Three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.
Given,
Three angles are 110°, 50° and 40°
Let the fourth angle be 'x'
We have,
Sum of all angles of a quadrilateral = 360°
110° + 50° + 40° = 360°
⟹ x = 360° - 200°
⟹ x = 160°
Therefore, the required fourth angle is 160°.
In a quadrilateral ABCD, the angles A, B, C and D are in the ratio of 1: 2: 4: 5. Find the measure of each angles of the quadrilateral.
Let the angles of the quadrilaterals be
A = x, B = 2x, C = 4x and D = 5x
Then,
A + B + C + D = 360°
⟹ x + 2x + 4x + 5x = 360°
⟹ 12x = 360°
⟹ x = 360°/12
⟹ x = 30°
Therefore, A = x = 30°
B = 2x = 60°
C = 4x = 120°
D = 5x = 150°
In a quadrilateral ABCD, CO and Do are the bisectors of ∠C and ∠D respectively. Prove that ∠COD = 1/2(∠A and ∠B).
In ΔDOC
∠1 + ∠COD + ∠2 = 180° [Angle sum property of a triangle]
⟹ ∠COD = 180 − (∠1 − ∠2)
⟹ ∠COD = 180 − ∠1 + ∠2
⟹ ∠COD = 180 − [1/2 LC + 1/2 LD] [∵ OC and OD are bisectors of LC and LD respectively]
⟹ ∠COD = 180 – 1/2(LC + LD) ... (i)
In quadrilateral ABCD
∠A + ∠B + ∠C + ∠D = 360° [Angle sum property of quadrilateral]
∠C + ∠D = 360° − (∠A + ∠B) .... (ii)
Substituting (ii) in (i)
⟹ ∠COD = 180 – 1/2(360 − (∠A + ∠B))
⟹ ∠COD = 180 − 180 +1/2(∠A + ∠B))
⟹ ∠COD = 1/2(∠A + ∠B))
The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.
Let the common ratio between the angles is 't'
So the angles will be 3t, 5t, 9t and 13t respectively.
Since the sum of all interior angles of a quadrilateral is 360°
Therefore, 3t + 5t + 9t + 13t = 360°
⟹ 30t = 360°
⟹ t = 12°
Hence, the angles are
3t = 3*12 = 36°
5t = 5*12 = 60°
9t = 9*12 = 108°
13t = 13*12 = 156°