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Solve the side length of the triangle

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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.

May 17, 2018

#1
+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

May 17, 2018

#1
+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