suppose two people are standing on a beach 50 meters apart. Each person looks at an island off in the distance, one has a 30 degree line of sight and the other has a 40 degree line of sight. How far is it from the beach to the island (find the shortest, straight line distance).
The distance from the beach to the island will form the altitude, A, of a triangle
Let x represent the distance that the person with the 30 degree line of sight is from where the altitude intersects the base
So...the distance that the person with the 40 degree line of sight to this intersection point is 50 -x
So we have this
tan 30 = A/ x
Rearrange as [ x = A / tan 30 ] ⇒
[ x = A / (1/√3) = √3A ]
And we have that
tan 40 = A / (50 - x) sub in for x and we get that
tan 40 = A / ( 50 - √3A)
(50 - √3A) * tan 40 = A
50sin 40 - √3A tan40 = A
50tan 40 = √3Atan 40 + A factor out A on the right
50tan 40 = A ( √3 tan 40 + 1) divide both sides by (sin 40 + 1)
A = 50tan 40 / [√3 tan 40 + 1 ] ≈ 17.1 (m)
EDIT TO CORRECT AN ERROR!!!
Slightly different approach using law of sines
x/sin30 = 50/sin10 yields x = 26.6044 (distance from 40 degree observer to island)
26.044/sin 90 = x /sin40 yields x = 17.1 m shortest distance to the beach....