Question

. In Fig. 9, PQ is a mirror, AB is the incident ray, and BC is the reflected ray, ABP = 2CBQ. If ABC = 46°, then find ABP. P Fig. 9 .​

Answer 1

Answer:

Answer:-

Given;

P & Q is a mirror (i.e; straight line) AB is the incident ray,

BC is the reflected ray,

ABP = 2CBQ &

ABC = 46⁰

To find :- ABP

Solution :- As we know that the angle of incidence is equals to the angle of reflection [(ie; angle i = angle r) First law of reflection]

｡ ⁰ ｡ APB = CBR

Let APB & CBP be x⁰Then,

ABC + (ABQ + CBP) = 180⁰ [Linear pairs]

46⁰ + x⁰ + x⁰ = 180⁰

=> 46⁰ + 2x⁰ = 180⁰

=> 2x⁰ = 180⁰ - 46⁰

=> 2x⁰ = 134⁰

=> x⁰ = 134⁰/2

。⁰ 。 x⁰ = 67⁰

Since; CBP = 67⁰

Therefore; ABP = (ABC + CBP) = 46⁰ + 67⁰ = 113⁰

HENCE ; Angle ABP is equals to 113⁰

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