Question

two adjacent angles of a parallelogram are (3a+10) and ( 3a-4) .Find all angles of the parallelogram

Answer 1

Hey !

I would be glad answering that !

As we know that sum of adjacent angles of a parallelogram is 180.

So, 3a + 10 + 3a - 4 = 180
6a +6 =180
6(a+6)=180
a+1=30
a = 29

So, the angles are 97, 83 , 97 and 83.

Hope I helped you !

Answer 2

Angles of parallelogram ABCD are 97°,83°,97°,and 83° respectively.

Given:

• Two adjacent angles of a parallelogram are (3a+10) and ( 3a-4)

To find:

• Find all angles of the parallelogram.

Solution:

Concept to be used:

• Sum of adjacent angles of a parallelogram is 180°.
• Opposite angles are equal.

Step 1:

Find the value of a.

ATQ,

$3a + 10 + 3a - 4 = 180$

or

$6a + 6 = 180$

or

$6a = 174$

or

$\bf a = 29$

Step 2:

Find the angles.

Let the parallelogram is ABCD.

$3a + 10 = 3 \times 29 + 10$

or

$= 87 + 10$

or

$= 97$

Let it is angle A,

$\bf \angle \: A = 97$

Opposite angle is $\bf \angle C = 97$

Another angle

$3a - 4 = 3 \times 29 - 4$

or

$= 87 - 4$

or

$= 83$

$\bf \angle B = \angle D = 83$

Thus,

Angles of parallelogram ABCD are 97°,83°,97°,and 83° respectively.

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