Trigonometry problem - A bee collecting nectar from flowers

Please see the attached file for figures. 

1) We have 

\[|BF_2|=| F_1A|=2.3 \cdot\cos (31)= 1.97,         |F_2A|=2.3 \cdot\sin (31)=1.18.\] 
Also 
\[\cos (51)=\frac{|BF_2|}{|F_2F_3|}  \Rightarrow  0.62=\frac{1.18}{|F_2F_3|}\]
\[\Rightarrow |F_2F_3|=\frac{1.97}{0.62}=3.17.\]
Hence the distance form the second flower to the third flower is about $3.2$ m.

(2) We have 
\[(*)   \sin (60)=\frac{|BC|}{|AB|} \Rightarrow  \frac{\sqrt{3}}{2}=\frac{3}{|AB|}\]
\[\Rightarrow |AB|=\frac{2 \cdot 3}{\sqrt{3}}=2\sqrt{3}.\]
So the length of hypotenuse is $2\sqrt{3}$. 

In the second figure we see that the two triangles ABC and AB'C' are similar. Hence
\[\sin (60)=\frac{|B'C'|}{|AB'|}=\frac{|BC|}{|AB|}.\]
This is why in (*) we have
\[\sin (60)=\frac{|BC|}{|AB|}.\]
So the formula in (*) is equivalnet to solving for similar triangles.

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Educational note:  In general if $\theta$ is one of the acute angles in a right-angle triangle, then the sine of theta is the ratio of the opposite side to the hypotenuse, the cosine is the ratio of the adjacent side to the hypotenuse, and the tangent is the ratio of the opposite side to the adjacent side.

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If we have $39$ degrees instead of $31$ degrees, the solution to Problem 1 will be midofied to 

1) We have 

\[|BF_2|=| F_1A|=2.3 \cdot\cos (39)= 1.78,         |BF_2|=|F_1A|=2.3 \cdot\sin (39)=1.44.\] 
Also 
\[\cos (51)=\frac{|BF_2|}{|F_2F_3|}  \Rightarrow  0.62=\frac{1.78}{|F_2F_3|}\]
\[\Rightarrow |F_2F_3|=\frac{1.78}{0.62}=2.87.\]
Hence the distance form the second flower to the third flower is about $2.9$ m.