+2440 There is a theorem called the Exterior Angle Theorem, which states that the measure of the exterior angle of a triangle equals the sum of the measure of the two nonadjacent angles (also known as remote interior angles).
Knowing this information above, we can set up an equation and solve for x:
| \(28x-1=19x+14+6x\) | Combine like terms on the right hand side of the equation. |
| \(28x-1=25x+14\) | Subtract 25x from both sides to eliminate two instances of the variable. |
| \(3x-1=14\) | Add 1 to both sides. |
| \(3x=15\) | Divide by 3 on both sides of the equation to isolate the variable. |
| \(x=5\) | However, solving for x is not actually answering the question; we need to solve for \(m\angle BCD\) |
| \(m\angle BCD=(28x-1)^{\circ}\) | Substitute the now known value for x, which is 3. |
| \(m\angle BCD=(28*3-1)^{\circ}\) | Simplify the measure of the angle. |
| \(m\angle BCD=83^{\circ}\) | |
+2440 There is a theorem called the Exterior Angle Theorem, which states that the measure of the exterior angle of a triangle equals the sum of the measure of the two nonadjacent angles (also known as remote interior angles).
Knowing this information above, we can set up an equation and solve for x:
| \(28x-1=19x+14+6x\) | Combine like terms on the right hand side of the equation. |
| \(28x-1=25x+14\) | Subtract 25x from both sides to eliminate two instances of the variable. |
| \(3x-1=14\) | Add 1 to both sides. |
| \(3x=15\) | Divide by 3 on both sides of the equation to isolate the variable. |
| \(x=5\) | However, solving for x is not actually answering the question; we need to solve for \(m\angle BCD\) |
| \(m\angle BCD=(28x-1)^{\circ}\) | Substitute the now known value for x, which is 3. |
| \(m\angle BCD=(28*3-1)^{\circ}\) | Simplify the measure of the angle. |
| \(m\angle BCD=83^{\circ}\) | |
