Can you guys help me with this question?

A submarine is at a depth of 200 feet. It is aiming a torpedo at a destroyer at an angle of 46 degrees. If the submarine dives at an angle of 55 degrees and travels forward on that angle another 139 feet, then what the new firing angle of the submarine should be? (Round to the nearest degree)

Ein U-Boot ist in einer Tiefe von 200 Fuß. Es zielt auf einen Torpedo bei einem Zerstörer in einem Winkel von 46 Grad. Wenn das Unterseeboot in einem Winkel von 55 Grad taucht und in diesem Winkel weitere 139 Fuß vorwärts fährt, was dann der neue Schusswinkel des U-Bootes sein sollte?

\({\color{blue}position1}\)

P₁(0;-200)

\({\color{blue}position \ destroyer}\)

\(m=\frac{y-y1}{x-x1}\)

\(tan46°=\frac{y+200}{x}\)

\(y=xd\times tan46°-200\)

\(xd=\frac{200}{tan46°}=193.138\)

Pd(193.138;0)

\({\color{blue}position2}\)

\(y2=-200-139\times sin55°\)

\(y2=-313.862\)

\(x2=139\times cos55°=79.727\)

P2(79.727;-313.862)

\({\color{blue}new\ firing \ angle \ α}\)

\(m=\frac{yd-y2}{xd-x2}\)

\(m=\frac{0+313.862}{193.138-79.727}\)

\(m=2.767474\)

\(m=tan\ α\)

\(α=arctan \ m= arctan \ 2.767474\)

α = 70.133168°       !

 A submarine is at a depth of 200 feet. It is aiming a torpedo at a destroyer at an angle of 46 degrees. If the submarine dives at an angle of 55 degrees and travels forward on that angle another 139 feet, then what the new firing angle of the submarine should be? (Round to the nearest degree)

Here, I'm assuming that the sub is diving at an angle of 55 degrees to the horizontal and traveling forward another 139 feet after it has already dived 200 feet

We first need to find the (diagonal) distance, d, the sub is from the ship after the first dive....this is given by :

d = 200/ sin 46  = about 278 ft

Next.....we need to figure the straight-line distance, s, from the sub to the ship after the first dive....we can use the Law of Cosines

s^2 = 278^2 + 139^2  - 2(278)(139)cos (46 + 55)

s =  √ [ 278^2 + 139^2  - 2(278)(139)cos (46 + 55) ]  = about  333.7 ft

And using the Law of Sines, the new firing angle can be calculated thusly :

sin θ / 278  = sin (46 + 55) / 333.7

arcsin [ 278 * sin (46 + 55) / 333.7]  = about 54.86°

And the new firing angle will be  =  180 - 55 - 54.86  =  about 70.14°   [ just as  asinus found !!!  ]

Here is an (approximate) diagram :

The sub is at point C after the two maneuvers.....the ship is at point B....the firing angle is BCE