Answer:
4.2 in
Explanation:
let us first visualize the sail as a triangular shape
the angle of the triangle from top is 40°
the height of the triangle is give as 5 in
we can apply SOH CAH TOA to solve for the base of the sail
the opposite = the base of the sail
the adjacent = the height of the sail= 5 in
therefore
Tan∅= Opp/Adj
Tan(40)= Opp/5
Opp= Tan(40)*5
Opp= 0.8390*5
Opp= 4.195 in
Hence the width of the sail is 4.2 in to the nearest tenths