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Three sides of triangle ABC has measure 3, 4, 6 .O is point such that angle OBC=OAB then find angle OAB

 Jul 18, 2017
 #1
avatar+7798 
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Three sides of triangle ABC has measure 3, 4, 6 .O is point such that angle OBC=OAB then find angle OAB

 

\(a=3\\ b=4\\ c=6 \)

 

\(b^2=a^2+c^2-2ac\ cos\ \beta\\ cos\ \beta=\frac{a^2+c^2-b^2}{2ac}\\ cos\ \beta=\frac{9+36-16}{48}=0.6041\overline{66}\\ \color{blue}\beta=52.831° \)

 

\(a^2=b^2+c^2-2bc\ cos\ \alpha\\ cos\ \alpha=\frac{b^2+c^2-a^2}{2bc}\\ cos\ \alpha=\frac{16+36-9}{48}=0.8958\overline{33}\\ \color{blue}\alpha=26.384°\)

 

\(\color{blue}\gamma=180°-\alpha-\beta=100.785°\)

 

Later it goes on ...

 

laugh  !

 Jul 18, 2017
 #2
avatar+7798 
+1

Three sides of triangle ABC has measure 3, 4, 6 .O is point such that angle angle OBC=angle OAB then find angle OAB.

 

It goes on.

 

Coordinate system x, y on the triangle.
A (0,0)
B (6,0)
O (xo,yo)

angle OAB = OBC = δ

 

Equation A

\(m=tan\ (δ)\)

 

\(tan\ (δ)=\color{blue}\frac{y-y_o}{x-x_o}\)                                Point-direction form


Equation B

\(m=tan\ (180°-\beta\ +\ δ)\)  

 

\(\beta=52.831°\)

 

\(tan\ (180°-52.831°\ +\ δ)=\frac{y-y_o}{x-x_o}=\frac{y-y_B}{x-x_B}\)   Two-point form

 

\(tan\ (127.169°\ +\ δ)={\color{blue}\frac{y-y_o}{x-x_o}}=\frac{y}{x-6}\) 

 

I have to think. Sorry.  crying

 Jul 19, 2017

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