+0  
 
0
216
8
avatar+124 

1)Find BC.

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

 

2)Find AC. 

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

 

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

 

4)Find AC.

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

 

5)Find sin B.

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

 

6)Find cos A.

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

 

7)In triangle ABC, sin A = 4/5. Find cos B.

Picture: https://latex.artofproblemsolving.com/c/8/3/c83240e209d830cd2275ed9f771185a1d1c17733.png

 

8)Find DB in the diagram below.

Picture: https://latex.artofproblemsolving.com/f/6/f/f6f917a5c86b363a5cdf8871300d061dcd54fc0b.png

 Jun 12, 2018
 #1
avatar+20847 
+1

1)

Find BC.

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

 

cos-rule:

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

 

laugh

 Jun 13, 2018
 #2
avatar+20847 
+1

2)

Find AC. 

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

 

sin-rule

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

 

laugh

 Jun 13, 2018
 #3
avatar+20847 
+1

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

 

sin-rule

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

 

laugh

 Jun 13, 2018
 #4
avatar+20847 
+1

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

 

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

 

laugh

 Jun 13, 2018
 #5
avatar+20847 
+1

6)

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

 

cos-rule

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

 

laugh

 Jun 13, 2018
 #6
avatar+20847 
+1

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

 

sin-rule

 

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

 

 

laugh

 Jun 13, 2018
 #7
avatar+20847 
+1

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

 

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

 

laugh

 Jun 13, 2018
 #8
avatar+20847 
0

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

 

cos-rule

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

 

laugh

 Jun 13, 2018

17 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.