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

The 4 angles formed in each case are:
`55^(0)`,
`55^(0)`,
`125^(0)` and
`125^(0)`.

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)`.

Answer 2

Angles are either 55° or 125°.

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)