UP Board Class 10 maths Paper set 822 (AV) solution 1(e)

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1. (e) In figure, O is the centre of circle. If  ∠ BAO = 25° and ∠ BCO = 20°  then the measure of             ∠ADC will be

(i). 20°
(ii) 25°
(iii) 45°
(iv) 90°
Solution: Answer (iii)  45°
In English                
  In the given ∠BAO = 25° 
                                 and  ∠OCB = 20°
                                  Let  ∠AOC = x°
                            ∠ABC = 1/2 ✕ ∠ AOC (angle is the remaining part of the circle)
                                        = 1/2  ✕ x =(1/2)x
ext ∠AOC =360°− ∠AOC ( angle at a point )
                   =(360− x)°
In Quadrilateral AOCB 
     We know that Sum of angles of a quadrilateral is 360°   
∠BAO +∠ABC +ext ∠AOC + ∠OCB = 360°
25° +  (1/2) x + (360− x)°+ 20°= 360°
                        405° −  (1/2) x = 360°
                                        (1/2) x = 405° − 360°
                                                   x =  90°
                                           ∠AOC = x° = 90°
                           ∠ABC = 1/2 ✕ ∠ AOC = 1/2 ✕ 90°= 45° 
                                          ∠ADC =∠ABC =  45° ( angle in same segment)
                                              ∠ADC =  45° 
In Hindi
1 (e).fp= esa] o`Rr dk dsUnz O gSA ;fn  ∠ BAO = 25° rFkk ∠ BCO = 20° rks   ∠ADC- dh eki gksxh 

 Solution:    ज्ञात है ∠BAO = 25° 
                                 और   ∠OCB = 20°
                                  ekuk ∠AOC = x°
                            ∠ABC = 1/2 ✕ ∠ AOC (,dkUrj [k.M ds fdlh ,d fcUnq ij vkUrfjr dks.k)
                                        = 1/2  ✕ x =(1/2)x
                         ext ∠AOC =360°− ∠AOC (,d fcUnq ij vkUrfjr dks.k) )
                                            =(360− x)°
prqHkqZt  AOCB  esa 
     ge tkurs gS fd prqHkqZt ds lHkh vUr%dks.kksa dk ;ksx 360 gksrk gSA 
     ∠BAO +∠ABC +ext ∠AOC + ∠OCB = 360°
     25° +  (1/2) x + (360− x)°+ 20°= 360°
                        405° −  (1/2) x = 360°
                                        (1/2) x = 405° − 360°
                                                   x =  90°
                                           ∠AOC = x° = 90°
                           ∠ABC = 1/2 ✕ ∠ AOC = 1/2 ✕ 90°= 45° 
                                          ∠ADC =∠ABC =  45° ( ,d gh o`Rr[k.M ds cjkcj dks.k)
                                              ∠ADC =  45° 

                             

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