Question

1. in the figure, m∠2 = 70. find the measure of each angle. a. ∠3 b. ∠5 c. ∠8 d. ∠1 e. ∠4 f. ∠6 g. using the same image above (but disregard the previous numbers), find the value of x if m∠1 = 10x + 7 and m∠8 = 12x - 13 3. find x. 5x + 10 6x - 5 4. x = 3x + 10 2x + 15 9. 3x + 20 5x - 8 x = type of angles: congruent or supplementary 10. 2x + 24 15x x = congruent or supplementary 11. find the indicated angles. m∠1 = m∠2 = m∠3 = m∠4 = m∠5 = m∠6 = m∠7 = m∠8 = m∠9 = 36 triangle congruence theorems determine which theorem can be used to prove that the triangles are congruent. if it is not possible to prove that they are congruent, write

ot possible\. if possible, state the congruence statement. 12. 13. 14. congruent: yes no △__≅△ by congruent: yes no △≅△ by congruent: yes no △≅△ by __

Explanation:

Step1: Identify angle - angle relationships

In the first part, if \(m\angle2 = 70^{\circ}\), and assuming \(r\parallel s\) and \(q\) is a transversal:

• \(\angle2\) and \(\angle3\) are vertical - angles. Vertical angles are congruent. So \(m\angle3=m\angle2 = 70^{\circ}\).
• \(\angle2\) and \(\angle5\) are corresponding angles. Corresponding angles of parallel lines are congruent. So \(m\angle5=m\angle2 = 70^{\circ}\).
• \(\angle2\) and \(\angle8\) are alternate - exterior angles. Alternate - exterior angles of parallel lines are congruent. So \(m\angle8=m\angle2 = 70^{\circ}\).
• \(\angle2\) and \(\angle1\) are supplementary (linear pair). So \(m\angle1 = 180 - m\angle2=180 - 70=110^{\circ}\).
• \(\angle2\) and \(\angle4\) are vertical angles. So \(m\angle4=m\angle2 = 70^{\circ}\).
• \(\angle2\) and \(\angle6\) are alternate - interior angles. So \(m\angle6=m\angle2 = 70^{\circ}\).

Step2: Solve for \(x\) when \(m\angle1 = 10x + 7\) and \(m\angle8 = 12x-13\)

Since \(\angle1\) and \(\angle8\) are supplementary (linear - pair of angles formed by a transversal intersecting two parallel lines), we have the equation \((10x + 7)+(12x-13)=180\).
First, combine like terms: \(10x+12x+7 - 13=180\), which simplifies to \(22x-6 = 180\).
Add 6 to both sides: \(22x=180 + 6=186\).
Divide both sides by 22: \(x=\frac{186}{22}=\frac{93}{11}\approx8.45\).

Step3: Solve for \(x\) in other angle - pair equations

For the pair \(5x + 10\) and \(6x-5\) (assuming they are vertical angles, so they are congruent):
Set up the equation \(5x + 10=6x-5\).
Subtract \(5x\) from both sides: \(10=x - 5\).
Add 5 to both sides: \(x = 15\).
For the pair \(3x + 10\) and \(2x+15\) (assuming they are vertical angles, so they are congruent):
Set up the equation \(3x + 10=2x+15\).
Subtract \(2x\) from both sides: \(x+10 = 15\).
Subtract 10 from both sides: \(x = 5\).
For the pair \(3x + 20\) and \(5x-8\) (assuming they are vertical angles, so they are congruent):
Set up the equation \(3x + 20=5x-8\).
Subtract \(3x\) from both sides: \(20 = 2x-8\).
Add 8 to both sides: \(28 = 2x\).
Divide both sides by 2: \(x = 14\).
For the pair \(2x + 24\) and \(15x\) (assuming they are vertical angles, so they are congruent):
Set up the equation \(2x+24 = 15x\).
Subtract \(2x\) from both sides: \(24 = 13x\).
Divide both sides by 13: \(x=\frac{24}{13}\approx1.85\).

Step4: Determine triangle congruence

For problem 12:
If we assume we have two right - angled triangles \(\triangle FGI\) and \(\triangle HGI\) with \(GI\) as a common side, and right - angles at \(\angle FIG\) and \(\angle HIG\), and if we know some other side lengths or angle measures. If we have the hypotenuse and one leg equal (HL - Hypotenuse - Leg theorem), we need more information to determine congruence. Without further information, we say "not possible".
For problem 13:
If we assume we have two triangles \(\triangle NOP\) and \(\triangle QOP\) with \(OP\) as a common side, and we know some side - side - side (SSS) or side - angle - side (SAS) or other congruence criteria information. If we have \(NO = QO\), \(NP = QP\) and \(OP\) is common, then \(\triangle NOP\cong\triangle QOP\) by SSS.
For problem 14:
If we assume we have two triangles \(\triangle ABC\) and \(\triangle EDC\) with \(\angle ACB=\angle ECD\) (vertical angles), and if we know some side - angle - side (SAS) or angle - side - angle (ASA) or other congruence criteria information. If we have \(AC = EC\) and \(BC = DC\), then \(\triangle ABC\cong\triangle EDC\) by SAS.

Answer:

a. \(m\angle3 = 70^{\circ}\)
b. \(m\angle5 = 70^{\circ}\)
c. \(m\angle8 = 70^{\circ}\)
d. \(m\angle1 = 110^{\circ}\)
e. \(m\angle4 = 70^{\circ}\)
f. \(m\angle6 = 70^{\circ}\)
g. \(x=\frac{93}{11}\)

1. \(x = 15\)
2. \(x = 5\)
3. \(x = 14\), Type of Angles: Vertical, Congruent
4. \(x=\frac{24}{13}\), Type of Angles: Vertical, Congruent
5. (No information given about the angles in the figure to find \(m\angle1,m\angle2,\cdots,m\angle9\))
6. Congruent: No, \(\triangle\) - not applicable by - not applicable
7. Congruent: Yes, \(\triangle NOP\cong\triangle QOP\) by SSS
8. Congruent: Yes, \(\triangle ABC\cong\triangle EDC\) by SAS