Solve question number ten fastly plz................
Answers
The value of x, y and z are as follows:
Figure (a): x = 50°, y = 65° and z = 65°
Figure (b): x = 80°, y = 60° and z = 40°
Figure (c): x = 60°, y = 50° and z = 70°
Step-by-step explanation:
Case 1: Figure (a)
Given:
PQ // RS
∠y = ∠z
We have,
∠QAC = ∠z = 65° …… [alternate angles]
∴ ∠z = ∠y = 65°
Now, in ∆ABC, using the angle sum property, we get
∠x + ∠y + ∠z = 180°
⇒ ∠x + (2 * 65°) = 180°
⇒ ∠x = 180°- 130°
⇒ ∠x = 50°
Thus, in figure (a) x = 50°, y = 65° and z = 65°
Case 2: Figure (b)
Given,
PQ // BC
∠ABC = 80°
∠ACB = 40°
We have,
∠ABC = ∠x = 80° …… [alternate angles]
And,
∠ACB = ∠z = 40° …….. [alternate angles]
Since x, y and z forms a linear pair
∴ ∠x + ∠y + ∠z= 180°
⇒ 80° + ∠y + 40° = 180°
⇒ ∠y = 180° - 120°
⇒ ∠y = 60°
Thus, in figure (b) x = 80°, y = 60° and z = 40°
Case 3: Figure (c)
Given,
OP//QR
∠ACR = 110°
∠OAB = 60°
We have,
∠ACB + ∠ACR = 180°……. [Linear Pair]
∠ACB = 180° - 110° = 70°
Also,
∠ACB = ∠z = 70° ……. [alternate angles]
and
∠OAB = ∠x = 60°
Now, in ∆ABC, using the angle sum property, we get
∠x + ∠y + ∠ACB = 180°
⇒ ∠y = 180° - (70° + 60°)
⇒ ∠y = 180° - 130°
⇒ ∠y = 50°
Thus, in figure (c) x = 60°, y = 50° and z = 70°
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