To find the angle B we can use the propertie that sya that the sum of the internal angles of a triangle is equal to 180º so:
![\measuredangle b+90º+40º,41^(\prime)=180](https://img.qammunity.org/2023/formulas/mathematics/college/pfw7q2o1cbw5l43axnca1hczyvbkct4tp9.png)
and we solve for angle b so:
![\begin{gathered} \measuredangle b=180º-90º-40º,41^(\prime) \\ \measuredangle b=49º,19^(\prime) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3zadksdpksuz1u95nkc39t7sz2i8ynvg7o.png)
So B is equal to: 49 degrees and 19 minutes
So now to find a we can use the trigonometric identitie of sin so:
![\begin{gathered} \sin (40.68)=(a)/(961) \\ a=961\cdot\sin (40.68) \\ a\approx626 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kxssa0u4omga9fxex7vrr2s9viry12plum.png)
and to find b we use the trigonometryc identitie of cos so:
![\begin{gathered} \cos (40.68)=(b)/(961) \\ b=961\cdot\cos (40.68) \\ b\approx729 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lhqq5mhzgz3j01863gf3fdvpr59dsdbkq9.png)