Question

Lines a and b are parallel. Line c is a transversal. Find the measures of all angles formed by a, b, and c, One of the angles is 70° bigger than the other

Answer 1

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 2

Answer:

The 4 angles formed in each case are: $55^{0}$, $55^{0}$, $125^{0}$ and $125^{0}$.

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 $x^{0}$, but the other is greater by $70^{0}$, so that =  ($70^{0}$ + $x^{0}$)

Thus, we have;

    $x^{0}$ + $x^{0}$ + ($70^{0}$ + $x^{0}$) + ($70^{0}$ + $x^{0}$) = $360^{0}$    ( the sum of angles at a point)

     $x^{0}$ + $x^{0}$ + $70^{0}$ + $x^{0}$ + $70^{0}$ + $x^{0}$ = $360^{0}$

$4x^{0}$  +  $140^{0}$ = $360^{0}$

$4x^{0}$ = $360^{0}$ - $140^{0}$

$4x^{0}$ = $220^{0}$

Divide both sides by 4,

$x^{0}$ = $55^{0}$

The other angle is calculated thus,

   ($70^{0}$ + $x^{0}$) = $70^{0}$ + $55^{0}$

                 = $125^{0}$

Thus the 4 angles formed in both cases have the values; $55^{0}$, $55^{0}$, $125^{0}$ and $125^{0}$.