Two divers are 50 m apart. Each diver sees a treasure chest on the sea floor. The treasure chest is vertically below the line between the divers. From the divers, the angles of depression to the treasure chest are 35° and 51°. To the nearest metre, how far is the treasure chest from each diver? Consider possible cases and show your work.
I'm really having trouble with this one. Thanks in advance!
+270 I draw diagram, it's quite fiddly, plz understand 
By trig Z rule,
Then angle between those is 180-35-51=94
Now you have 3 angles and one side.
You can use cosine rule or sine rule to get it.
Because sine rule is more simple, I will use sine rule here.
$$\frac{50}{sin94}=\frac{x}{sin35}=\frac{y}{sin51}$$
where x and y being opposite side of each angles.
$$x=\frac{50sin35}{sin94}$$
$$x=28.7 (3 s.f.)$$
Hope it helped! 
+270 I draw diagram, it's quite fiddly, plz understand
By trig Z rule,
Then angle between those is 180-35-51=94
Now you have 3 angles and one side.
You can use cosine rule or sine rule to get it.
Because sine rule is more simple, I will use sine rule here.
$$\frac{50}{sin94}=\frac{x}{sin35}=\frac{y}{sin51}$$
where x and y being opposite side of each angles.
$$x=\frac{50sin35}{sin94}$$
$$x=28.7 (3 s.f.)$$
Hope it helped!
