# A spotlight on the ground 3 feet away from a 5 feet tall man cast a 15 feet shadow on a wall 6 feet away from the man. Find the angle in which light is placed?

Feb 25, 2018

At an angle of ${59.04}^{0}$ the light was placed.

#### Explanation:

Let $S$ be the spot light on the ground , $A B = 5$ ft tall man at

$S A = 3$ feet away from $S$ and $C D = 15$ ft tall shadow at

$A C = 6$ feet away from man $\therefore S C = S A + A C = 3 + 6 = 9$

feet. So the triangles formed $\Delta S A B \mathmr{and} \Delta S C D$

are similar triangles.Let angle cast by the light be$\angle S = \theta$

then , $\tan \theta = \frac{A B}{S A} = \frac{5}{3} \mathmr{and} \tan \theta = \frac{C D}{S C} = \frac{15}{9}$

$\therefore \theta = {\tan}^{-} 1 \left(\frac{5}{3}\right) = {59.04}^{0}$ or

$\theta = {\tan}^{-} 1 \left(\frac{15}{9}\right) = {59.04}^{0}$. Therefore at an angle of

${59.04}^{0}$ the light was placed. [Ans]

Feb 25, 2018

${59.04}^{\circ}$

#### Explanation:

Both the man and the shadow form proportional right triangles, so we can really use either one to solve your problem. Let's use the man.

The triangle is formed by the spotlight, the man's feet on the ground and his head (It's helpful to draw a picture). If we're looking at the light as an angle ($\theta$), then $\theta$ can be found using the opposite side (the height of the man, 5ft) and the adjacent side (the distance from the light to the man, 3ft). Since the tangent ratio is $\frac{o p p o s i t e}{a \mathrm{dj} a c e n t}$, we can use the inverse tangent ratio to come up with the angle.

${\tan}^{-} 1 \left(\frac{5}{3}\right) = {59.04}^{\circ}$

It also works if you use the larger triangle (spotlight, base of wall, top of shadow)

${\tan}^{-} 1 \left(\frac{15}{9}\right) = {59.04}^{\circ}$

Hope that helps.