ANSWER
[tex]5(\sin \text{ 160}\degree\text{ - sin 62}\degree)[/tex]STEP-BY-STEP EXPLANATION:
Given information
[tex]10\text{ sin 49}\degree\cos \text{ 111}\degree[/tex][tex]\text{Let }\alpha\text{ =49}\degree\text{ and }\beta\text{ = 111}\degree[/tex]Recall that, the product as a sum containing only sine and cosine can be written below as
[tex]sin\alpha\cos \beta\text{ = }\frac{1}{2}\lbrack\sin (\alpha\text{ + }\beta)\text{ + (sin }\alpha\text{ - }\beta)\rbrack[/tex]The next step is to substitute the given value into the above formula
[tex]\begin{gathered} 10\sin \text{ 49}\degree\cos 111\degree\text{ = 10 }\times\frac{1}{2}\lbrack\sin (49\text{ + 111) + sin(}49\text{ - 111)} \\ 10\sin 49\degree\cos 111\degree\text{ = 5\lbrack{}sin(160}\degree)\text{ + sin (-62}\degree) \\ 10\text{ sin49}\degree\cos \text{ 111}\degree\text{ = 5 (sin 160}\degree\text{ - sin 62}\degree) \end{gathered}[/tex]
