Given
The sides of a triangle,
a=25cm, b=32cm & c=37cm.
To solve the oblique triangle ABC.
Step-by-step explanation:
It is given that,
The sides of a triangle,
a=25cm, b=32cm & c=37cm.
That implies,
By using cosine law,
![\begin{gathered} a^2=b^2+c^2-2bc\cos A \\ 25^2=32^2+37^2-2*32*37\cos A \\ 625=1024+1369-2368\cos A \\ 2368\cos A=1768 \\ \cos A=(1768)/(2368) \\ \cos A=0.7466 \\ A=\cos^(-1)(0.7466) \\ A=41.7\degree \\ A=42\degree \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3z4xp23tkinm3bdnta6wrzi8uo6vbsynuv.png)
Also,
![\begin{gathered} (\sin A)/(a)=(\sin B)/(b) \\ (\sin41.7\degree)/(25)=(\sin B)/(32) \\ (0.6652)/(25)=(\sin B)/(32) \\ \sin B=(21.287)/(25) \\ \sin B=0.8515 \\ B=\sin^(-1)(0.8515) \\ B=58.37\degree \\ B=58\degree \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yugm9yo361ijyguym0x3o3w0imtyvnk2am.png)
Therefore,
![\begin{gathered} \angle C=180-(\angle A+\angle B) \\ \angle C=180-(42+58) \\ \angle C=80\degree \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ezzvevmdtclxsmtks8gbzbp6dkw46o72ui.png)
Hence,
The answers are,
![\angle A=42\degree,\angle B=58\degree,\angle C=80\degree](https://img.qammunity.org/2023/formulas/mathematics/college/v6eh0cmqrc7mrn4c559ciyvyxe7rzloaf1.png)