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So, which options are the correct ratios?

A  no...that is TAN R

B no  the terms need to be squared to equal 1

C Yes    Cos p = adj/hyp = QP/hyp   and sine r = opp/hyp = QP/hyp

D No   sin and cos are both between 1 and -1 inclusive and would never add to 90

First we need to find the hypotenuse using the Pythagorean theorem.

5² + 12²  =  hypotenuse²

25 + 144  =  hypotenuse²

169  =  hypotenuse²

13  =  hypotenuse

sin A  =  side opposite A / hypotenuse  =  12 / 13

cos A  =  side adjacent to A / hypotenuse  =  5 / 13

sin B  =  side opposite B / hypotenuse  =  5 / 13

tan A  =  side opposite A / side adjacent to A  =  12 / 5

tan B  =  side opposite B / side adjacent to B  =  5 / 12

Ah shoot, my bad! I got mixed up >.<

f / r² + a   =   n + k

                                    Subtract  a  from both sides of the equation.

f / r²   =   n + k - a

                                    Multiply both sides of the equation by  r² .

f   =   r²(n + k - a)

                                    Divide both sides of the equation by  (n + k - a)

f / (n + k - a)  =  r²

                                    Take the ± square root of both sides.

±√[ f / (n + k - a) ]  =  r

r  =  ±√[ f / (n + k - a) ]

ax² + bx + c  =  0     is called a quadratic equation. If you solve this equation for  x  you'll get

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)    This is the quadratic formula.

Nevermind I got the answer! :D

Thanks

All I see is "What is the quadratic formula" so I'm guessing you just want to know what that is (if there is an image, i just can't see it)

The quadratic formula is ax^2 + bx + c = 0

Oh no, of course not, Manuel, what would make you think that?

\(\quad(x^4y^7)^{\frac23}\\ =\\ \quad(x^4)^{\frac23}(y^7)^{\frac23}\\ =\\ \quad x^{4\cdot\frac23}y^{7\cdot\frac23}\\ =\\ \quad x^\frac83y^\frac{14}3 \\ =\\ \quad x^\frac63x^\frac23y^\frac{12}3y^\frac23\\ =\\ \quad x^2x^\frac23y^4y^\frac23 \\ =\\ \quad x^2y^4(xy)^\frac23\\ =\\ \quad x^2y^4\sqrt[3]{(xy)^2} \\ =\\ \quad x^2y^4\sqrt[3]{x^2y^2}\)

Cos T   has  the same value  as  Cos P  =  10/15  = 2/3

So

Cos T   =  2/3  =  6/x

Cross-multiply

2x   = 3 * 6

2x  =  18     divide both sides by 2

x  =  9   

Remember that   sin (x)  =  cos (90 - x)

So

sin (23)  =  cos (90 - 23)  =  cos (67)    ....so  x    =   67°

so you are calling me retardo?

I do wonder at the marvels of the sub-human brain. How did you ever figured out logging on to the forum?  Someone must have showed you several dozen times. I’m sure that’s how the other retardos learned.

Hyp  =

sqrt ( 15^2 + 25^2)  =    sqrt ( 850)  =  29.15  ft.

The best thing to do is to draw a mock figure

           R

P                  Q

PQ  is the hypotenuse

RQ  is the leg adjacent to angle Q

So.....cos Q  =   adjacent  / hypotenuse   =  RQ / PQ  =  QR / PQ

Woo I was right!! 

Thank you :D

Note that   3-4-5  is a well-known RT

Any multiple  of this set is also a RT

So

5 (3 - 4 - 5)   =    15 - 20 - 25

The other side is

sqrt ( 360^2 - 160^2)  ≈  322.49 ft

So.....the perimeter is  2  (160 + 322.49)  = 964.98 ft

Ooops.....I did !  Thanx Chris.....       15/3 = 3 in my world today.

Ty :D