Q.7: In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.
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Answer❀✿°᭄
Let Z be the number
Z = 13x + 11 where x is the quotient when Z is divided by 13
Z = 17y + 9 where y is the quotient when Z is divided by 17
13x + 11 = 17y + 9
13x + 2 = 17y since x and y are quotients they should be whole numbers . Since y has to be a whole number the left hand side should be multiple of 17
The least possible value of x satisfying the condition is 9 and y will be 7
The answer is 13*9 + 11 = 128 or it is 17*7 + 9 = 128
This is the least number possible. There will be multiple answers and will increase in multiples 17*13 = 221 like 349 , 570, etc
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Answer:
Draw BL⊥PQ and CM⊥RS. As PQ∥RS, so, BL∥CM.
The alternate interior angles are equal, so,
∠LBC=∠MCB (1)
It is known that the angle of reflection is equal to the angle of incidence, therefore,
∠ABL=∠LBC (2)
And,
∠MCB=∠MCD (3)
From equation (1), (2) and (3),
∠ABL=∠MCD (4)
Add equation (1) and (4),
∠LBC+∠ABL=∠MCB+∠MCD
∠ABC=∠BCD
Since, these are the interior angles and are equal, hence, AB∥CD.
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