PLS HELP

1)

Find BC.

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

cos-rule:

\(\begin{array}{|rcll|} \hline BC^2 &=& 6^2 +(7\sqrt{2})^2-2\cdot 6 \cdot7\sqrt{2}\cdot \cos(45^{\circ}) \quad & | \quad \cos(45^{\circ}) = \frac{\sqrt{2}}{2} \\ BC^2 &=& 36 + 49\cdot 2 -2\cdot 6 \cdot7\sqrt{2}\cdot \frac{\sqrt{2}}{2} \\ BC^2 &=& 36 + 98 -2\cdot 6 \cdot7 \cdot \frac{2}{2} \\ BC^2 &=& 36 + 98 -2\cdot 6 \cdot7 \\ BC^2 &=& 50 \\ BC &=&\sqrt{ 50} \\ BC &=&\sqrt{ 5^2\cdot 2} \\ BC &=& 5\sqrt{2} \\ \hline \end{array}\)

2)

Find AC. 

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

sin-rule

\(\begin{array}{|rcll|} \hline \dfrac{AC}{\sin(45^{\circ})} &=& \dfrac{30} { \sin\Big(180^{\circ}-(75^{\circ}+45^{\circ}) \Big) } \quad & | \quad \small{\sin\Big(180^{\circ}-(75^{\circ}+45^{\circ}) \Big) = \sin(75^{\circ}+45^{\circ})} \\\\ \dfrac{AC}{\sin(45^{\circ})} &=& \dfrac{30} { \sin(75^{\circ}+45^{\circ}) } \\\\ \dfrac{AC}{\sin(45^{\circ})} &=& \dfrac{30} { \sin(120^{\circ}) } \\\\ AC &=& \dfrac{30\cdot \sin(45^{\circ})} { \sin(120^{\circ}) } \quad & | \quad \sin(45^{\circ}) = \frac{\sqrt{2}} {2} \\\\ AC &=& \dfrac{30\cdot \frac{\sqrt{2}} {2}} { \sin(120^{\circ}) } \quad & | \quad \sin(120^{\circ} = \frac{\sqrt{3}} {2} \\\\ AC &=& \dfrac{30\cdot \frac{\sqrt{2}} {2}} { \frac{\sqrt{3}} {2} } \\\\ AC &=& \dfrac{30\cdot \sqrt{2} } { \sqrt{3} } \cdot \frac{ \sqrt{3} } {\sqrt{3}} \\\\ AC &=& \dfrac{30\cdot \sqrt{2} \sqrt{3} } { 3 } \\\\ AC &=& 10\cdot \sqrt{6} \\ AC &=& 24.4948974278 \\ \hline \end{array}\)

3)
In triangle ABC, CA = 4sqrt2, CB = 4sqrt3, and angle A = 60 degrees. What is B in degrees?

sin-rule

\(\begin{array}{|rcll|} \hline \dfrac{\sin(B)} {4\sqrt{2}} &=& \dfrac{\sin(60^{\circ})} {4\sqrt{3}} \\\\ \dfrac{\sin(B)} { \sqrt{2}} &=& \dfrac{\sin(60^{\circ})} { \sqrt{3}} \quad & | \quad \sin(60^{\circ}) = \frac{ \sqrt{3} } {2} \\\\ \dfrac{\sin(B)} {\sqrt{2}} &=& \dfrac{ \sqrt{3} } {2\sqrt{3}} \\\\ \dfrac{\sin(B)} {\sqrt{2}} &=& \dfrac{ 1 } {2 } \\\\ \sin(B) &=& \dfrac{ \sqrt{2} } {2 } \\\\ \mathbf{B} & \mathbf{=} & \mathbf{45^{\circ}} \\ \hline \end{array}\)

5)
Find sin B.
Picture:https://latex.artofproblemsolving.com/4/7/a/47a312412eff41c09e4ce21e5edbb7e52e8b151c.png

\(\begin{array}{|rcll|} \hline AD^2 + 6^2 &=& 10^2 \\ AD^2 & = & 10^2 - 6^2 \\ AD^2 & = & 64 \\ \mathbf{AD} & \mathbf{=} & \mathbf{ 8 } \\\\ AB^2 &=& 15^2 + AD^2 \quad & | \quad AD^2 = 64 \\ AB^2 &=& 15^2 + 64 \\ AB^2 &=& 289 \\ \mathbf{AB} & \mathbf{=} & \mathbf{17 } \\\\ \sin(B) &=& \dfrac {AD} {AB} \\\\ \mathbf{\sin(B)} & \mathbf{=} & \mathbf{ \dfrac {8} {17}} \\ \hline \end{array}\)

6)

Find cos A.
Picture: https://latex.artofproblemsolving.com/6/4/a/64ae47768bae6b24ab792c44a92c5de244955cad.png

cos-rule

\(\begin{array}{|rcll|} \hline 2^2 &=& 3^2+4^2-2\cdot 3 \cdot 4\cdot \cos(A) \\ 4 &=& 9+ 16 -24\cdot \cos(A) \\ 4 &=& 25 -24\cdot \cos(A) \quad & | \quad +24\cdot \cos(A) \\ 4 +24\cdot \cos(A) &=& 25 \quad & | \quad -4 \\ 24\cdot \cos(A) &=& 25 -4 \\ 24\cdot \cos(A) &=& 21 \quad & | \quad : 24 \\ \cos(A) &=& \dfrac{21} {24} \\\\ \mathbf{\cos(A)} & \mathbf{=} & \mathbf{ \dfrac{7} {8} } \\ \hline \end{array}\)

