Height of lighthouse based on angle difference I have a question in my maths book: A lookout in a lighthouse tower can see two ships approaching the coast. Their angles of depression are 25° and 30°. If the ships are 100 m apart, show that the height of the lighthouse, to the nearest metre, is 242 metres. I have no clue how to solve it, please can someone give me a step-by-step process of how to work it out??
 A: 
HERE LET AB=H (HEIGHT OF LIGHTHOUSE)
Here lets apply trigonometry in $\triangle ABC$
$$tan\angle ACB=\frac{AB}{BC}$$
$$tan30=\frac{AB}{x}$$
$$\rightarrow AB=H=tan30(x)\tag1$$
Applying the same thing in $\triangle ABD$ we can get
$$tan\angle ADB=\frac{AB}{100+x}$$
$$tan25=\frac{AB}{100+x}$$
$$\rightarrow AB=H=tan25(100+x)\tag2$$
As $(1)=(2)$
We can easily get 
$$x=420.09m\tag3$$ using $tan25=0.466$ and $tan30=0.577$
Now again coming back to $\triangle ABC$ and using equation $(1)$
$$H=420.09(tan30)$$ 
$$H=242.5m \sim 242m$$
A: Let x be the height of the lighthouse, let d be the distance between the first boat and the base of the lighthouse
To begin, using the fact that we know 'x' will be the same for both triangles created. Using basic trigonometry, we come to the point where x = d/tan(60) = (d+100)/tan(65)
From this we begin to find x: Note: bolded italics is an explanation of what is occurring
d⋅tan(65) = (d+100)⋅tan(60)
d⋅tan(65) = (d+100)⋅√(3) [tan(60)=√(3)]
d⋅tan(65) = d√(3) + 100√(3) [Expand brackets]
d⋅tan(65) - d√(3) =  100√(3) [- d√(3)]
d (tan(65) - √(3)) = 100√(3) [Factor d]
d = 100√(3) / (tan(65) - √(3)) [/tan(65) - √(3)]
d ≈ 419.94m
Now knowing what the distance from base of lighthouse to first ship is, we can calculate the height using trigonometry
tan(60)= 419.94 / x
x = 419.94 / tan(60)
x ≈ 242.45m
In the same fashion:
tan(65)= (419.94 + 100) / x
x = 519.94 / tan(65)
x ≈ 242.45m
∴ The height of the lighthouse is 242.45m
