in the given figure angle ABD =x, angle CBD=y angle BCD=z AB is parallel to DC x=4/3y, y=3/8z find the value of x,y,z
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Solution:-
Given : AB || DC ; x = 4/3y and y = 3/8z.
AB || DC,
⇒ ∠ ABD = ∠ BDC (Alternate angles)
In Δ BDC,
x + y + z = 180°
⇒ 4/3y + 3/8z + z = 180°
⇒ First we will solve the 4/3y.
⇒ 4/3y (since he value of y = 3/8z)
⇒ 4/3 × 3/8z
⇒ 12/24z
⇒ 1/2z
Now, 1/2z + 3/8z + z = 180°
⇒ After taking L.C.M. we get,
⇒ (4z + 3z + 8z)/8 = 180°
⇒ 15z/8 = 180°
⇒ 15z = 180 × 8
⇒ z = 1440/15
⇒ z = 96°
y = 3/8z
y = (3 × 96)/8
⇒ y = 288/8
⇒ y = 36°
x= 4/3y
⇒ x = (4 × 36)/3
⇒ x = 144/3
⇒ x = 48°
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