Question

1. ( mangle a=)____ in ( \triangle abc,mangle a=(3x + 15)^{circ},mangle b = 4x^{circ}) and ( mangle c=(6x-30)^{circ}). what type of triangle is ( \triangle abc)? ( mangle b=)__ ( mangle c=)__ 4. ( mangle a=)__ in ( \triangle abc,mangle a=(3x + 4)^{circ},mangle b=(2x-4)^{circ}) and ( mangle c=(4x-18)^{circ}). what type of triangle is ( \triangle abc)? ( mangle b=)__ ( mangle c=)____

Explanation:

Problem 3

Step1: Use the angle - sum property of a triangle

The sum of the interior angles of a triangle is \(180^{\circ}\). So, \(m\angle A + m\angle B+m\angle C=180^{\circ}\).
Substitute \(m\angle A=(3x + 15)^{\circ}\), \(m\angle B = 4x^{\circ}\), and \(m\angle C=(6x-30)^{\circ}\) into the equation:
\((3x + 15)+4x+(6x-30)=180\)

Step2: Simplify the left - hand side of the equation

Combine like terms: \((3x+4x + 6x)+(15 - 30)=180\)
\(13x-15 = 180\)

Step3: Solve for \(x\)

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

Step4: Find the measures of the angles

• For \(m\angle A\):

Substitute \(x = 15\) into \(m\angle A=(3x + 15)^{\circ}\)
\(m\angle A=(3\times15 + 15)^{\circ}=(45+15)^{\circ}=60^{\circ}\)

• For \(m\angle B\):

Substitute \(x = 15\) into \(m\angle B = 4x^{\circ}\)
\(m\angle B=4\times15^{\circ}=60^{\circ}\)

• For \(m\angle C\):

Substitute \(x = 15\) into \(m\angle C=(6x-30)^{\circ}\)
\(m\angle C=(6\times15-30)^{\circ}=(90 - 30)^{\circ}=60^{\circ}\)

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

Step1: Use the angle - sum property of a triangle

The sum of the interior angles of a triangle is \(180^{\circ}\). So, \(m\angle A + m\angle B+m\angle C=180^{\circ}\)
Substitute \(m\angle A=(3x + 4)^{\circ}\), \(m\angle B=(2x-4)^{\circ}\), and \(m\angle C=(4x-18)^{\circ}\) into the equation:
\((3x + 4)+(2x-4)+(4x-18)=180\)

Step2: Simplify the left - hand side of the equation

Combine like terms: \((3x+2x + 4x)+(4-4-18)=180\)
\(9x-18 = 180\)

Step3: Solve for \(x\)

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

Step4: Find the measures of the angles

• For \(m\angle A\):

Substitute \(x = 22\) into \(m\angle A=(3x + 4)^{\circ}\)
\(m\angle A=(3\times22 + 4)^{\circ}=(66 + 4)^{\circ}=70^{\circ}\)

• For \(m\angle B\):

Substitute \(x = 22\) into \(m\angle B=(2x-4)^{\circ}\)
\(m\angle B=(2\times22-4)^{\circ}=(44 - 4)^{\circ}=40^{\circ}\)

• For \(m\angle C\):

Substitute \(x = 22\) into \(m\angle C=(4x-18)^{\circ}\)
\(m\angle C=(4\times22-18)^{\circ}=(88 - 18)^{\circ}=70^{\circ}\)

Since \(m\angle A=m\angle C = 70^{\circ}\) and \(m\angle B = 40^{\circ}\), \(\triangle ABC\) is an isosceles triangle.

Answer:

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

Problem 4