The diagram represents a right-angled triangle and we are given the following sides
AB=6
AC=4.9
BC=?
METHOD 1
We can write an expression for BC using the Pythagoras theorem below:
![\begin{gathered} AB^2=AC^2+BC^2 \\ BC^2=AB^2-AC^2 \\ BC=\sqrt[]{AB^2-AC^2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vygkrr6wosadh5thy8thspcxutmefe0d1l.png)
Therefore the expression is
![BC=\sqrt[]{AB^2-AC^2}](https://img.qammunity.org/2023/formulas/mathematics/college/4zmwdj2tj8dkmjehw8f3qotr7fv43oie8d.png)
The expression can be further solved to get BC
![\begin{gathered} AB^2=AC^2+BC^2 \\ 6^2=4.9^2+BC^2 \\ 36=24.01+BC^2 \\ BC^2=36-24.01 \\ BC^2=11.99 \\ BC^2=\sqrt[]{11.99} \\ BC=3.46 \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4po8jg4fmizm9fb9xm6arp54njvwf0tynt.png)
METHOD 2
The second expression can be gotten using SOHCAHTOA:
Since we are given the opposite side and the hypotenuse we would make use of
CAH
![\begin{gathered} \cos \theta=\frac{\text{Adj}}{Hyp} \\ \cos \theta=(BC)/(6) \\ cross\text{ multiply} \\ BC=6\cos \theta \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qbmobhyddl9gio9vmfhuxisr3wnla063dc.png)
Therefore the expression is
![BC=6\cos \theta](https://img.qammunity.org/2023/formulas/mathematics/college/3fao90mwxg7ywk4x0csnhfk6z985bl9w6l.png)
The expression can be further solved to get BC
![\begin{gathered} BC=6\cos 55 \\ BC=6*0.5736 \\ BC=3.44 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/t5fz9juqbv6mklrbpaz1sr5ieqfm4kd2kr.png)
We can clearly see that both methods give a very similar answer. Therefore the expressions are correct