In the given figure, if ∠BCD = 25°, ∠BAQ =110° and ∠ACR = 125°, then find the values of x, y and z.
Since, BCR is a straight line.
∴ ∠BCD + ∠DCA + ∠ACR = 180°
=>25° + y + 125° =180°
=>y = 180° -150°
=>y = 30…(i)
∠CAD + ∠DAQ = 180°
[linear pair axiom]
∠CAD + 110° =180°
=> ∠CAD = 180° -110°
=> ∠CAD =70° ,…(ii)
In ∆ABC, ∠ABC +∠BCA + ∠BAC = 180°
[by angle sum property of a triangle]
∴ x + 25° + y + ∠CAD = 180°
=> x + 25° + 30°+ 70° =180°
[from Eqs. (i) and (ii)]
=> x = 180° -125° = 55°
Now, in ∆BCD, ∠ADC =∠CBD + ∠BCD
[by exterior angle property of a triangle]
=> z = re + 25° = 55° + 25° = 80° [∵ x= 55°]
Hence, x = 55°, y = 30° and z = 80°.