Question

In AKLM, LM = 5 and M

Answer 1

To answer this question, we can use trigonometric ratios. We have that the triangle is as follows:

As we can see we need to find the hypotenuse, and we can see that 5 is the opposite side to the angle 45.

We also know that:

`\sin (\theta)=(opposite)/(hypotenuse)`

Therefore, we can use this trigonometric ratio to solve this question. Then we have:

`\begin{gathered} \sin (45^(\circ))=(5)/(hypotenuse)=(5)/(KL) \\ \sin (45^(\circ))=(5)/(KL) \end{gathered}`

Now, we can multiply both sides by KL as follows:

`\begin{gathered} KL\cdot\sin (45^(\circ))=KL\cdot(5)/(KL) \\ KL\cdot\sin (45^(\circ))=5 \\ KL\cdot(\sin (45^(\circ)))/(\sin (45^(\circ)))=(5)/(\sin (45^(\circ))) \\ KL=(5)/(\sin (45^(\circ))) \end{gathered}`

Notice that we also divided both sides by sin(45°). Then we have:

`\begin{gathered} \sin (45^(\circ))=\frac{\sqrt[]{2}}{2} \\ KL=\frac{5}{\frac{\sqrt[]{2}}{2}}\Rightarrow KL=5\cdot\frac{2}{\sqrt[]{2}}=\frac{10}{\sqrt[]{2}}\cdot\frac{\sqrt[]{2}}{\sqrt[]{2}}=\frac{10\sqrt[]{2}}{2}=5\sqrt[]{2} \end{gathered}`