let's notice something, angles α and β are both in the I Quadrant, and on the first quadrant the x-coordinate/cosine and y-coordinate/sine are both positive.
![\bf \textit{Sum and Difference Identities} \\\\ cos(\alpha - \beta)= cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(\alpha)=\cfrac{\stackrel{opposite}{15}}{\stackrel{hypotenuse}{17}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BSum%20and%20Difference%20Identities%7D%20%5C%5C%5C%5C%20cos%28%5Calpha%20-%20%5Cbeta%29%3D%20cos%28%5Calpha%29cos%28%5Cbeta%29%20%2B%20sin%28%5Calpha%29sin%28%5Cbeta%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20sin%28%5Calpha%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B15%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B17%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%20side%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D)
![\bf \pm\sqrt{17^2-15^2}=a\implies \pm\sqrt{64}=a\implies \pm 8 = a\implies \stackrel{I~Quadrant}{\boxed{+8=a}} \\\\[-0.35em] ~\dotfill\\\\ cos(\beta)=\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ 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}](https://tex.z-dn.net/?f=%5Cbf%20%5Cpm%5Csqrt%7B17%5E2-15%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B64%7D%3Da%5Cimplies%20%5Cpm%208%20%3D%20a%5Cimplies%20%5Cstackrel%7BI~Quadrant%7D%7B%5Cboxed%7B%2B8%3Da%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20cos%28%5Cbeta%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B3%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B5%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Bopposite%20side%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-a%5E2%7D%3Db%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D)
![\bf \pm\sqrt{5^2-3^2}=b\implies \pm\sqrt{16}=b\implies \pm 4=b\implies \stackrel{\textit{I~Quadrant}}{\boxed{+4=b}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cpm%5Csqrt%7B5%5E2-3%5E2%7D%3Db%5Cimplies%20%5Cpm%5Csqrt%7B16%7D%3Db%5Cimplies%20%5Cpm%204%3Db%5Cimplies%20%5Cstackrel%7B%5Ctextit%7BI~Quadrant%7D%7D%7B%5Cboxed%7B%2B4%3Db%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf cos(\alpha - \beta)=\stackrel{cos(\alpha)}{\left( \cfrac{8}{17} \right)}\stackrel{cos(\beta)}{\left( \cfrac{3}{5} \right)}+\stackrel{sin(\alpha)}{\left( \cfrac{15}{17} \right)}\stackrel{sin(\beta)}{\left( \cfrac{4}{5} \right)}\implies cos(\alpha - \beta)=\cfrac{24}{85}+\cfrac{60}{85} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill cos(\alpha - \beta)=\cfrac{84}{85}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20cos%28%5Calpha%20-%20%5Cbeta%29%3D%5Cstackrel%7Bcos%28%5Calpha%29%7D%7B%5Cleft%28%20%5Ccfrac%7B8%7D%7B17%7D%20%5Cright%29%7D%5Cstackrel%7Bcos%28%5Cbeta%29%7D%7B%5Cleft%28%20%5Ccfrac%7B3%7D%7B5%7D%20%5Cright%29%7D%2B%5Cstackrel%7Bsin%28%5Calpha%29%7D%7B%5Cleft%28%20%5Ccfrac%7B15%7D%7B17%7D%20%5Cright%29%7D%5Cstackrel%7Bsin%28%5Cbeta%29%7D%7B%5Cleft%28%20%5Ccfrac%7B4%7D%7B5%7D%20%5Cright%29%7D%5Cimplies%20cos%28%5Calpha%20-%20%5Cbeta%29%3D%5Ccfrac%7B24%7D%7B85%7D%2B%5Ccfrac%7B60%7D%7B85%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20cos%28%5Calpha%20-%20%5Cbeta%29%3D%5Ccfrac%7B84%7D%7B85%7D~%5Chfill)
Answer:
move each point of the triangle 5 units to the right and one unit up.
G is at -3,-1
-3+5 = 2
-1+1 = 0
2,0 is the coordinates of G' or the new G
T is at -1,-1
-1+5 = 4
-1+1 = 0
4,0 is the coordinates of T' or the new T
Last of all, B is at -3,-5
-3+5 = 2
-5+1 = -4
2,-4 is the coordinates of the B' or the new B.
Step-by-step explanation:
Answer:
a. (-3, 2).
b. √65
c. (x + 3)^2 + (y - 2)^2 = 65
Step-by-step explanation:
a. The center is the midpoint of the diameter PQ.
= (-10+4)/2, (-2+6)/2
= (-3, 2).
b. The radius is the distance from the center to a point on the circle.
Take the point (4, 6):
The radius = √((-3-4)^2 + (2-6)^2)
= √65.
c. The equation of the circle is:
Using the standard form
(x - h)^2 + (y - k)^2 = r^2 where (h, k) is the center and r = the radius:
it is (x - (-3)^2 + (y - 2) = 65
= (x + 3)^2 + (y - 2)^2 = 65.
