Given tan g = 5/12, find cos g:
5
tan g = ———
12
The tangent function is, by definition, the quotient between sine and cosine:
sin g 5
———— = ———
cos g 12
Product of them extremes = product of the means
12 · sin g = 5 · cos g
Square both sides:
(12 · sin g)² = (5 · cos g)²
12² · sin² g = 5² · cos² g
144 · sin² g = 25 · cos² g
But sin² g = 1 – cos² g. Substitute it for sin² g into the equation above, and you have
144 · (1 – cos² g) = 25 · cos² g
Multiply out the brackets, and then isolate cos² g:
144 – 144 · cos² g = 25 · cos² g
144 = 25 · cos² g + 144 · cos² g
144 = 169 · cos² g
Divide both sides by 169:
144
cos² g = ———
169
12²
cos² g = ———
13²
cos² g = (12/13)²
Now, take the square root of both sides:
cos g = ± √(12/13)²
12
cos g = ± ———
13
The sign of cos g depends on which quadrant the angle g lies. As tan g = 5/12, which is positive, then g lies either in the 1st or the 3rd quadrant:
• If g lies in the 1st quadrant, then
5
cos g > 0 ⇒ cos g = ——— ✔
13
• If g lies in the 3rd quadrant, then
5
cos g < 0 ⇒ cos g = – ——— ✔
13
I hope this helps. =)
Tags: <em>trigonometric relation tangent cosine sine tan cos sin trig trigonometry</em>
4 is the answer to the question you asked
It is (a) -5/3
Slope could also mean gradient, so the gradient of a line is always the number in front of x in the equation y=Mx+c ... Hope it was helpful:)
Answer:
65.57 should be the correct answer!
Step-by-step explanation:
Here is the whole work I did, hope this helps!
100 - 17 = 83
79 x 0.83 = 65.57
Answer:
What you want to do is solve for m in whichever equation it's easier to so in, then substitute that in to the other equation. For instance:
X=3m-4
Y=6-m
Solve the 2nd equation to get m=6-y
Plug that into the first equation to get x=3(6-y)-4
X=18-3y-4
X-14=-3y
14-x=3y
Y=(14-x)/3
