a) Diagram below
b) AC = 28.3 km
1) Considering what's been given we can sketch out:
• Note that we're considering that the bearing is always taken in a clockwise direction from North.
,• Since DiNozzo headed East then we can state the leg BC is perpendicular, therefore we have a right triangle.
b) To find out AC we need to find out at least one angle.
We can make it using some properties and trigonometric ratios.
[tex]\begin{gathered} \tan (C)=\frac{15}{24} \\ C=\tan ^{-1}(\frac{15}{24}) \\ m\angle C=32.00 \end{gathered}[/tex]To find out the distance AC, we can write out the Law of Cosines to find this out:
[tex]\begin{gathered} b^2=a^2+c^2-2ac\cdot\cos (B) \\ b^2=15^2+24^2-2\cdot15\cdot_{}24\cdot\cos (90) \\ b^2=225+576-0 \\ b^2=801 \\ \sqrt[]{b^2}=\sqrt[]{801} \\ b=28.30 \end{gathered}[/tex]3) Hence, the distance of AC, i.e. b, is 28.3 km approximately 28 km
