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:
[tex]\begin{gathered} AB^2=AC^2+BC^2 \\ BC^2=AB^2-AC^2 \\ BC=\sqrt[]{AB^2-AC^2} \end{gathered}[/tex]
Therefore the expression is
[tex]BC=\sqrt[]{AB^2-AC^2}[/tex]
The expression can be further solved to get BC
[tex]\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}[/tex]
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
[tex]\begin{gathered} \cos \theta=\frac{\text{Adj}}{Hyp} \\ \cos \theta=\frac{BC}{6} \\ cross\text{ multiply} \\ BC=6\cos \theta \end{gathered}[/tex]
Therefore the expression is
[tex]BC=6\cos \theta[/tex]
The expression can be further solved to get BC
[tex]\begin{gathered} BC=6\cos 55 \\ BC=6\times0.5736 \\ BC=3.44 \end{gathered}[/tex]
We can clearly see that both methods give a very similar answer. Therefore the expressions are correct
