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Given:
Vertices of JKLM are J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2).
To find:
The perimeter P of a parallelogram JKLM.
Solution:
Distance formula:
Using distance formula, we get
Similarly,
Now, perimeter P of ▱JKLM is
Therefore, the perimeter P of ▱JKLM is 19.2 units.
Answer:
sin(B) = 12/13
cos(B) = 5/13
tan(B) = 12/5
csc(B) = 13/12
sec(B) = 13/5
cot(B) = 5/12
Step-by-step explanation:
If ABC is a right triangle, assuming that ∠C = 90°, then the segment AB =13 is the hypotenuse and the other two sides are:
The six trigonometric functions of angle B are: