Answer:
<em>Fencing required = 1586 m</em>
Step-by-step explanation:
The given statements can be thought of a triangle
as shown in the diagram attached.
A be the 1st vertex, B be the 2nd vertex and C be the 3rd vertex.
Distance between 1st and 2nd vertex, AB = 435 m
Distance between 2nd and 3rd vertex, AC = 656 m
![\angle A =49^\circ](https://tex.z-dn.net/?f=%5Cangle%20A%20%3D49%5E%5Ccirc)
To find:
Fencing required for the triangular field.
Solution:
Here, we know two sides of a triangle and the angle between them.
To find the fencing or perimeter of the triangle, we need the third side.
Let us use <em>Cosine Rule </em>to find the third side.
Formula for cosine rule:
![cos A = \dfrac{b^{2}+c^{2}-a^{2}}{2bc}](https://tex.z-dn.net/?f=cos%20A%20%3D%20%5Cdfrac%7Bb%5E%7B2%7D%2Bc%5E%7B2%7D-a%5E%7B2%7D%7D%7B2bc%7D)
Where
a is the side opposite to ![\angle A](https://tex.z-dn.net/?f=%5Cangle%20A)
b is the side opposite to ![\angle B](https://tex.z-dn.net/?f=%5Cangle%20B)
c is the side opposite to ![\angle C](https://tex.z-dn.net/?f=%5Cangle%20C)
![\Rightarrow cos 49^\circ = \dfrac{656^{2}+435^{2}-BC^{2}}{2\times 435\times 656}\\\Rightarrow BC^2 = 430336+189225-2 (435)(656)cos49^\circ\\\Rightarrow BC^2 = 430336+189225-570720\times cos49^\circ\\\Rightarrow BC^2 =619561-570720\times cos49^\circ\\\Rightarrow BC \approx 495\ m](https://tex.z-dn.net/?f=%5CRightarrow%20cos%2049%5E%5Ccirc%20%3D%20%5Cdfrac%7B656%5E%7B2%7D%2B435%5E%7B2%7D-BC%5E%7B2%7D%7D%7B2%5Ctimes%20435%5Ctimes%20656%7D%5C%5C%5CRightarrow%20BC%5E2%20%3D%20430336%2B189225-2%20%28435%29%28656%29cos49%5E%5Ccirc%5C%5C%5CRightarrow%20BC%5E2%20%3D%20430336%2B189225-570720%5Ctimes%20cos49%5E%5Ccirc%5C%5C%5CRightarrow%20BC%5E2%20%3D619561-570720%5Ctimes%20cos49%5E%5Ccirc%5C%5C%5CRightarrow%20BC%20%5Capprox%20495%5C%20m)
Perimeter of the triangle = Sum of three sides = AB + BC + AC
Perimeter of the triangle = 435 + 495 + 656 = <em>1586 m</em>
<em></em>
<em>Fencing required = 1586 m</em>