The projector fully illuminates screen A, 60 cm high, located 80 cm from the projector.

The projector fully illuminates screen A, 60 cm high, located 80 cm from the projector. Screen B is located at a distance of 240 cm from screen A. What is the maximum height of screen B to be fully illuminated.

The projector emits light with a scattering angle equal to the apex angle of an equilateral triangle with a base of 60 cm and a height of 80 cm.

Since the triangle is equilateral, therefore the apex is the height, and the tangent at the apex of this angle can be found as the ratio of half the base to the height:

60/2/80 = 30/80 = 3/8.

Let the screen be x cm high.Therefore, half the tangent of the scattering angle will be equal to:

x / 2/240 = 3/8.

x / 2 = 240 * 3/8 = 30 * 3 = 90;

x = 90 * 2 = 180.

Answer: 180 cm.