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°.