Answer:
5. m∠C ≈ 34°
6. AC ≈ 29.7 metres
Explanation:
5. We need to use the Law of Sines here, which says that for a triangle with sides a, b, and c and angles A, B, and C:
![(a)/(sinA) =(b)/(sinB) =(c)/(sinC)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/7gvwd9mghdt3nhwk1jqp0qovul3vooz9uc.png)
Here, a = 25, ∠A = 93°, and c = 14, so we want to find ∠C:
![(a)/(sinA) =(b)/(sinB) =(c)/(sinC)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/7gvwd9mghdt3nhwk1jqp0qovul3vooz9uc.png)
![(25)/(sin93) =(14)/(sinC)](https://img.qammunity.org/2021/formulas/mathematics/high-school/3bh9vfgzo57ybf8kgya226amn6baygadhu.png)
sinC ≈ 0.56
∠C ≈ 34°
6. We need to use the Law of Cosines here, which says that for a triangle with sides a, b, and c and angles A, B, and C:
![c^2=a^2+b^2-2abcosC](https://img.qammunity.org/2021/formulas/mathematics/high-school/xvpm57czgiztfm3sp7awor8em2gimwj2dw.png)
![b^2=a^2+c^2-2accosB](https://img.qammunity.org/2021/formulas/mathematics/high-school/djhbi4ziotqqq2nvkb0oj8fzt97nr6owiz.png)
![a^2=b^2+c^2-2bccosA](https://img.qammunity.org/2021/formulas/mathematics/high-school/byc8y35y5j6bgznw4jtd67nl1nmn9hfphd.png)
Here, a = 23.2, c = 18.1, ∠B = 91°, and b = x. We want to find b, so:
![b^2=a^2+c^2-2accosB](https://img.qammunity.org/2021/formulas/mathematics/high-school/djhbi4ziotqqq2nvkb0oj8fzt97nr6owiz.png)
≈ 880.5
x ≈ 29.7 metres