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Help me with this maths question plz also explain why thanks

https://snag.gy/NiUh6u.jpg

YEEEEEET  Jan 6, 2018

Best Answer 

 #1
avatar+6945 
+1

 

m∠RAB   =   160° - 90°   =   70°

m∠ABR   =   360° - 90° - 220°   =   50°

m∠ARB   =   180° - 70° - 50°   =   60°

m∠TAR   =   180° - 160°   =   20°

 

Now we can use the law of sines to find the length of AR .

 

\(\frac{AR}{\sin 50°}\,=\,\frac{12}{\sin60°} \\~\\ AR\,=\,\frac{12\sin50°}{\sin60°}\)

 

Now we can use the law of cosines to find the length of TR.

 

\(TR^2\,=\,AT^2+AR^2-2(AT)(AR)\cos20° \\~\\ TR^2\,=\,5^2+(\frac{12\sin50°}{\sin60°})^2-2(5)(\frac{12\sin50°}{\sin60°})\cos20° \\~\\ TR^2\,=\,25+\frac{144\sin^250°}{\sin^260°}-\frac{120\sin50°\cos20°}{\sin60°} \)

                                                                                 Take the positive sqrt of both sides.

\(TR\,=\,\sqrt{25+\frac{144\sin^250°}{\sin^260°}-\frac{120\sin50°\cos20°}{\sin60°}} \\~\\ TR\,\approx\,6.158\quad\text{km}\)

hectictar  Jan 6, 2018
edited by hectictar  Jan 6, 2018
Sort: 

6+0 Answers

 #1
avatar+6945 
+1
Best Answer

 

m∠RAB   =   160° - 90°   =   70°

m∠ABR   =   360° - 90° - 220°   =   50°

m∠ARB   =   180° - 70° - 50°   =   60°

m∠TAR   =   180° - 160°   =   20°

 

Now we can use the law of sines to find the length of AR .

 

\(\frac{AR}{\sin 50°}\,=\,\frac{12}{\sin60°} \\~\\ AR\,=\,\frac{12\sin50°}{\sin60°}\)

 

Now we can use the law of cosines to find the length of TR.

 

\(TR^2\,=\,AT^2+AR^2-2(AT)(AR)\cos20° \\~\\ TR^2\,=\,5^2+(\frac{12\sin50°}{\sin60°})^2-2(5)(\frac{12\sin50°}{\sin60°})\cos20° \\~\\ TR^2\,=\,25+\frac{144\sin^250°}{\sin^260°}-\frac{120\sin50°\cos20°}{\sin60°} \)

                                                                                 Take the positive sqrt of both sides.

\(TR\,=\,\sqrt{25+\frac{144\sin^250°}{\sin^260°}-\frac{120\sin50°\cos20°}{\sin60°}} \\~\\ TR\,\approx\,6.158\quad\text{km}\)

hectictar  Jan 6, 2018
edited by hectictar  Jan 6, 2018
 #2
avatar+85819 
+1

Nice, hectictar.....!!!

 

 

cool cool cool

CPhill  Jan 6, 2018
 #3
avatar+6945 
+1

Thanks CPhill!! That is weird...is the picture blocked for you? I don't think it was blocked last night!

hectictar  Jan 6, 2018
 #4
avatar+85819 
+1

Yep......blocked for me as well.......odd  !!!

 

cool cool cool

CPhill  Jan 6, 2018
 #5
avatar+85819 
+1

It has suddenly reappeared......ghosts in the machine....maybe????

 

cool cool cool

CPhill  Jan 6, 2018
 #6
avatar+6945 
+1

LOL! I just tried re-uploading it...seemed to work!  laughlaugh

hectictar  Jan 6, 2018

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