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