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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

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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

Guest Jul 18, 2017
#1
+7485
0

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 ...

!

asinus  Jul 18, 2017
#2
+7485
+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.

asinus  Jul 19, 2017