Answer:
![Area = 538.5 m^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/6p3mda7oe8177nwaakusarfz0lgfjmys1o.png)
Explanation:
Given:
∆XVW
m < X = 50°
m < W = 63°
XV = w = 37 m
Required:
Area of ∆XVW
Solution:
Find side length XW using Law of Sines
![(v)/(sin(V)) = (w)/(sin(W))](https://img.qammunity.org/2021/formulas/mathematics/high-school/8ategatedcaq0m91scj3l2z77dhxn2flj1.png)
W = 63°
w = XV = 37 m
V = 180 - (50+63) = 67°
v = XW = ?
![(v)/(sin(67)) = (37)/(sin(63))](https://img.qammunity.org/2021/formulas/mathematics/high-school/vlyzwpsy6g1axlug0p0f5q3xcz0223wi82.png)
Cross multiply
![v*sin(63) = 37*sin(67)](https://img.qammunity.org/2021/formulas/mathematics/high-school/eyrwszfirq02k5nw0lqt247a85gezp1f8d.png)
Divide both sides by sin(63) to make v the subject of formula
![(v*sin(63))/(sin(63)) = (37*sin(67))/(sin(63))](https://img.qammunity.org/2021/formulas/mathematics/high-school/vx8z1p7o3vamjcytaddj8cx10tblaz90a3.png)
![v = (37*sin(67))/(sin(63))](https://img.qammunity.org/2021/formulas/mathematics/high-school/88zt2c6b99h9cdsjmkuq1vx50cnf2iduck.png)
(approximated to nearest whole number)
![XW = v = 38 m](https://img.qammunity.org/2021/formulas/mathematics/high-school/hhv3nxfvyolhxwrltwn1lq1ht9lrje3kza.png)
Find the area of ∆XVW
![area = (1)/(2)*v*w*sin(X)](https://img.qammunity.org/2021/formulas/mathematics/high-school/uond3m1c4ws1x810g6tet5qzv90tgdrfrn.png)
![= (1)/(2)*38*37*sin(50)](https://img.qammunity.org/2021/formulas/mathematics/high-school/wp3ud3u68dggjzphdygp8n45can9x233sz.png)
![= (38*37*sin(50))/(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/kdyj4ofitwc6nqra6jnlx0zv1yrnbrfebi.png)
(to nearest tenth).