Question

in the above figure PQ is a mirror, AB is the incident ray and BC is the reflected ray. if angle ABC =46°, then angle ABP IS .....
(a) 44°
(b)67°
(c) 13°
(d) 62°
Can u pls tell this question's answer like step by step with statements​

Answer 1

Answer:

ABP=67

Step-by-step explanation:

Draw normal BN perpendicular to mirror PQ

According to laws of reflection,

I=r

We know that I+r=46

So, I=46/2

I=23

l+ABP=90

ABP=90-23

ABP=67

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Answer 2

SoluTion :-

$\bigstar\;\boxed{\bf Angle\ of\ incidence = Angle\ of\ reflection}}}}}$

∴ ABP = CBQ

Now,

$\mathrm {\angle \ ABP + \angle \ ABC + \angle \ CBQ=180^{o}}\\\\\\\mathrm {\angle \ ABP + \angle \ ABP + 46^{o}=180^{o}\ \ \{\because \angle \ ABP = \angle \ CBQ\}}\\\\\\\mathrm {2 \angle \ ABP = 180^{o}-46^{o}}\\\\\\\mathrm {2 \angle \ ABP = 134^{o}}\\\\\\\mathrm {\angle \ ABP = \frac{134}{2} =67^{o}}$

∠ ABP = 67°