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=)______

1. ( 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 triangle angle - sum property

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

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

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

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

Step1: Use triangle angle - sum property

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

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

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

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