Question

of the parallelogram.
3. The measures of two adjacent angles of a parallelogram are such that one is 30°more
than the other, find the measure of each angle of the parallelogram.
D
C С​

Answer 1

Answer:

∠A = 75°

∠B = 105°

∠C = 75°

= 75°∠D = 105°

Step-by-step explanation:

As per the provided information in the given question, we have :

• The measures of two adjacent angles of a parallelogram are such that one is 30° more
• than the other.

We are asked to calculate the measure of each angle of the parallelogram.

Let ABCD be a parallelogram. ∠A and ∠B are the adjacent angles of the parallelogram. Let us suppose the measure of ∠B as x°. So, ∠B will become (30 + x)° according to the question.

As the opposite angles of the parallelogram are equal. So, ∠C = ∠A and ∠D = ∠B.

Now, as it is known to us that the sum of all the interior angles of the quadrilateral is 360°. So, the sum of all the interior angles of the parallelogram will also be 360°. Writing it in the form of an equation,

$\\ \twoheadrightarrow \sf{\quad { \angle A + \angle B + \angle C + \angle D = 360^\circ }}$

Substituting the values, we get,

$\\ \twoheadrightarrow \sf{\quad { x^\circ + (30 +x)^\circ + x^\circ + (30 + x)^\circ = 360^\circ }}$

Removing the brackets.

$\\ \twoheadrightarrow \sf{\quad { x^\circ + 30^\circ +x^\circ + x^\circ + 30^\circ + x^\circ = 360^\circ }}$

Performing addition of the like terms.

$\\ \twoheadrightarrow \sf{\quad { 60^\circ + 4x^\circ = 360^\circ }}$

Transposing 60° from L.H.S to R.H.S, its sign will get changed.

$\\ \twoheadrightarrow \sf{\quad { 4x^\circ = 360^\circ -60^\circ }}$

Performing subtraction in R.H.S.

$\\ \twoheadrightarrow \sf{\quad { 4x^\circ = 300^\circ }}$

Transposing 4° from L.H.S to R.H.S, its arithmetic operator will get changed.

$\\ \twoheadrightarrow \sf{\quad { x^\circ = \cancel{\dfrac{300^\circ}{4}} }}$

Dividing 300 by 4.

$\\ \twoheadrightarrow \sf{\quad { \textbf{ \textsf{x}}^\circ = \textbf{ \textsf{75}}^\circ}}$

Therefore,

⠀⠀⠀⠀★ $\angle\textbf{ \textsf{A}} = \textbf{ \textsf{75}}^\circ$

⠀⠀⠀⠀★ $\angle\textbf{ \textsf{C}} = \textbf{ \textsf{75}}^\circ$

Now,

$\\ \twoheadrightarrow \sf{\quad { \angle B=(x + 30)^\circ }}$

Substitute the value of x.

$\\ \twoheadrightarrow \sf{\quad { \angle B=(75 + 30)^\circ }}$

Performing addition.

$\\ \twoheadrightarrow \sf{\quad { \angle \textbf{ \textsf{B}} = \textbf{ \textsf{105}}^\circ}}$

And, ∠D = ∠B.

Therefore,

⠀⠀⠀⠀★ $\angle\textbf{ \textsf{B}} = \textbf{ \textsf{105}}^\circ$

⠀⠀⠀⠀★ $\angle\textbf{ \textsf{D}} = \textbf{ \textsf{105}}^\circ$

Answer 2

☀️ Given that, the measures of two adjacent angles of a parallelogram are such that one is 30° more than other and we need to find measure of each angle.

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━

❍ Let's say that, one of the adjacent angles is ( x° ) and other is ( x + 30° ).

As we know that,

• Adjacent angles of a parallelogram are supplementary [ sum is 180°]
• Opposite angles of a parallelogram are equal

Angles of the parallelogram -

∠A  = x°

∠B = x + 30°

∠C = x°    

∠D = x + 30°  

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━

Finding the value of x :-

•  ∠A + ∠B = 180°

⤳ x + x + 30° = 180°

⤳ 2x + 30° = 180°

⤳ 2x = 180° - 30°

⤳ 2x = 150°

⤳ x = 150°/2

⤳ x = 75°

Finding each angle :-

→  ∠A = x = 75°

→  ∠B = x + 30 = 75 + 30 = 105°

→  ∠C = x = 75°

→  ∠D = x + 30 = 75 + 30 = 105°