![\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad \qquad % cosine cos(\theta)=\cfrac{adjacent}{hypotenuse} \\ \quad \\ % tangent tan(\theta)=\cfrac{opposite}{adjacent} \qquad \qquad % secant sec(\theta)=\cfrac{hypotenuse}{adjacent}\\\\ -----------------------------\\\\ sec(\theta)=\cfrac{6}{5}\cfrac{\leftarrow hypotenuse}{\leftarrow adjacent} \\\\\\ ](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Bhypotenuse%7D%0A%5Cqquad%20%5Cqquad%20%0A%25%20cosine%0Acos%28%5Ctheta%29%3D%5Ccfrac%7Badjacent%7D%7Bhypotenuse%7D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%25%20tangent%0Atan%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Badjacent%7D%0A%5Cqquad%20%5Cqquad%20%0A%25%20secant%0Asec%28%5Ctheta%29%3D%5Ccfrac%7Bhypotenuse%7D%7Badjacent%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Asec%28%5Ctheta%29%3D%5Ccfrac%7B6%7D%7B5%7D%5Ccfrac%7B%5Cleftarrow%20hypotenuse%7D%7B%5Cleftarrow%20adjacent%7D%0A%5C%5C%5C%5C%5C%5C%0A)
![\bf \textit{so.. using the pythagorean theorem, we get} \\\\\\ c^2=a^2+b^2\implies \pm \sqrt{c^2-a^2}=b\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite \end{cases} \\\\\\ \textit{that simply means that }\pm \sqrt{6^2-5^2}=b\implies \pm \sqrt{11}=b](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bso..%20using%20the%20pythagorean%20theorem%2C%20we%20get%7D%0A%5C%5C%5C%5C%5C%5C%0Ac%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%20%5Csqrt%7Bc%5E2-a%5E2%7D%3Db%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ac%3Dhypotenuse%5C%5C%0Aa%3Dadjacent%5C%5C%0Ab%3Dopposite%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bthat%20simply%20means%20that%20%7D%5Cpm%20%5Csqrt%7B6%5E2-5%5E2%7D%3Db%5Cimplies%20%5Cpm%20%5Csqrt%7B11%7D%3Db)
but.. which is it, the positive or negative version of "b"? well.... we know tangent is < 0, that's another way of saying, tangent is negative
now.. .tangent is opposite over adjacent... for tangent to be negative, the sign of those two must differ
so.. where does that happens? well, it happens on the 2nd and 4th quadrants, so..... which quadrant then?
well, we know the hypotenuse is always positive, is just the radius anyway, and in this case is 6
but the adjacent is positive 5, that means the adjacent is positive, thus the opposite must be negative, and that happens on the 4th quadrant
so that means
![\bf -\sqrt{11}=b](https://tex.z-dn.net/?f=%5Cbf%20-%5Csqrt%7B11%7D%3Db)
so... now, you have all three sides, the hypotenuse, the adjacent, and the opposite, so, just fill those in, in the ratios for cosine, sine and tangent
Answer:
3x=9.5, you would divide 3 on each side to get x by itself, so (3x)/3=9.5/3, which would lead to be x=9.5/3 or x= 3 and 1/6th, 3 + 1/6
Step-by-step explanation:
Sorry, it's late, and I'm a bad explainer.
The error is adding (2x-12) with x and 30. This is wrong because you are adding the angles inside the triangle and you are assuming that (2x - 12) is the unlabeled angle INSIDE the triangle, when it is the exterior angle/outside of the triangle.
A straight line is also 180°.
(2x - 12) + ? = 180
30 + x + ? = 180
If you look at the equations, and put parentheses around 30 + x, (30 + x) and (2x - 12) should be the SAME NUMBER. So you could set them equal to each other to find x. (or you could also look at the picture and see that they both need/are missing the same angle)
2x - 12 = 30 + x
x = 42
Now you plug 42 into the exterior angle equation
2(42) - 12 = 84 - 12 = 72°
Answer:
The ratio of apple to orange is 3 to 7 because if you simplify 15 to 21, you get that answer.
Step-by-step explanation:
Ok so find 1's
remember that 4/4=1 also x/x=1 those are ones
also remember
if you had
10/30 that equals 10/10 times 1/3=1 times 1/3
so find ones
4/10=2/2 times 2/5=1 times 2/5=2/5