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# Position of the where the light shines- solve problem (physics)?

Can someone please show how to solve this and explain? I'm studying for my physics exam and am having trouble because my professor does not lecture much :( Physics is hard...

A light shines on a mirror. The position of the mirror relative to the light source is 12m away and 82.29443° above horizontal. The light bounces off the mirror and passes directly over the light source onto a wall. The beam moving from the mirror to the wall travels a distance of 15m at an angle of 24.86977° above horizontal. What is the position of where the light shines on the wall relative to the original source?

Thanks so much:)

Ps the answer is 21.8m at 56.601511 above horizontal toward the wall

### 1 Answer

- NCSLv 71 month agoFavorite Answer
Forget about the light and mirror. This is just a geometry problem where they've given you three points defining a triangle.

The angle between the incoming and reflected light ray at the mirror is

Θ = 82.29443° + 24.86977° = 107.16420º

law of cosines to find the distance from light to wall:

c² = (12m)² + (15m)² - 2*12m*15m*cos107.1642º = 475.24 m²

c = 21.8 m

and the law of sines to find the angle:

sinφ / 15m = sin107.1642 / 21.8m

where φ is the internal angle of the triangle at the light

sinφ = 0.657

φ = 41.1º

and so the angle from horizontal from the light to the spot on the wall is

α = 180º - 41.1º - 82.29443º = 56.601511º

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