+0  
 
0
286
2
avatar

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
avatar+7348 
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 ...

 

laugh  !

asinus  Jul 18, 2017
 #2
avatar+7348 
+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

asinus  Jul 19, 2017

11 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.