![\bf cos\left[tan^{-1}\left(\frac{12}{5} \right)+ tan^{-1}\left(\frac{-8}{15} \right) \right]\\ \left. \qquad \qquad \quad \right.\uparrow \qquad \qquad \qquad \uparrow \\ \left. \qquad \qquad \quad \right.\alpha \qquad \qquad \qquad \beta \\\\\\ \textit{that simply means }tan(\alpha)=\cfrac{12}{5}\qquad and\qquad tan(\beta)=\cfrac{-8}{5} \\\\\\ \textit{so, we're really looking for }cos(\alpha+\beta)](https://tex.z-dn.net/?f=%5Cbf%20cos%5Cleft%5Btan%5E%7B-1%7D%5Cleft%28%5Cfrac%7B12%7D%7B5%7D%20%20%5Cright%29%2B%20tan%5E%7B-1%7D%5Cleft%28%5Cfrac%7B-8%7D%7B15%7D%20%20%5Cright%29%20%5Cright%5D%5C%5C%0A%5Cleft.%20%5Cqquad%20%20%5Cqquad%20%20%5Cquad%20%20%20%5Cright.%5Cuparrow%20%5Cqquad%20%5Cqquad%20%20%5Cqquad%20%20%5Cuparrow%20%5C%5C%0A%5Cleft.%20%5Cqquad%20%20%5Cqquad%20%20%5Cquad%20%20%20%5Cright.%5Calpha%20%5Cqquad%20%5Cqquad%20%20%5Cqquad%20%20%5Cbeta%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bthat%20simply%20means%20%7Dtan%28%5Calpha%29%3D%5Ccfrac%7B12%7D%7B5%7D%5Cqquad%20and%5Cqquad%20tan%28%5Cbeta%29%3D%5Ccfrac%7B-8%7D%7B5%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bso%2C%20we%27re%20really%20looking%20for%20%7Dcos%28%5Calpha%2B%5Cbeta%29)
now.. hmmm -8/15 is rather ambiguous, since the negative sign is in front of the rational, and either 8 or 15 can be negative, now, we happen to choose the 8 to get the minus, but it could have been 8/-15
ok, well hmm so, the issue boils down to
![\bf tan(\theta)=\cfrac{opposite}{adjacent}\qquad thus \\\\\\ tan(\alpha)=\cfrac{12}{5}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a} \\\\\\ \textit{so, what is the hypotenuse "c"?}\\ \textit{ well, let's use the pythagorean theorem} \\\\\\ c=\sqrt{a^2+b^2}\implies c=\sqrt{25+144}\implies c=\sqrt{169}\implies \boxed{c=13}\\\\ -----------------------------\\\\ \textit{this simply means }\boxed{cos(\alpha)=\cfrac{5}{13}\qquad \qquad sin(\alpha)=\cfrac{12}{13} }](https://tex.z-dn.net/?f=%5Cbf%20tan%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Badjacent%7D%5Cqquad%20thus%0A%5C%5C%5C%5C%5C%5C%0Atan%28%5Calpha%29%3D%5Ccfrac%7B12%7D%7B5%7D%5Ccfrac%7B%5Cleftarrow%20opposite%3Db%7D%7B%5Cleftarrow%20%20adjacent%3Da%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bso%2C%20what%20is%20the%20hypotenuse%20%22c%22%3F%7D%5C%5C%0A%5Ctextit%7B%20well%2C%20let%27s%20use%20the%20pythagorean%20theorem%7D%0A%5C%5C%5C%5C%5C%5C%0Ac%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B25%2B144%7D%5Cimplies%20c%3D%5Csqrt%7B169%7D%5Cimplies%20%5Cboxed%7Bc%3D13%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ctextit%7Bthis%20simply%20means%20%7D%5Cboxed%7Bcos%28%5Calpha%29%3D%5Ccfrac%7B5%7D%7B13%7D%5Cqquad%20%5Cqquad%20sin%28%5Calpha%29%3D%5Ccfrac%7B12%7D%7B13%7D%0A%7D)
now, let's take a peek at the second angle, angle β
![\bf tan(\beta)=\cfrac{-8}{15}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a} \\\\\\ \textit{again, let's find "c", or the hypotenuse} \\\\\\ c=\sqrt{15^2+(-8)^2}\implies c=\sqrt{289}\implies \boxed{c=17}\\\\ -----------------------------\\\\ thus\qquad \boxed{cos(\beta)=\cfrac{15}{17}\qquad \qquad sin(\beta)=\cfrac{-8}{17}}](https://tex.z-dn.net/?f=%5Cbf%20tan%28%5Cbeta%29%3D%5Ccfrac%7B-8%7D%7B15%7D%5Ccfrac%7B%5Cleftarrow%20opposite%3Db%7D%7B%5Cleftarrow%20adjacent%3Da%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bagain%2C%20let%27s%20find%20%22c%22%2C%20or%20the%20hypotenuse%7D%0A%5C%5C%5C%5C%5C%5C%0Ac%3D%5Csqrt%7B15%5E2%2B%28-8%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B289%7D%5Cimplies%20%5Cboxed%7Bc%3D17%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Athus%5Cqquad%20%5Cboxed%7Bcos%28%5Cbeta%29%3D%5Ccfrac%7B15%7D%7B17%7D%5Cqquad%20%5Cqquad%20sin%28%5Cbeta%29%3D%5Ccfrac%7B-8%7D%7B17%7D%7D)
now, with that in mind, let's use the angle sum identity for cosine
Answer:
Я не говорю по-русски, но я думаю, что у нее 1/64 домашнее задание после 6 часов.
Step-by-step explanation:
Каждый час она завершает половину работы предыдущего часа.
c*y*k*a b*l*y*a*t comrade
Answer:
0.00114
Step-by-step explanation:
Divide length value by 5280
Answer:
I would but...
Step-by-step explanation:
where the question at
Answer:
w = (S - a) / l
Step-by-step explanation:
Given:
S = a + lw
Find a formula for w in terms of s
S = a + lw
Subtract a from both sides
S - a = a - a + lw
S - a = lw
Divide both sides by l
(S - a) / l = lw / l
(S - a) / l = w
Therefore,
w = (S - a) / l
It can also be written as (S - a ) ÷ l