Question

solve all four problems for \x\. add your results together to prove your triangle excellence!
level 4

1. find the measure of \\( \angle a b c \\).
2. find the measure of \\( \angle a c b \\).
3. find the measure of \\( \angle c d e \\).
4. find the sum of the measures of \\( \angle d c e \\) and \\( \angle a b e \\).

what is the sum of your 4 answers? *

Explanation:

Problem 13:

Step1: Use the triangle - angle sum property

The sum of angles in a triangle is \(180^{\circ}\). So, \((x - 2)+(2x + 7)+4x=180\).

Step2: Simplify the left - hand side

Combine like terms: \(x-2 + 2x+7 + 4x=(x + 2x+4x)+(-2 + 7)=7x + 5\).
The equation becomes \(7x+5 = 180\).

Step3: Solve for \(x\)

Subtract 5 from both sides: \(7x=180 - 5=175\). Then \(x=\frac{175}{7}=25\).
\(\angle ABC=(x - 2)^{\circ}=(25 - 2)^{\circ}=23^{\circ}\).

Problem 14:

Step1: Use the triangle - angle sum property

The sum of angles in a triangle is \(180^{\circ}\). So, \(10x+(4x + 20)+(x + 30)=180\).

Step2: Simplify the left - hand side

Combine like terms: \(10x+4x+20+x + 30=(10x+4x+x)+(20 + 30)=15x+50\).
The equation becomes \(15x+50 = 180\).

Step3: Solve for \(x\)

Subtract 50 from both sides: \(15x=180 - 50 = 130\). Then \(x=\frac{130}{15}=\frac{26}{3}\).
\(\angle ACB=(4x + 20)^{\circ}=4\times\frac{26}{3}+20=\frac{104}{3}+20=\frac{104 + 60}{3}=\frac{164}{3}\approx54.67^{\circ}\).

Problem 15:

Step1: Use the vertical - angles and triangle - angle sum property

Since \(\angle CED=\angle AEB = 5x^{\circ}\) (vertical angles). In \(\triangle AEB\), using the angle - sum property of a triangle: \((4x - 7)+(x - 3)+5x=180\).

Step2: Simplify the left - hand side

Combine like terms: \(4x-7+x - 3+5x=(4x+x+5x)+(-7 - 3)=10x-10\).
The equation becomes \(10x-10 = 180\).

Step3: Solve for \(x\)

Add 10 to both sides: \(10x=180 + 10=190\). Then \(x = 19\).
\(\angle CDE=(43)^{\circ}\) (alternate - interior angles, since \(CD\parallel AB\) and using the property of congruent triangles formed by the intersecting lines).

Problem 16:

Step1: Use the exterior - angle property

The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles.
For \(\angle DCE\): The exterior angle at \(A\) gives \(126=5x+\angle ABC\). For \(\angle ABE\): The exterior angle at \(B\) gives \(\angle ABE = 5x + 16\).
First, find \(x\) from the triangle with exterior angle \(126^{\circ}\). The interior angle adjacent to \(126^{\circ}\) is \(180 - 126=54^{\circ}\).
Using the triangle - angle sum property (assuming the triangle has angles \(5x\), \(54\), and \(\angle ABC\)). But using the exterior - angle property directly:
The sum of \(\angle DCE\) and \(\angle ABE\):
\(\angle DCE=126 - 5x\) and \(\angle ABE=180 - 16=164-5x\) (using linear - pair and angle - sum relations).
Another way:
The sum of \(\angle DCE\) and \(\angle ABE\):
\(\angle DCE+\angle ABE=(180 - 5x)+(180 - 16)=344-5x\).
Using the exterior - angle at \(A\): \(126 = 5x+(180-(180 - 16))\) (simplify to find \(x\)).
Or, using the fact that the sum of \(\angle DCE\) and \(\angle ABE\):
\(\angle DCE = 126\) (exterior angle property for the triangle with \(5x\) and the non - adjacent interior angle) and \(\angle ABE=164\) (linear - pair with \(16^{\circ}\)).
\(\angle DCE+\angle ABE=126 + 164=290^{\circ}\).

Answer:

The sum of the four answers (\(23+54.67 + 43+290\)) is approximately \(410.67\).