+343 1) We will use Law of cosines:
\((CB)^2=(CA)^2+(BA)^2-2(CA)(BA)cosA<=>2^2=3^2+4^2-(2\times3\times4cosA)<=>-21=-24cosA<=>cosA=\frac{7}{8}\)
So \(cosA=\frac{7}{8}\) or \(cosA=0.875\)
2)We will do pythagorean theorem at ACD : \((AC)^2= (DC)^2 + (AC)^2<=> 10^2=6^2 + (AC)^2<=>(AC)^2=10^2-6^2<=>AC=\sqrt{100-36}=\sqrt{64}=8\)
So \(AC=8 \)
\(sinB=\frac{AC}{BC}=\frac{8}{15}\) or \(sinB=0.533...\)
3) We will use Law of cosines (like 1):
\((BC)^2=(AB)^2+(AC)^2-2(AB)(AC)cos150<=>174=(CA)^2+81+15.588CA<=>(CA)^2+15.588CA-66=0\)
CA=x => \(x^2+15.588x-66=0<=>x=\frac{-15.588±\sqrt{506.985744}}{2}\)
\(x1=3.4641...\) , \(x2=-19.0521...\)
x2<0 We want distance witch always > 0
So \( AC=x1=3.4641...\)
BUT It's more easy to solve it like CPhill solution with roots!
This is a different way to solve it.
Hope I help you!
+128474 We can actually find an exact value for AC on the last one....we have....
(7√3)^2 = 9^2 + AC^2 - 2(9)(AC)cos (150)
147 = 81 + AC^2 - 18AC * [ -√3 / 2]
66 = AC^2 + 9√3 AC let AC = x and we have that
66 =x^2 + 9√3x rearrange as
x^2 + 9√3x - 66 = 0
Using the quadratic formula
x = -9√3 + √ [ (9√3)^2 - 4(-66) ] -9√3 + √[ 243 + 264]
_________________________ = _________________ =
2 2
-9√3 + √507 -9√3 + √[169 * 3 ] -9√3 + 13√3
___________ = ___________________ = __________ =
2 2 2
4√3
___ = 2√3 = AC
2
