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My textbook says that theta = ~1.02 radians. The equation is on the left. How did they arrive at that answer?

 Sep 15, 2014

Best Answer 

 #1
avatar+33653 
+5

θ is the angle whose cosine is 3*√(6)/14 so

θ = cos-1(3*√(6)/14)

 

$${\mathtt{theta}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6}}}}}{{\mathtt{14}}}}\right)} \Rightarrow {\mathtt{theta}} = {\mathtt{58.339\: \!117\: \!225\: \!405^{\circ}}}$$

180° is pi radians, so in radians we have


$${\mathtt{theta}} = {\frac{{\mathtt{58.339\: \!117\: \!225\: \!405}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{180}}}} \Rightarrow {\mathtt{theta}} = {\mathtt{1.018\: \!209\: \!678\: \!290\: \!256\: \!2}}$$

 

or θ ≈ 1.02 radians

 Sep 15, 2014
 #1
avatar+33653 
+5
Best Answer

θ is the angle whose cosine is 3*√(6)/14 so

θ = cos-1(3*√(6)/14)

 

$${\mathtt{theta}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6}}}}}{{\mathtt{14}}}}\right)} \Rightarrow {\mathtt{theta}} = {\mathtt{58.339\: \!117\: \!225\: \!405^{\circ}}}$$

180° is pi radians, so in radians we have


$${\mathtt{theta}} = {\frac{{\mathtt{58.339\: \!117\: \!225\: \!405}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{180}}}} \Rightarrow {\mathtt{theta}} = {\mathtt{1.018\: \!209\: \!678\: \!290\: \!256\: \!2}}$$

 

or θ ≈ 1.02 radians

Alan Sep 15, 2014

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