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I have a triangle where angle A is 32°  and i know AC side is 8,2 cm and the BC side is 6,7 cm the mission is to solve the last remaining side which would be AB but theres a condition, the triangle area must be less than 10cm^2.

Guest May 17, 2018

Best Answer 

 #1
avatar+7336 
+2

 

By the Law of Sines,

 

\(\frac{\sin B}{8.2}=\frac{\sin 32°}{6.7}\\~\\ \sin B=\frac{8.2\sin 32°}{6.7}\\~\\ \begin{array}{lcl} B=\arcsin(\frac{8.2\sin 32°}{6.7})&\qquad\text{or}\qquad&B=180°-\arcsin(\frac{8.2\sin 32°}{6.7})\\~\\ B\approx40.433°&\text{or}&B\approx139.567° \end{array}\)

 

First let's use the first possible value of  B .  If  B = 40.433° ,  then...

 

C  =  180° - 32° - 40.433°

C  =  107.567°

 

And by the Law of Sines,

 

\(\frac{AB}{\sin107.567°}=\frac{6.7}{\sin32°}\\~\\ AB=\frac{6.7\sin107.67°}{\sin32°}\\~\\ AB\approx12.054\)

 

Second let's use the second possible value of  B .  If  B = 139.567° ,  then...

 

C  =  180° - 32° - 139.567°

C  =  8.433°

 

And by the Law of Sines,

 

\(\frac{AB}{\sin8.433°}=\frac{6.7}{\sin32°}\\~\\ AB=\frac{6.7\sin8.433°}{\sin32°}\\~\\ AB\approx1.854\)

 

So the two possible values of  AB  are   AB  ≈  12.054   and   AB  ≈  1.854

 

If   AB  =  12.054   then

the area of  △ABC  =  (1/2)(8.2)(12.054 sin 32° )

the area of  △ABC  ≈  26.189

 

If  AB  =  1.854   then

the area of  △ABC  =  (1/2)(8.2)(1.854 sin 32° )

the area of  △ABC  ≈  4.028

 

So....     AB  ≈  1.854

hectictar  May 17, 2018
 #1
avatar+7336 
+2
Best Answer

 

By the Law of Sines,

 

\(\frac{\sin B}{8.2}=\frac{\sin 32°}{6.7}\\~\\ \sin B=\frac{8.2\sin 32°}{6.7}\\~\\ \begin{array}{lcl} B=\arcsin(\frac{8.2\sin 32°}{6.7})&\qquad\text{or}\qquad&B=180°-\arcsin(\frac{8.2\sin 32°}{6.7})\\~\\ B\approx40.433°&\text{or}&B\approx139.567° \end{array}\)

 

First let's use the first possible value of  B .  If  B = 40.433° ,  then...

 

C  =  180° - 32° - 40.433°

C  =  107.567°

 

And by the Law of Sines,

 

\(\frac{AB}{\sin107.567°}=\frac{6.7}{\sin32°}\\~\\ AB=\frac{6.7\sin107.67°}{\sin32°}\\~\\ AB\approx12.054\)

 

Second let's use the second possible value of  B .  If  B = 139.567° ,  then...

 

C  =  180° - 32° - 139.567°

C  =  8.433°

 

And by the Law of Sines,

 

\(\frac{AB}{\sin8.433°}=\frac{6.7}{\sin32°}\\~\\ AB=\frac{6.7\sin8.433°}{\sin32°}\\~\\ AB\approx1.854\)

 

So the two possible values of  AB  are   AB  ≈  12.054   and   AB  ≈  1.854

 

If   AB  =  12.054   then

the area of  △ABC  =  (1/2)(8.2)(12.054 sin 32° )

the area of  △ABC  ≈  26.189

 

If  AB  =  1.854   then

the area of  △ABC  =  (1/2)(8.2)(1.854 sin 32° )

the area of  △ABC  ≈  4.028

 

So....     AB  ≈  1.854

hectictar  May 17, 2018

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