Problem with Sine Law: Find the length of the inner bisector of the corner at the top A.
- Thread starter Aysun
- Start date
This is the image.
And I think this should be the solution:
\(\displaystyle \angle\)CAB=180 \(\displaystyle ^\circ\) - (\(\displaystyle \beta\) + \(\displaystyle \gamma\))
\(\displaystyle \Rightarrow\) \(\displaystyle \angle\)CAL=\(\displaystyle \angle\)LAB=\(\displaystyle \frac{180 ^\circ - (\beta+\gamma)}{2}\)
here we use the sines law:
\(\displaystyle \frac{AL}{sin\gamma}\)=\(\displaystyle \frac{CL}{\frac{sin180 ^\circ - (\beta+\gamma))}{2}}\)
\(\displaystyle \Rightarrow\) \(\displaystyle \frac{AL}{sin\gamma}\)=\(\displaystyle \frac{2CL}{sin( \beta+\gamma)}\)
\(\displaystyle \Rightarrow\) AL=\(\displaystyle \frac{2CL.sin \gamma}{sin(\beta+\gamma)}\)
And if I use the sines law for the triangle ALB:
AL=\(\displaystyle \frac{2LB.sin\beta}{sin(\beta+\gamma)} \)
Is it right ? I'm sorry if I have made any mistakes.My English is not that good.
And I think this should be the solution:
\(\displaystyle \angle\)CAB=180 \(\displaystyle ^\circ\) - (\(\displaystyle \beta\) + \(\displaystyle \gamma\))
\(\displaystyle \Rightarrow\) \(\displaystyle \angle\)CAL=\(\displaystyle \angle\)LAB=\(\displaystyle \frac{180 ^\circ - (\beta+\gamma)}{2}\)
here we use the sines law:
\(\displaystyle \frac{AL}{sin\gamma}\)=\(\displaystyle \frac{CL}{\frac{sin180 ^\circ - (\beta+\gamma))}{2}}\)
\(\displaystyle \Rightarrow\) \(\displaystyle \frac{AL}{sin\gamma}\)=\(\displaystyle \frac{2CL}{sin( \beta+\gamma)}\)
\(\displaystyle \Rightarrow\) AL=\(\displaystyle \frac{2CL.sin \gamma}{sin(\beta+\gamma)}\)
And if I use the sines law for the triangle ALB:
AL=\(\displaystyle \frac{2LB.sin\beta}{sin(\beta+\gamma)} \)
Is it right ? I'm sorry if I have made any mistakes.My English is not that good.