Question

I'm put at a dilemma on my project need help

Answer 1

see the figure below with letters to better understand the problem

step 1

Find out the length BC in the right triangle BDC

we have a 45-90-45 degrees triangle

$\begin{gathered} cos(45^o)=\frac{BC}{DC}=\frac{BC}{6} \\ cos(45^o)=\frac{\sqrt{2}}{2} \end{gathered}$

Find out the value of BC

$\begin{gathered} \frac{BC}{6}=\frac{\sqrt{2}}{2} \\ BC=3\sqrt{2} \end{gathered}$

step 2

In the right triangle ABC

$\begin{gathered} sin30^o=\frac{BC}{AC} \\ \\ AC=\frac{BC}{sin30^o} \end{gathered}$

we have

$\begin{gathered} BC=3\sqrt{2} \\ sin30^o=\frac{1}{2} \end{gathered}$

substitute

$\begin{gathered} AC=\frac{3\sqrt{2}}{\frac{1}{2}}=6\sqrt{2} \\ AC=8.5\text{ m} \end{gathered}$

The answer Part A is 8.5 meters

Part B

Find out the difference

8.5-6=2.5 meters

The answer Part B is 2.5 meters