Question

find the measure of all angles of a parallelogram if one angle is 48° less than the twice of the smallest angle.​

Answer 1

Answer:

the answer to the question is 76° & 104°

Step-by-step explanation:

Let the angles of parallelogram ABCD be A ,B, C and D

Let angle A be x and the smallest angle (angle B) be y

In parallelogram opposite angles are equal

thus angle A = angle C = x

angle B = angle D = y

According to the question

x = 2y - 48°. ......(1)

According to angle sum property of a quadrilateral

sum of all angles of a quadrilateral = 360°

angle( A +B +C +D) = 360°

x + y + x + y = 360°

2y - 48° + y + 2y - 48° + y = 360°. (From 1)

6y - 96° = 360°

6y = 360° + 96°

6y = 456°

y = 456° /6

y= 76°

putting the value of y in (1)

x = 2(76)-48

=152-48

=104

thus angle A = angle C = x = 104°

angle B = angle D = y = 76°

Answer 2

Answer:

76°,104°,76° and 104°

Step-by-step explanation:

Let x° be the smallest angle.

then, the 2nd angle=(2x-48)°

x+2x-48=180°. (co-interior angles)

=>3x-48=180

=>3x=180+48

=>3x=228

=>x=228÷3

=>x=76°

Hence,smallest angle=76°

2nd angle=(2*76-48)=104°

3rd angle=1st angle=76°

4th angle=2nd angle=104°