)+4x+(6x-30)=180\).
Simplify the left - hand side: \(3x+15 + 4x+6x-30=(3x + 4x+6x)+(15 - 30)=13x-15\).
So, \(13x-15 = 180\).

Step2: Solve for \(x\)

Add \(15\) to both sides of the equation \(13x-15 = 180\): \(13x=180 + 15=195\).
Divide both sides by \(13\): \(x=\frac{195}{13}=15\).

Step3: Find the measures of the angles

• \(m\angle A=(3x + 15)^{\circ}\). Substitute \(x = 15\): \(m\angle A=(3\times15 + 15)^{\circ}=(45+15)^{\circ}=60^{\circ}\).
• \(m\angle B=4x^{\circ}\). Substitute \(x = 15\): \(m\angle B=4\times15^{\circ}=60^{\circ}\).
• \(m\angle C=(6x-30)^{\circ}\). Substitute \(x = 15\): \(m\angle C=(6\times15-30)^{\circ}=(90 - 30)^{\circ}=60^{\circ}\).

Since \(m\angle A=m\angle B=m\angle C = 60^{\circ}\), \(\triangle ABC\) is an equilateral triangle.

For the second problem:

Step1: Use the triangle - angle sum theorem

The sum of the interior angles of a triangle is \(180^{\circ}\). For \(\triangle ABC\) with \(m\angle A=(3x + 4)^{\circ}\), \(m\angle B=(2x-4)^{\circ}\), and \(m\angle C=(4x-18)^{\circ}\), we have the equation \((3x + 4)+(2x-4)+(4x-18)=180\).
Simplify the left - hand side: \(3x + 4+2x-4+4x-18=(3x+2x + 4x)+(4 - 4-18)=9x-18\).
So, \(9x-18 = 180\).

Step2: Solve for \(x\)

Add \(18\) to both sides of the equation \(9x-18 = 180\): \(9x=180 + 18=198\).
Divide both sides by \(9\): \(x=\frac{198}{9}=22\).

Step3: Find the measures of the angles

• \(m\angle A=(3x + 4)^{\circ}\). Substitute \(x = 22\): \(m\angle A=(3\times22 + 4)^{\circ}=(66+4)^{\circ}=70^{\circ}\).
• \(m\angle B=(2x-4)^{\circ}\). Substitute \(x = 22\): \(m\angle B=(2\times22-4)^{\circ}=(44 - 4)^{\circ}=40^{\circ}\).
• \(m\angle C=(4x-18)^{\circ}\). Substitute \(x = 22\): \(m\angle C=(4\times22-18)^{\circ}=(88 - 18)^{\circ}=70^{\circ}\).

Since \(m\angle A=m\angle C = 70^{\circ}\), \(\triangle ABC\) is an isosceles triangle.

Answer:

1. \(m\angle A = 60^{\circ}\), \(m\angle B = 60^{\circ}\), \(m\angle C = 60^{\circ}\), \(\triangle ABC\) is an equilateral triangle.
2. \(m\angle A = 70^{\circ}\), \(m\angle B = 40^{\circ}\), \(m\angle C = 70^{\circ}\), \(\triangle ABC\) is an isosceles triangle.