Anonymous asked in Science & MathematicsPhysics · 1 month ago

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

  • NCS
    Lv 7
    1 month ago
    Favorite 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

    If you find this helpful, please award Favorite Answer!

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