Guyzzzz anybody help me wid solutions Document attached wid 5 questions ? Urgently required answer wid easy way.
Answers
Answer:
Q1) ∠COD = 180° - 116° = 64°
ABCD is a rectangle, so ΔCOD is isosceles. This means x = y.
Sum of angles in ΔCOD is 180° => x + y + 64° = 180°
=> 2x + 64° = 180° => 2x = 116° => x = 58° => y = 58°.
AB parallel to CD => z = y => z = 58°
Q2) Angles in ratio 11 : 18 : 21 : 22
=> they are 11x, 18x, 21x, 22x for some value x
sum of angles in quadrilateral = 360°
=> 11x + 18x + 21x + 22x = 360°
=> 72x = 360°
=> x = 5°
So the angles are
11x = 55°, 18x = 90°, 21x = 105°, 22x = 110°
Q3) sum of angles in quadrilateral = 360°
=> 2x + ( 2x + 15 ) + ( 4x - 12 ) + ( 3x - 0.5 ) = 360
=> 11x + 2.5 = 360
=> 11x = 357.5
=> x = 32.5
The angles are
2x = 65°, 2x + 15 = 80, 4x - 12 = 118°, 3x - 0.5 = 97°
Q4a) AD is parallel to BC
=> ∠B = 180° - ∠A = 180° - 50° = 130°
ABCD is a parallelogram,
so ∠C = ∠A = 50° and ∠D = ∠B = 130°
Q4b) ABCD is a parallelogram
=> opposite vertices are equal
=> 3x - 5 = 2x + 35
=> 3x - 2x = 35 + 5
=> x = 40
These two angles are then 3x - 5 = 3×40 - 5 = 120 - 5 = 115°
The other two angles are each 180° - 115° = 65°
Q5)
(i) ∠ACB is not uniquely determined.
(ii) ∠ABC is not uniquely determined, but ∠ABC = 180° - 40° - ∠ACB = 140° - ∠ACB
(iii) ∠ADC is not uniquely determined, but ∠ADC = ∠ABC = 140° - ∠ACB
(iv) ∠ACD = ∠BAC = 40° (Z rule since AB and DC are parallel)
(v) ∠CAD is not uniquely determined, but ∠CAD = ∠ACB (Z rule).
Q6) ∠ADB = ∠DBC ( Z rule, since AD and BC are parallel)
=> ∠ADB = 36°
Sum of angles in ΔADO = 180°
=> ∠OAD = 180° - ∠ADO - ∠AOD = 180° - 36° - 64° = 80°
Also
∠OCD = 64° - 28° (exterior angle of a triangle)
= 36°