Given m∠2 = 80°, m∠1 = 4 · m∠3, and ∠4 ≅ ∠ 8, find m∠1, m∠3, m∠4, m∠5, m∠6, m∠7, and m∠8.

m∠1= m∠5= 100°, m∠2= m∠6= 80°, m∠4=m∠8= 155°, m∠3= m∠7= 25°

Step-by-step explanation:

Supposing ∠1,∠2  make up a straight angle.Let ∠1, ∠5 and ∠2 and ∠6 be corresponding angles on a transversal  drawn on two  parallel lines. ∠3,∠4 and∠5 ∠6  be interior angles

If m ∠= 80° them m∠ 1= 100°

As ∠1= 4. m∠3 therefore

100°= 4.m∠3

m∠3= 100/4= 25°

Again supposing m∠3+ m∠4= 180°

m∠4= 180°- 25°= 155°

m∠4 =m∠8= 155°

m∠8 + m∠7= 180°

m∠7= 180°- 155°= 25°

Let  ∠5 and ∠1 be corresponding angles so m∠5= m∠1= 100°

and m∠6= m∠2= 80°

or

step-by-step explanation:

we need to solve for 'x' the following equation:  

we have that:

⇒

⇒ ⇒

factorizing, we get that:

  ⇒

then, the solutions are: or

nothing extraneous has been found!