First, Lte's draw a picture of our problem:
which is equivalent to the following picture:
where h denotes the altitute and x is the horizontal distance from the near side of the island to the airplane.
Now, from triangle ACD, we have that
[tex]tan68=\frac{1600+x}{h}[/tex]
By distributing h, this last equation is equivalent to
[tex]tan68=\frac{1600}{h}+\frac{x}{h}\text{ ...\lparen A\rparen}[/tex]
Now, from triangle BCD,we have
[tex]tan65=\frac{x}{h}[/tex]
By substituting this result into equation (A), we have
[tex]tan68=\frac{1600}{h}+tan65[/tex]
Then, by moving tan65 to the left hand side, we get
[tex]tan68-tan65=\frac{1600}{h}[/tex]
or equivalently,
[tex]\frac{1600}{h}=tan68-tan65[/tex]
By taking reciprocals, we get
[tex]\frac{h}{1600}=\frac{1}{tan68-tan65}[/tex]
so, by moving 1600 to the right hand side, we get
[tex]h=\frac{1600}{tan68-tan65}[/tex]
which gives
[tex]h=\frac{1600}{2.47508-2.14450}[/tex]
Then, the altitude is given by
[tex]h=4839.9792\text{ ft}[/tex]
Therefore, by rounding to the nearest whole number, the answer is 4840 feet.