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# Find AC, Find sin B, & Find cos A. Due in two more days

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Find cos A Find sinB Find AC Jun 5, 2019

#1
+2

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.

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Jun 5, 2019
edited by Dimitristhym  Jun 5, 2019
edited by Dimitristhym  Jun 5, 2019
edited by Dimitristhym  Jun 5, 2019
#4
+2

Actually, due to Pythagoras, AD=8 & AB=17. So the answer is 8/17. (For #2)

Aopshelp  Jun 6, 2019
#2
+1

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

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   Jun 5, 2019
#3
+2

Yes i calculate 150 rads :p

My bad I will edit it.

Dimitristhym  Jun 5, 2019