Respuesta :
Given
The sides of a triangle,
a=25cm, b=32cm & c=37cm.
To solve the oblique triangle ABC.
Explanation:
It is given that,
The sides of a triangle,
a=25cm, b=32cm & c=37cm.
That implies,
By using cosine law,
[tex]\begin{gathered} a^2=b^2+c^2-2bc\cos A \\ 25^2=32^2+37^2-2\times32\times37\cos A \\ 625=1024+1369-2368\cos A \\ 2368\cos A=1768 \\ \cos A=\frac{1768}{2368} \\ \cos A=0.7466 \\ A=\cos^{-1}(0.7466) \\ A=41.7\degree \\ A=42\degree \end{gathered}[/tex]Also,
[tex]\begin{gathered} \frac{\sin A}{a}=\frac{\sin B}{b} \\ \frac{\sin41.7\degree}{25}=\frac{\sin B}{32} \\ \frac{0.6652}{25}=\frac{\sin B}{32} \\ \sin B=\frac{21.287}{25} \\ \sin B=0.8515 \\ B=\sin^{-1}(0.8515) \\ B=58.37\degree \\ B=58\degree \end{gathered}[/tex]Therefore,
[tex]\begin{gathered} \angle C=180-(\angle A+\angle B) \\ \angle C=180-(42+58) \\ \angle C=80\degree \end{gathered}[/tex]Hence,
The answers are,
[tex]\angle A=42\degree,\angle B=58\degree,\angle C=80\degree[/tex]