Question #46824

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Answer 1 · 2017-10-15T07:30:00

$90°$

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We know that the reflected ray from the mirror, the mirror and the horizontal will make a triangle.

Sum of angles of a triangle $=180°$

Let's call that triangle $∆ABC$

We know that $A= theta$

As $angle i =theta, angler=theta$

And, $angler +angleABC=90°$
[With the normal]

$therefore angle ABC+theta=90°$
$=>angleABC=90°-theta---(1)$

$angle ABC + theta + angle BCA = 180°$

We want to find $angleBCA$ .

So, $angle BCA = 180°-(theta+angleABC)$

Substituting from (1),
$angle BCA = 180°-(cancel(theta)+90cancel(-theta))$

$therefore angleBCA=180-90=90°$

Since $angleBCA=90°$ , the angle between the reflected ray and the horizontal would be $90°$