Question

$\sf \large \red{\underline{ Question:-}}$

5. The measures of two adjacent angles of a parallelogram are in the ratio 3:2. Find the measure of each of the angles of the parallelogram.

$\\\\\sf \large \red{\underline{Given:-}}$
The measures of two adjacent angles of a parallelogram are in the ratio 3:2.

$\\\\\sf \large \red{\underline{To \: Find:-}}$

Find the measure of each of the angles of the parallelogram.

$\\\\\sf \large \red{\underline{Solution :- }}$
$\text{ \sf suppose the angles be equal to 3x and 2x.}$

$\boxed{ \sf \orange{ we \: have \: ardjacent \: angles \: of \: a \: parallelogram \: = 180}}$

$\\ \sf \underline{ \green{putting \: all \: values : }}$
$\: \\ \sf \to \: 3x + 2 x = 180\: \\ \\ \sf \to \: \: \: \: \: \: \: \: \: \: \:5x = 180 \\ \\ \: \sf \to \: \: \: \: \: \: \: \: \: \: \:x \: = \frac{180}{5} \\ \\ \sf \to \: \: \: \: \: \: \: \: \: \: \:x \: = \cancel{ \frac{180}{5} } \\ \\ \sf \to \: \: \: \: \: \: \: \: \: \: \purple{x = 36}$

$\sf \to \: 3x \\ \sf \to \: 3 \times 36 \\ \sf \to \red{108 }\\ \\ \\ \sf \to \: 2x \\ \sf \to \: 2 \times 36 \\ \sf \to \orange{72}$

$\sf \large\underline{ \blue{verification }} \huge \dag$
$\\ \\ \sf \to 3x + 2x = 180 \\ \\ \sf \to \: 3 \times 36 +2 \times 36 = 180 \\ \\ \sf \to \: 108 + 72 = 180 \\ \\ \sf \to \:180 = 180 \\ \\ \large \underline{ \pink{ \sf \: hence \: verified}} \huge \dag$

Solve Please !!!


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Answer 1

$\huge\red{Answer}$

Let the measures of two adjacent angles, ∠A and ∠B, of parallelogram ABCD are in the ratio of 3:2.

Let ∠A = 3x and ∠B = 2x

We know that the sum of the measures of adjacent angles is 180º for a parallelogram.

∠A + ∠B = 180º

3x + 2x = 180º

5x = 180º

$x = \frac{180}{5}$

=36°

∠A = ∠C = 3x = 108º (Opposite angles)

∠B = ∠D = 2x = 72º (Opposite angles)

Thus, the measures of the angles of the parallelogram are 108º, 72º, 108º, and 72º.

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