In Exercises 6 and 7, find sin J, sin K, cos J, cos K, tan J and tan K. Write each answer as a fraction and as a decimal rounded to four places.

Refer to the diagram below for the answers

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Explanation:

The rules I'm using are

The opposite side is the leg furthest from the reference angle. For example, if J is the reference angle, then leg LK is the opposite side. The adjacent side would be JL for reference angle J. The opposite and adjacent swap when you swap the reference angle.

The hypotenuse is always the longest side and always opposite the largest angle (always opposite the 90 degree angle).

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An example calculation would be:

sin(angle) = opposite/hypotenuse

sin(J) = LK/JL

sin(J) = 30/34

sin(J) = 0.8824

I'm referring to the triangle with sides 16,30,34. The other trig ratios are handled in pretty much the same way (use the rules mentioned above).

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841

i do not know the answer

step-by-step explanation: