Answer:
Step-by-step explanation:
area=(20 ×20)÷2=200 sq . units
<h3>The distance between two landmarks is 123 meters</h3>
<em><u>Solution:</u></em>
We have to find the distance between two landmarks
<em><u>Use the law of cosines</u></em>
The third side of a triangle can be found when we know two sides and the angle between them

Here, angle between 90 meters and 130 meters is 65 degrees
From figure,
a = 90
b = 130
c = d
Therefore,

Thus, the distance between two landmarks is 123 meters
Assuming linear.
if k is at (3,4) and j is at (-8,7)
set. (y1,x1) and(y2,x2)
[(y2-y1),(x2-x1)]
[(-8-3),(7-4)]
distance from k to j is (-11,3)
now take coordinates of j and distance from. k->j and add them to get coordinate for L
[(-11+(-8)),(7+3)]= (-19,10)=L
The question doesn't provide enough information to get an accurate answer. Please provide more information or provide acceptable assumptions.
AB = BC
3x-4=5x-10
-4=2x-10
6=2x
3=x
AB = 3x-4
AB = 3 (3)-4
AB = 9-4
AB =5
Therefore, AB is 5.
BC = 5x-10
BC = 5(3)-10
BC = 15-10
BC= 5