In a quadrilateral ABCD, ZA = 66°, ZB = ZC +16°, Z C = 6(x + 4) 'and
ZD=(Z A – 16°). Calculate ZB and Z D.
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Step-by-step explanation:
Step-by-step explanation:
Solution :-
We know that the sum of the opposite angles of a cyclic quadrilateral is 180°.
Therefore,
∠A + ∠C = 180°
and ∠B + ∠C = 180°
Now, ∠A + ∠C = 180°
⇒ (2x - 1) + (4x - 7) = 180°
⇒ 2(x + y) = 166
⇒ x + y = 166/2
⇒ x + y = 83 ..... (i)
And, ∠B + ∠C = 180°
⇒ (y + 5) + (4x - 7) = 180°
⇒ 4x + y = 180° ..... (i)
On subtracting (i) from (ii), we get
⇒ 3x = 182 - 83
⇒ 3x = 99
⇒ x = 99/3
⇒ x = 33
Putting x's value in Eq (i), we get
⇒ x + y = 83
⇒ 33 + y = 83
⇒ y = 83 - 33
⇒ y = 50
Here, x = 33 and y = 50
∠A = 2x - 1 = 2(33) - 1 = 65°
∠B = y + 5 = 50 + 5 = 55°
∠C = 2y + 15 = 2(50) + 15 = 115°
∠D = 4x - 7 = 4(33) - 7 = 125°
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