# The pond

We can see the pond at an angle 65°37'. Its end points are 155 m and 177 m away from the observer. What is the width of the pond?

Result

c =  180.836 m

#### Solution:

$a=155 \ \text{m} \ \\ b=177 \ \text{m} \ \\ \ \\ A=65 + \dfrac{ 37 }{ 60 }=\dfrac{ 3937 }{ 60 } \doteq 65.6167 \ ^\circ \ \\ \ \\ c=\sqrt{ a^2 + b^2 -2 \cdot \ a \cdot \ b \cdot \ \cos( A ^\circ \rightarrow\ \text{rad}) }=\sqrt{ a^2 + b^2 -2 \cdot \ a \cdot \ b \cdot \ \cos( A ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ ) }=\sqrt{ 155^2 + 177^2 -2 \cdot \ 155 \cdot \ 177 \cdot \ \cos( 65.616666666667 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ ) }=180.83555=180.836 \ \text{m}$

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