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!

Guest Dec 11, 2014

#1**+13 **

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!

flflvm97 Dec 11, 2014

#1**+13 **

Best Answer

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!

flflvm97 Dec 11, 2014