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?

2 Answers
Feb 25, 2018

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

Explanation:

Let S be the spot light on the ground , AB=5 ft tall man at

SA=3 feet away from S and CD= 15 ft tall shadow at

AC=6 feet away from man :.SC=SA+AC=3+6=9

feet. So the triangles formed DeltaSAB and Delta SCD

are similar triangles.Let angle cast by the light be/_S=theta

then , tan theta=(AB)/(SA)=5/3 and tan theta=(CD)/(SC)=15/9

:. theta=tan^-1 (5/3)=59.04^0 or

theta=tan^-1 (15/9)=59.04^0. Therefore at an angle of

59.04^0 the light was placed. [Ans]

Feb 25, 2018

59.04^@

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 (opposite)/(adjacent), we can use the inverse tangent ratio to come up with the angle.

tan^-1(5/3)=59.04^@

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

tan^-1(15/9)=59.04^@

Hope that helps.