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1.Find BC

https://latex.artofproblemsolving.com/f/e/4/fe45662f573f87e30f9f30d0b1ef1f94b4c22fc2.png

 

2.Find AC

https://latex.artofproblemsolving.com/6/8/d/68dd918aac5d85a0b2a5b3b9c6a79eabd3669f01.png

 

3.Find AC

https://latex.artofproblemsolving.com/9/1/d/91d520533f0fb8cb325932013b2aebf0db9bde54.png

 

4. Find sin B

https://latex.artofproblemsolving.com/4/7/a/47a312412eff41c09e4ce21e5edbb7e52e8b151c.png

 

5.Find cos A

https://latex.artofproblemsolving.com/6/4/a/64ae47768bae6b24ab792c44a92c5de244955cad.png


 

 May 18, 2020
 #1
avatar
0

That's a lot of trig.

 May 18, 2020
 #2
avatar+24949 
+1

1.

Find BC

 

cos-rule

\(\begin{array}{|rcll|} \hline BC^2 &=& (7\sqrt{2})^2+6^2-2*7\sqrt{2}*6\cos(45^\circ) \quad | \quad \cos(45^\circ)=\dfrac{\sqrt{2} }{2} \\\\ BC^2 &=& 49*2+6^2-2*7\sqrt{2}*6\dfrac{\sqrt{2} }{2} \\\\ BC^2 &=& 98+36-2*7*6 \\\\ BC^2 &=& 98+36-84 \\ BC^2 &=& 50 \\ BC^2 &=& 2*25 \\ \mathbf{BC} &=& \mathbf{5\sqrt{2}} \\ \hline \end{array} \)

 

 

laugh

 May 18, 2020
 #3
avatar+24949 
+1

2.

Find AC

 

 

\(\angle C = 180^\circ - (75^\circ+45^\circ) \\ \angle C = 60^\circ \)

 

\(\begin{array}{|rcll|} \hline \dfrac{AC}{\sin(45^\circ)} &=& \dfrac{30}{\sin(60^\circ)} \\\\ AC &=& \dfrac{30\sin(45^\circ)}{\sin(60^\circ)} \quad | \quad \sin(45^\circ)=\dfrac{\sqrt{2} }{2},\ \sin{60^\circ}=\dfrac{\sqrt{3} }{2} \\\\ AC &=& 30*\dfrac{\sqrt{2}}{2} *\dfrac{2}{\sqrt{3}} \\\\ AC &=& \dfrac{30\sqrt{2}}{\sqrt{3}} \\\\ AC &=& \dfrac{30\sqrt{2}}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\\\ AC &=& \dfrac{30\sqrt{2}\sqrt{3}}{3} \\\\ \mathbf{AC} &=& \mathbf{10\sqrt{6}} \\ \hline \end{array} \)

 

laugh

 May 18, 2020
edited by heureka  May 18, 2020
 #4
avatar+24949 
+1

3.

Find AC

 

cos-rule

\(\begin{array}{|rcll|} \hline (7\sqrt{3})^2 &=& 9^2 + AC^2 -2*9*AC\cos(150^\circ) \quad | \quad \cos(150^\circ)=-\dfrac{\sqrt{3} }{2} \\\\ 49*3 &=& 81 + AC^2 + 2*9*AC\dfrac{\sqrt{3} }{2} \\\\ 147 &=& 81 + AC^2 + 9*AC \sqrt{3} \\ 66 &=& AC^2 + 9\sqrt{3}AC \\ \mathbf{AC^2 + 9\sqrt{3}AC -66} &=& \mathbf{0} \\\\ AC &=& \dfrac{ -9\sqrt{3} \pm\sqrt{81*3-4*(-66) } } {2} \\\\ AC &=& \dfrac{ -9\sqrt{3} \pm\sqrt{81*3+264 } } {2} \\\\ AC &=& \dfrac{ -9\sqrt{3} \pm\sqrt{507} } {2} \\\\ AC &=& \dfrac{ -9\sqrt{3} \pm\sqrt{13^2*3} } {2} \\\\ AC &=& \dfrac{ -9\sqrt{3} \pm 13\sqrt{3} } {2} \\\\ AC &=& \dfrac{ -9\sqrt{3} \mathbf{+} 13\sqrt{3} } {2} \\\\ AC &=& \dfrac{ 4\sqrt{3} } {2} \\\\ \mathbf{AC} &=& \mathbf{2\sqrt{3} } \\ \hline \end{array}\)

 

laugh

 May 18, 2020
 #5
avatar+24949 
+1

4.

Find sin B

 

\(\begin{array}{|rcll|} \hline \cos(C) &=& \dfrac{6}{10} \\\\ \mathbf{\cos{C}} &=& \mathbf{\dfrac{3}{5}} \\ \hline \sin(C) &=& \sqrt{1-\cos^2(C)} \\ \sin(C) &=& \sqrt{1-\dfrac{9}{25}} \\ \sin(C) &=& \sqrt{\dfrac{16}{25}} \\ \mathbf{\sin(C)} &=& \mathbf{ \dfrac{4}{5} } \\ \hline \end{array}\)

 

cos-rule

\(\begin{array}{|rcll|} \hline AB^2 &=& 10^2 + (15+6)^2 -2*10(15+6)\cos(C) \quad | \quad \cos(C)=\dfrac{3}{5} \\\\ AB^2 &=& 100 + 441 -2*10*21*\dfrac{3}{5} \\ AB^2 &=& 541 -2*2*21*3 \\ AB^2 &=& 541 -252 \\ AB^2 &=& 289 \\ \mathbf{AB} &=& \mathbf{17} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \dfrac{\sin(B)}{10} &=& \dfrac{\sin(C)}{AB} \\\\ \sin(B) &=& \dfrac{10\sin(C)}{AB} \quad | \quad \sin(C)=\dfrac{4}{5},\ AB = 17 \\\\ \sin(B) &=& \dfrac{10*\dfrac{4}{5}}{17} \\\\ \sin(B) &=& \dfrac{2*4}{17} \\\\ \mathbf{ \sin(B) } &=& \mathbf{ \dfrac{8}{17} } \\ \hline \end{array}\)

 

laugh

 May 18, 2020
 #6
avatar+24949 
+1

5.

Find cos A

 

cos-rule

\(\begin{array}{|rcll|} \hline 2^2 &=& 3^2+4^2-2*3*4*\cos(A) \\\\ 4 &=& 9+16-24\cos(A) \\ 4 &=& 25-24\cos(A) \\ 24\cos(A) &=& 25- 4 \\ 24\cos(A) &=& 21 \\ \cos(A) &=& \dfrac{21}{24} \\ \mathbf{\cos(A)} &=& \mathbf{\dfrac{3}{8} } \\ \hline \end{array}\)

 

laugh

 May 18, 2020

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