Question

(b)
In the given figure, ABCD is a rectangle, whose diagonals intersect at 'o.
[3]
Diagonal AC is produced to E and ZDCE = 145º.
Find:

angleCAB
(ii) angleAOB
(iii) angleACB​

Answer 1

∠CAB =   35° , ∠AOB = 110°  , ∠ACB  = 55°

Step-by-step explanation:

AC is extended to E

=> ACE is a straight line

=> ∠DCE + ∠DCA  = 180°

=> 145° + ∠DCA  = 180°

=> ∠DCA  = 35°

∠CAB = ∠DCA   ( as AB || CD )

=> ∠CAB =   35°

∠CAB + ∠ACB  = 90°

=> ∠ACB  = 55°

∠AOB = 180° - 2(35°)

=> ∠AOB = 110°

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