Note: form the question, he ran from point A to B to C
Hence we will need to find DB, BC
From triangle CBD
[tex]\begin{gathered} \text{Opp = }DB \\ \text{Adj = 60} \\ \theta\text{ = 27} \\ \text{ Using the Trig function} \\ \tan \theta\text{ = }\frac{\text{Opp}}{\text{Adj}} \\ \tan \text{ 27 = }\frac{DB}{60} \\ By\text{ cross multiply} \\ DB\text{ = tan 27 }\times\text{ 60} \\ DB\text{ = 30.57meter} \\ \end{gathered}[/tex]
DB = 30.57
Step2: we will need to find CB still from triangle CBD
[tex]\begin{gathered} CB\text{ = Hyp} \\ \text{Adj= 60 } \\ \theta\text{ = 27} \\ \cos \theta\text{ = }\frac{\text{Adj}}{\text{Hyp}} \\ \cos \text{ 27 = }\frac{60}{CB} \\ 0.8910\text{ =}\frac{60}{CB} \\ 0.8910\times CB\text{= 60} \\ CB\text{ = }\frac{60}{0.8910} \\ CB\text{ =67.3meter} \end{gathered}[/tex]
CB = 67.3meter
STEP 3
From traingle ADC
From the triangle, we will need to find AB, AC
Step 4: we will need to find AC from still from triangle ABC
[tex]\begin{gathered} \text{Hyp = AC} \\ Adj\text{ = 60} \\ \theta\text{ = 43} \\ \cos \text{ }\theta\text{ =}\frac{Adj}{\text{Hyp}} \\ \cos \text{ 43 = }\frac{60}{AC} \\ 0.731\text{ = }\frac{60}{AC} \\ 0.731\text{ }\times\text{ AC = 60} \\ AC\text{ = }\frac{60}{0.731} \\ AC\text{ = 82.039} \\ \end{gathered}[/tex]
The Dishes distance. is the sum of all our answer
82.039 + 67.3 + 55.95 + 30.57 = 233.859meter
