Check the picture below.
notice that the triangle ADH, since the segment AL is an angle bisector, meaning it cuts the angle A in two equal halves, then the triangle ADH is only using half of A.
Answer:
Distance= 6.6 miles
Bearing= N 62.854°W
Step-by-step explanation:
Let's determine angle b first
Angle b=20° (alternate angles)
Using cosine rule
Let the distance between the liner and the port be x
X² =8.8²+2.4²-2(8.8)(2.4)cos20
X²= 77.44 + 5.76-(39.69)
X²= 43.51
X= √43.51
X= 6.596
X= 6.6 miles
Let's determine the angles within the triangle using sine rule
2.4/sin b = 6.6/sin20
(2.4*sin20)/6.6= sin b
0.1244 = sin b
7.146= b°
Angle c= 180-20-7.146
Angle c= 152.854°
For the bearing
110+7.146= 117.146
180-117.146= 62.854°
Bearing= N 62.854°W
D = rt
/r /r
D/r = T
*sample text*
The first equation given 
We have to add 1 and 5 to the right side first. We will get,

To get rid of 6 from the right side we have to subtract 6 from both sides.



To find n we have to move -2 to the other side by dividing both side by -2.



So we have got the required answer for the first question.
The solution is n = 0.
The second equation given,

First we have to move 7x to the left side by subtracting it from both sides.



Now we have to move -2 to the right side by adding 2 to both sides.



We have got the required answer for the second question.
The solution is x = -7.
The third equation given,

We have to get rid of that negative sign from both sides. As we have negative sign to both sides we can cancel it out. We will get,

Now we have to move 4 to left side by subtracting it from both sides.




So we have got the required answer .
The solution is x = 4.