A 200-lb force (\(P\)) directed vertically downward is applied to the toggle vise at \(C\). Knowing that link \(BD\) is 6 in. long and that \(a = 4\) in., determine the horizontal force exerted on block \(E\). (Round the final answer to two decimal places.)**Diagram Explanation:**The diagram depicts a toggle mechanism involving several components:- Link \(AC\) and \(DE\) are connected, forming a mechanical linkage.- The vertical force \(P\) is applied at point \(C\), shown by a downward red arrow.- The length from point \(A\) to point \(B\) is labeled as \(a = 4\) in.- Link \(BD\) is 6 in. long.- There is an angle of 15 degrees at point \(A\).- Block \(E\) is depicted at the end of the mechanism, opposing the applied force.**Question:**The horizontal force exerted on block \(E\) is \(\_\_\_\_\_\_\_\) lb. →

Given Data : Length of AB= 4 in Length of BC= 6 in length of BD= 6 in Force (P)= 200 lbBF=aSinθ BF=4Sin15° BF=1.0352 Sinβ=BF6 β=9.93° AF=aCos15°=3.86 in FD=BD Cosβ=6×Cos9.93 FD=5.91 in so, AD=AF+FD AD=9.77 in ∑MA=0 200×10Cos15°-FBDSinβ×AD=0 FBD=1146.64 lb FBDx=FBD Cos β (FBD)x=1129.46 lb The horizontal force on block E is 1129.46 lb