This is a very classical problem. I have a boat traveling north across a $1$km wide river at a velocity of $1.5$km/h and the current downstream (west) is $0.8$km/h. Then the resultant velocity is of magnitude $1.7$km/h and direction N$28.07$E. I am asked to find how far downstream the boat will drift before reaching the other side. I got two solutions with different answers, which one is wrong and why?
If I use the distance formula: $$t=\frac{d}{v}=\frac{1}{1.5}=\frac{2}{3} \rightarrow d = v*t=0.8*\frac{2}{3}=0.53$$ so it will be $0.53$km downstrean.
But using trig functions (as in here): $$\tan(28.07)=\frac{1}{d} \rightarrow d=1.87$$ I get $1.87$km. What's going on?
Your first method gives the correct answer. However, you used the variables $d$ and $v$ in two different ways.
Let $d_n$ denote the distance the boat travels northward; let $d_w$ denote the distance the boat travels towards the west due to the river's current; let $v_n$ denote the boat's speed toward the north; let $v_c$ denote the speed of the current toward the west. Then the time it takes for the boat to cross the river is $$t = \frac{d_n}{v_n} = \frac{1~\text{km}}{1.5~\frac{\text{km}}{\text{h}}} = \frac{2}{3}~\text{h}$$ The distance the boat travels downstream is $$d_w = v_ct = \left(0.8~\frac{\text{km}}{\text{h}}\right)\left(\frac{2}{3}~\text{h}\right) \approx 0.53~\text{km}$$
As for your second method, you worked out the magnitude of the resultant velocity correctly. However, since the current carries the boat downstream, it will be diverted towards the west, as shown in the diagram.
$$\tan\theta = \frac{0.8~\frac{\text{km}}{\text{h}}}{1.5~\frac{\text{km}}{\text{h}}} = \frac{8}{15}$$ Hence, $$\theta = \arctan\left(\frac{8}{15}\right) \approx 28.07^\circ$$ so the direction of the boat is $\approx 28.07^\circ$ west of north.
Let $d_n$ denote the distance the boat travels northward; let $d_w$ denote the distance the boat travels towards the west due to the river's current.
Since the boat travels $1~\text{km}$ northward as it crosses the river, the distance it drifts downstream is $$d_w = d_n\tan\theta = 1~\text{km}\left(\frac{8}{15}\right) \approx 0.53~\text{km}$$