Page 12 - Ans Sub

From the figure we know that ∠AEF and ∠KFE are alternate angles
So, we get
0
∠AEF + ∠KFE = 180
0
p + ∠KFE = 180
So, we get
0
∠KFE = 180 - p ……. (2)
By adding both the equations (1) and (2) we get
0
∠KFG + ∠KFE = 180 - p + r
From the figure ∠KFG + ∠KFE can be written as ∠EFG
0
∠EFG = 180 - p + r
We know that ∠EFG = q
0
q = 180 - p + r
It can be written as
0
p + q – r = 180
0
Therefore, it is proved that p + q – r = 180

15. 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.

Solution:
Draw ray BL ⊥PQ and CM ⊥ RS

∵ PQ || RS ⇒ BL || CM
[∵ BL || PQ and CM || RS]
Now, BL || CM and BC is a transversal.
∴∠LBC = ∠MCB …(1) [Alternate interior angles]
Since, angle of incidence = Angle of reflection

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