Trig Word problem, Challenging?

Question

A ship is moving due west at 8 knots. You are in a speed boat at a distance of 10 nautical miles from the ship. Your bearing as seen from the ship is 130 degrees. You need to catch up with the ship. So you take off at a speed of 16 knots and a bearing of ???? degrees, and you reach the ship in ???? minutes. FYI Bearing Starts at 90 degrees and move couterclockwise

Bearing goes Clockwise.

Answer 1

(Ok so this is a relative-velocity problem in 2 dimensions.
The fiddly part is setting up the distances and velocities in the coord system. After that it's fairly straightforward.)

> S is moving west at 8 knots.
v_S = -8i + 0j in Cartesian coords

>B is at a distance of 10 nautical miles from the ship.
>B is 40°E of S from Ship, or 50°S of E.
So, initial relative displacement vector s(0)_BS :
s(0)_BS = 10(cos(-50),sin(-50))
s(0)_BS = 10(cos50, -sin50)

Your bearing as seen from the ship is 130°. You need to catch up with the ship.

B heads speed of 16 knots and a bearing of α degrees
> bearing of α degrees => direction vector is (-sin α, cos α)
> [Note: this is as per your original statement that bearing should be measured counterclockwise instead of clockwise]
v_B = 16(-sin α, cos α) = -16sin α i + 16cos α j

So the relative velocity is
v_BS = v_B - v_S
= 16(-sin α, cos α) - (-8,0)
= (8-16sin α, 16cos α)

So the relative displacement vector at any time t (hrs) is:
s(t)_BS = s(0)_BS + v_BS * t
= 10(cos50, -sin50) + t(8-16sin α, 16cos α)
= (10cos50 + t(8-16sin α), -10sin50 + 16tcos α)

We will choose bearing angle α such that both the x- and y-components of s(t)_BS reach zero (i.e. boat intercepts ship) at the same time ti.
So we have a pair of simultaneous eqns in unknowns ti, α:

10cos50 + ti(8-16sin α) = 0 [s_x]
-10sin50 + ti(16cos α) = 0 [s_y]

[s_y] => ti = 10sin50/16cos α
[s_x] => 10cos50 + ti(8-16sin α) = 0
(10sin50)(8-16sin α)/16cos α = -10cos50
(8-16sin α)/16cos α = -cot50
1 -2sin α = -2cot50 cos α

Maybe transform variable to s=sin α, thus cos α = √(1-s²):
1 -2s = -2cot50 √(1-s²)

Note we have to isolate the radical term on the right, so we can square it:
1 -4s +4s² = -2cot50 (1-s²)
4s² -4s + 1 = 2cot50 (s²-1)
(4-2cot50)s² -4s + (1+2cot50) = 0
s = [ 4±√(-4)² + 4(4-2cot50)(1+2cot50) ] / 2(4-2cot50)
= [ 4±√((-4)² + 4(4-2cot50)(1+2cot50)) ] / 2(4-2cot50)
= [ 4±√(16 + 8(2-cot50)(1+2cot50)) ] / 4(2-cot50)
= [ 4±√(16 + 8(2+3cot50 - 2cot²50)) ] / 4(2-cot50)
= [ 4±√(16 + 16 +24cot50 -16cot²50)) ] / 4(2-cot50)
= [ 4±√(32 +24cot50 -16cot²50)) ] / 4(2-cot50)
= [ 1±√(2 + 1.5cot50 -cot²50)) ] / (2-cot50)
= [ 1±√((cot50)(1.5 -cot50) + 2) ] / (2-cot50)
= [ 1±√2.554 ] / (2-cot50)
= [ 1±1.5983 ] / 1.161
= 2.238 or -0.515 (extraneous)

So: s=sin α = -0.515
=> bearing α = arcsin(+0.515) = 31°
> since we chose bearing of α degrees => direction vector is (-sin α, cos α)
> you confused this by saying bearing was measured counterclockwise instead of clockwise, anyway you get the idea

> and you reach the ship in ? minutes.

Finally to get intercept time you plug α back into:
ti = 10sin50/16cos α
= 0.5587 hrs
= 33min 31sec