plz answer both questions
Answers
Sum of interior angles of a quadrilateral is 360.
I applied the above result in two problems and found the required angles.
Answer:
6.∠A = 72° , ∠C = 84° , ∠B = 96° and ∠D = 108°
7. x = 24 , ∠DAB = 76° and ∠ADB = 54°.
Step-by-step explanation:
6.
Given:
ABCD is a Quadrilateral.
AB || CD
∠A : ∠D = 2 : 3 and ∠C : ∠B = 7 : 8
To find: Measure of all angles.
let ∠A = 2x , ∠D = 3x , ∠C = 7y , ∠B = 8y
We know that when 2 lines are parallel then sum of the angle on the same side of the traversal is 180°.
when AD is traversal,
∠A + ∠D = 180°
2x + 3x = 180
5x = 180
x = 36
∠A = 2 × 36 = 72° and ∠D = 3 × 36 = 108°
when CB is traversal,
∠C + ∠B = 180°
7y + 8y = 180
15x = 180
x = 12
∠C = 7 × 12 = 84° and ∠B = 8 × 12 = 96°
Therefore, ∠A = 72° , ∠C = 84° , ∠B = 96° and ∠D = 108°
7.
Given:
Figure is given with all angles marked.
To find: Value of x , ∠DAB , ∠ADB
We know that sum of all interior angle of quadrilateral is 360°
(3x + 10) + (5x + 8) + (3x + 4) + (50) + (x) = 360
3x + 5x + 3x + x + 10 + 8 + 4 + 50 = 360
12x + 72 = 360
12x = 360 - 72
12x = 288
x = 24
∠DAB = 3x + 4 = 3(24) + 4 = 72 + 4 = 76°
Angle sum property of triangle is 180°
So,
∠CDB + ∠DCB + ∠CBD = 180°
∠CDB + 5x + 8 + x = 180°
∠CDB + 5(24) + 8 + 24 = 180°
∠CDB + 120 + 32 = 180°
∠CDB = 180 - 152
∠CDB = 28
⇒ ∠ADB = ∠ADC - ∠CDB = 3x + 10 - 28 = 3(24) - 18 = 72 - 18 = 54°
Therefore, x = 24 , ∠DAB = 76° and ∠ADB = 54°.