Answer:
Angles are either 55° or 125°.
Step-by-step explanation:
See the attached diagram.
Let aa' and bb' are two parallel straight lines and cc' is a transversal that meets aa' at o and bb' at o' points.
Now, ∠coa' + ∠coa =180° ..... (1)
Assume by the condition given ∠coa' = x and ∠coa = x+70
Hence, from equation (1), 2x + 70 = 180
⇒ x = 55°
Then ∠coa' =55° and ∠coa = 70+55 = 125°
So, ∠o'oa' = 125° as ∠coa = ∠ o'oa' {Opposite angles}
Again, ∠aoo' = 55° as ∠coa' = ∠aoo' {Opposite angles}
Now, ∠coa' = ∠oo'b' {Corresponding angles} = 55°
and ∠bo'c' = ∠oo'b' = 55° {Opposite angles}
Again ∠oo'b = ∠coa = 125° {Corresponding angles}
and ∠b'o'c' = ∠oo'b =125° {Opposite angles}
(Answer)
Answer:
The 4 angles formed in each case are: [tex]55^{0}[/tex], [tex]55^{0}[/tex], [tex]125^{0}[/tex] and [tex]125^{0}[/tex].
Step-by-step explanation:
Line c being transversal implies that it forms 4 angles with lines a and b individually of which 2 in each case are opposite angles, thus are equal.
Let one of the angles be represented by [tex]x^{0}[/tex], but the other is greater by [tex]70^{0}[/tex], so that = ([tex]70^{0}[/tex] + [tex]x^{0}[/tex])
Thus, we have;
[tex]x^{0}[/tex] + [tex]x^{0}[/tex] + ([tex]70^{0}[/tex] + [tex]x^{0}[/tex]) + ([tex]70^{0}[/tex] + [tex]x^{0}[/tex]) = [tex]360^{0}[/tex] ( the sum of angles at a point)
[tex]x^{0}[/tex] + [tex]x^{0}[/tex] + [tex]70^{0}[/tex] + [tex]x^{0}[/tex] + [tex]70^{0}[/tex] + [tex]x^{0}[/tex] = [tex]360^{0}[/tex]
[tex]4x^{0}[/tex] + [tex]140^{0}[/tex] = [tex]360^{0}[/tex]
[tex]4x^{0}[/tex] = [tex]360^{0}[/tex] - [tex]140^{0}[/tex]
[tex]4x^{0}[/tex] = [tex]220^{0}[/tex]
Divide both sides by 4,
[tex]x^{0}[/tex] = [tex]55^{0}[/tex]
The other angle is calculated thus,
([tex]70^{0}[/tex] + [tex]x^{0}[/tex]) = [tex]70^{0}[/tex] + [tex]55^{0}[/tex]
= [tex]125^{0}[/tex]
Thus the 4 angles formed in both cases have the values; [tex]55^{0}[/tex], [tex]55^{0}[/tex], [tex]125^{0}[/tex] and [tex]125^{0}[/tex].
