Question

In figure. 5 AB||CD. Find the value of x and the angles marked as 1,2,3 and 4.
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Answer 1

From the given figure,

• ∠ 1 and ∠ 4 are alternate interior angles. So, their measure must be same.

• ∠3 and ∠ 2 are also alternate interior angles. So, their measure must be same.

• ∠ 1 and ∠ 2 are co-interior angles. So, the sum of their measure must be 180°.

Solution,

$\sf { \implies \angle 1 + \angle 2 = {180}^{\circ} }$

[ Sum of pair of co-interior angles is 180°.]

$\sf { \implies 3x + 5 + 7x + 5 = {180}^{\circ} }$

Collecting all like terms.

$\sf { \implies 3x + 7x + 5 + 5 = {180}^{\circ} }$

Performing addition.

$\sf { \implies 10x + 10 = {180}^{\circ} }$

Transposing 10 from LHS to RHS.

$\sf { \implies 10x = {180}^{\circ}-10 }$

Performing substraction.

$\sf { \implies 10x = {170}^{\circ}}$

Transposing 10 from LHS to RHS in order to find the value of x.

$\sf { \implies x =\dfrac{ {170}^{\circ}}{10} }$

Performing division.

$\sf { \implies x = 17 }$

Therefore,

Measure of ∠ 1 and ∠ 4 :

→ ∠ 1 = (3x+ 5)°

→ ∠ 1 = [3(17) + 5]°

→ ∠ 1 = [51 + 5]°

→ ∠ 1 = 56 °

$\implies$ $\boxed {\bf \red { \angle 1 = {56}^{\circ} }}$

Also,

$\implies$ $\boxed {\bf \red { \angle 4 = {56}^{\circ} }}$

[ Since, alternate interior angles are equal.]

Measure of ∠ 2 and ∠ 3 :

→ ∠ 2 = (7x+ 5)°

→ ∠ 2 = [7(17) + 5]°

→ ∠ 2 = [119 + 5]°

→ ∠ 2 = 124 °

$\implies$ $\boxed {\bf \red { \angle 2 = {124}^{\circ} }}$

Also,

$\implies$ $\boxed {\bf \red { \angle 3 = {124}^{\circ} }}$

[ Since, alternate interior angles are equal.]

_______________________________

Verification:

As we know that,

• The sum of all the angles lie on the straight line in 180°.

So,

Sum of ∠ 1 and ∠ 4 ; ∠ 2 and ∠ 3 must be 180°.

LHS:

→ ∠ 1 + ∠ 4

→ 56° + 124°

→ 180°

RHS:

→ 180°

LHS = RHS

Similarly,

LHS

→ ∠ 2 + ∠ 3

→ 56° + 124°

→ 180°

RHS:

→ 180°

Hence, verified!

Answer 2

We know, sum of allied angles or co-interior angles = ${180}^{\degree}$

∴ $\angle 1 + \angle 2 = {180}^{\degree}$

⇒ $(7x + 5) + (3x + 5) = {180}^{\degree}$

⇒ $7x + 5 + 3x + 5 = {180}^{\degree}$

⇒ $10x = {180}^{\degree} - {10}^{\degree} = {170}^{\degree}$

⇒ $x = {17}^{\degree}$

So, $\angle 2 = \angle 3 = [7(17^{\degree}) + 5] = 124^{\degree}$

(Alternate interior angles)

Also, $\angle 1 = \angle 4$

(Alternate interior angles)

⇒ $\angle 1 = \angle 4 = [3(17^{\degree}) + 5] = 56^{\degree}$

∴ $\angle 1 = \angle 4 = 56^{\degree}$ and $\angle 2 = \angle 3 = 124^{\degree}$