7)
In triangle ABC, sin A = 4/5. Find cos B.
Picture: https://latex.artofproblemsolving.com/c/8/3/c83240e209d830cd2275ed9f771185a1d1c17733.png

sin-rule

\(\begin{array}{|rcll|} \hline \dfrac{\sin(B)} {10} &=& \dfrac{\sin(A)}{17} \\\\ \sin(B) &=& \dfrac{10\cdot \sin(A)}{17} \quad & | \quad \sin(A) = \dfrac {4}{5} \\\\ \sin(B) &=& \dfrac{10\cdot \dfrac {4}{5}}{17} \\\\ \sin(B) &=& \dfrac{2\cdot 4 }{17} \\\\ \mathbf{\sin(B)} & \mathbf{=} & \mathbf{ \dfrac{8}{17} } \\\\ \cos(B) &=& \sqrt{1-\sin^2(B)} \\ \\ \cos(B) &=& \sqrt{1- \left(\dfrac{8}{17} \right)^2 } \\ \\ \cos(B) &=& \sqrt{1- \dfrac{8^2}{17^2} } \\ \\ \cos(B) &=& \dfrac{ \sqrt{17^2-8^2} } {17} \\ \\ \cos(B) &=& \dfrac{ \sqrt{225} } {17} \\ \\ \mathbf{\cos(B)} & \mathbf{=} & \mathbf{ \dfrac{ 15 } {17} } \\ \hline \end{array}\)

8)
Find DB in the diagram below.
Picture: https://latex.artofproblemsolving.com/f/6/f/f6f917a5c86b363a5cdf8871300d061dcd54fc0b.png

\(\begin{array}{|rcll|} \hline \angle C &=& 180^{\circ} - ( 30^{\circ} + 45^{\circ}) \\ \angle C &=& 105^{\circ} \\\\ \angle ACD &=& \angle C - 60^{\circ} \\ \angle ACD &=& 105^{\circ} - 60^{\circ} \\ \angle ACD &=& 45^{\circ} \\\\ \dfrac{DC} {\sin(30^{\circ})} &=& \dfrac{2}{\sin(ACD)} \\\\ \dfrac{DC} {\sin(30^{\circ})} &=& \dfrac{2}{\sin(45^{\circ}) } \\\\ DC &=& \dfrac{2\cdot \sin(30^{\circ}) }{\sin(45^{\circ}) } \quad & | \quad \sin(30^{\circ}) = \dfrac{1}{2} \\\\ DC &=& \dfrac{2\cdot \dfrac{1}{2} }{\sin(45^{\circ}) } \\\\ DC &=& \dfrac{1}{\sin(45^{\circ}) } \quad & | \quad \sin(45^{\circ}) = \dfrac{\sqrt{2}}{2} \\\\ DC &=& \dfrac{1}{\dfrac{\sqrt{2}}{2} } \\\\ DC &=& \dfrac{2}{ \sqrt{2} }\cdot \dfrac{\sqrt{2}}{\sqrt{2}} \\\\ DC &=& \dfrac{2\sqrt{2}}{2 } \\\\ \mathbf{DC} & \mathbf{=} & \mathbf{ \sqrt{2} } \\\\ \dfrac{DB} {\sin(60^{\circ})} &=& \dfrac{\sqrt{2}}{\sin(45^{\circ})} \\\\ DB &=& \dfrac{\sqrt{2}\cdot \sin(60^{\circ})}{\sin(45^{\circ})} \quad & | \quad \sin(45^{\circ}) = \dfrac{\sqrt{2}}{2} \\\\ DB &=& \dfrac{\sqrt{2}\cdot \sin(60^{\circ})}{\dfrac{\sqrt{2}}{2}} \\\\ DB &=& 2\cdot \sin(60^{\circ}) \quad & | \quad \sin(60^{\circ}) = \dfrac{\sqrt{3}}{2} \\\\ DB &=& 2\cdot \dfrac{\sqrt{3}}{2} \\\\ \mathbf{DB} & \mathbf{=} & \mathbf{ \sqrt{3} } \\ \hline \end{array}\)

4)
Find AC.
Picture: https://latex.artofproblemsolving.com/9/1/d/91d520533f0fb8cb325932013b2aebf0db9bde54.png

cos-rule

\(\begin{array}{|rcll|} \hline \mathbf{(7\sqrt{3})^2 }& \mathbf{=}& \mathbf{ 9^2 + AC^2 -2\cdot 9 \cdot AC \cdot \cos(150^{\circ}) } \\ 49\cdot 3 &=& 81 + AC^2 - 18 \cdot AC \cdot \cos(150^{\circ}) \quad & | \quad \small{\cos(150^{\circ}) = -\dfrac{\sqrt{3}}{2}} \\ 49\cdot 3 &=& 81 + AC^2 + 18 \cdot AC \cdot \dfrac{\sqrt{3}}{2} \\ 147 &=& 81 + AC^2 + 9 \cdot \sqrt{3} \cdot AC \quad & | \quad - 81 \\ 66 &=& AC^2 + 9 \sqrt{3} \cdot AC \\ AC^2 + 9 \sqrt{3} \cdot AC - 66 &=& 0 \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm \sqrt{ (9\sqrt{3})^2 - 4\cdot (-66) } } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm \sqrt{ (81\cdot 3 + 264 } } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm \sqrt{507} } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm \sqrt{13^2\cdot 3} } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} \pm 13\sqrt{3} } {2} \\\\ AC &=& \dfrac{-9 \sqrt{3} + 13\sqrt{3} } {2} \quad & | \quad \text{AC}> 0\ !\\\\ AC &=& \dfrac{4\sqrt{3} } {2} \\\\ \mathbf{AC} & \mathbf{=} & \mathbf{ 2\sqrt{3} } \\ \hline \end{array}\)