Jeff is standing on the 10 m diving platform. His coach is on the ground to the left of him, while his parents are on the ground to the right of him. His coach is looking at him at an angle of 37 degrees with the ground, while his parents are looking at him at an angle of 32 degrees with the ground. If the angle at the base of the platform between Jeff’s coach and Jeff’s parents is 100 degrees, how far apart are his coach and his parents to the nearest meter?

Since Jeff's coach is at the left of the platform and the platform at an angle of elevation of 37° and Jeff's parent's are on the right of the platform with the platform at an angle of elevation of 32°. The line of sight of Jeff's coach, the distance between Jeff's coach and parents and the line of sight of Jeff's parents form a triangle.

Since the platform is 10 m high, it is a perpendicular to the base of the triangle.

This perpendicular bisects the original triangle into two right angled triangles - one on the left and one on the right - with base angle 37° and 32° respectively.

Using trigonometric ratios with the opposite side to each angle the height of the platform from the ground (10 m) and the opposite angles 37° and 32° respectively, then the length of each base is gotten from

tan37° = 10/L and tan32° = 10/L' where L and L' are the lengths of each base of the two right angled triangles respectively.

So, L = 10/tan37° = 10/0.7536 = 13.3 m and L' = 10/tan32° = 10/0.6249 = 16 m.

The length d = L + L' is the distance between Jeff's coach and Jeff's parents

So, d = L + L'

= 13.3 m + 16 m

= 29.3 m

≅ 29 m to the nearest meter.

Note that we do not need the third angle in the triangle which is 100° the angle of separation between Jeff's coach and Jeff's parents which is at the base of the platform.