a ship leave port p and sails on a bearing N50 degree E to Port Q 15km away. then it sails on a bearing of S45 degree E to Port R,20km away

Port r is 100° from Port p and 26km from Port p

Step-by-step explanation:

Lets note the dimension.

From p to q = 15 km = a

From q to r = 20 km= b

Angle at q = 50° + 45°

Angle at q = 95°

Ley the unknown distance be x

Distance from p to r is the unknown.

The formula to be applied is

X²= a²+ b² - 2abcosx

X²= 15² + 20² - 2(15)(20)cos95

X²= 225+400-(-52.29)

X²= 677.29

X= 26.02

X is approximately 26 km

To know it's direction from p

20/sin p = 26/sin 95

Sin p= 20/26 * sin 95

Sin p = 0.7663

P= 50°

So port r is (50+50)° from Port p

And 26 km far from p

dunno

step-by-step explanation:

50x5=$250

step-by-step explanation: