1. bearings

I don't need solutions, I just need someone to help me visualize these questions so I can set up the problems.

1. A pilot wishes to fly an airfield S20šE of his present position. If the average airspeed of the plane is 520 km/h and the wind is from N80šE at 46 km/h, what direction does he need to steer?

2. A destroyer detects a submarine 8 nautical miles due east traveling northeast at 20 knots. If the destroyer has a top speed of 30 knots, at what heading should it travel to intercept the submarine?

2. Hello, ty2391!

1. A pilot wishes to fly an airfield S20šE of his present position. If the average
airspeed of the plane is 520 km/h and the wind is from N80šE at 46 km/h,
what direction does he need to steer?
Code:
                  O
W - - - - - - * - - - - - - - - - - - E
10° *    *  *  θ
* 46      *     * 520
B *              *        *
*           *a          *
*        *              * A
*     *         *
*  *    *
*
P
$\displaystyle \angle EOA = \theta, \quad |OA| = 520, \quad \overrightarrow{OA} \:=\:\langle 520\cos\theta,\:\text{-}520\sin\theta\rangle$

$\displaystyle \angle WOB = 10^o,\quad |OB| = 46, \quad \overrightarrow{OB} \:=\:\langle\text{-}46\cos10^o,\:\text{-}46\sin10^o\rangle$

$\displaystyle \angle EOP = 70^o, \quad |OP| = a, \quad \overrightarrow{OP} \:=\:\langle a\cos70^o,\:\text{-}a\sin70^o\rangle$

We want: .$\displaystyle \overrightarrow{OA} + \overrightarrow{OB} \:=\:\overrightarrow{OP}$

. . $\displaystyle \begin{array}{cccccccc}520\cos\theta - 46\cos10^o &=& a\cos70^o & \Rightarrow & a &=&\dfrac{520\cos\theta - 46\cos10^o}{\cos70^o} & {\color{blue}[1]} \\ \\[-3mm] \text{-}520\sin\theta - 46\sin10^o &=&\text{-}a\sin70^o & \Rightarrow & a &=&\dfrac{520\sin\theta + 46\sin10^o}{\sin70^o} & {\color{blue}[2]}\end{array}$

Equate [1] and [2]: .$\displaystyle \frac{520\cos\theta - 46\cos10^o}{\cos70^o} \:=\:\frac{520\sin\theta + 46\sin10^o}{\sin70^o}$

. . $\displaystyle 520\sin70^o\cos\theta - 46\sin70^o\cos10^o \:=\:520\cos70^o\sin\theta + 46\cos70^o\sin10^o$

. . .$\displaystyle 520\sin70^o\cos\theta - 520\cos70^o\sin\theta \;=\;46\sin70^o\cos10^o + 46\sin10^o\cos70^o$

. . . . $\displaystyle 520(\sin70^o\cos\theta - \cos70^o\sin\theta) \;=\;46(\sin70^o\cos10^o + \sin10^o\cos70^o)$

. . . . . . . . . . . . . $\displaystyle 520\sin(70^o-\theta) \;=\;46\sin(70^o+10^o)$

And you can solve for $\displaystyle \theta.$