Question

The angles of depression of two ships from the top of the light house are 45° and 30° towards east. If the ships are 200 m apart, find the height of the lighthouse.
please answer with figure.

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Answer 1

Answer:

$h\approx 273.21 \text{ meters}$

Step-by-step explanation:

Please refer to the attachment.

In the attachment, h is the height of the lighthouse and x is the distance from the lighthouse to Ship A.

Since the angle of depression from the top of the lighthouse to Ship A is 45°, this means that the angle of elevation from Ship A to the top of the lighthouse is 45°.

Likewise, the angle of elevation from Ship B to the top of the lighthouse is 30°.

So, we will form two right triangles: the smaller, 45-45-90 triangle, and the larger 30-60-90 triangle.

Remember that in 45-45-90 triangles, the two legs are congruent.

Therefore, we can write that:

$h=x$

Next, in 30-60-90 triangles, the longer leg is always √3 times the shorter leg.

In our 30-60-90 triangle, the shorter leg is given by:

$\text{Shorter leg}=h$

And the longer leg is given by:

$\text{Longer leg}=x+200$

So, the relationship between the shorter leg and longer leg is:

$\sqrt3h = (x+200)$

And since we know that h is equivalent to x, we can write:

$\sqrt3h=(h+200)$

Now, we just have to solve for h. We can subtract h from both sides:

$\sqrt3h-h=200$

Factoring out the h yields:

$h(\sqrt3-1)=200$

Therefore:

$\displaystyle h=\frac{200}{\sqrt3-1}$

Approximate. So, the height of the lighthouse is approximately:

$h \approx 273.2050 \text{ meters}$

Answer 2

Answer:

273 meters

Step-by-step explanation:

See image attached for the diagram I used to represent this scenario.

The distance between the ships, at angles 30 and 45, is 200 meters. The distance between the left ship and the lighthouse is x meters.

We can use trigonometric ratios to solve this problem. We can use the tangent ratio $\big{(} \frac{\text{opposite}}{\text{adjacent}} \big{)}$ to create an equation with the two angles.

• $\displaystyle \text{tan(45)} = \frac{h}{x}$
• $\displaystyle \text{tan(30)} = \frac{h}{x+200} }$

Let's take these two equations and solve for x in both of them.

$\textbf{Equation I}$

• $\displaystyle \text{tan(45)} = \frac{h}{x}$  

tan(45) = 1, so we can rewrite this equation.

• $\displaystyle 1=\frac{h}{x}$

Multiply x to both sides of the equation.

• $\displaystyle x = h$

$\textbf{Equation II}$

• $\displaystyle \text{tan(30)} = \frac{h}{x+200} }$

Multiply x + 200 to both sides and divide h by tan(30).

• $\displaystyle \text{x + 200} = \frac{h}{\text{tan (30)}}$  

Subtract 200 from both sides of the equation.

• $\displaystyle \text{x} = \frac{h}{\text{tan (30)}} - 200$

Simplify h/tan(30).

• $x=\sqrt{3}h - 200$  

$\textbf{Equation I = Equation II}$

Take Equation I and Equation II and set them equal to each other.

• $h=\sqrt{3}h-200$

Subtract √3 h from both sides of the equation.

• $h-\sqrt{3}h=-200$

Factor h from the left side of the equation.

• $h(1-\sqrt{3}) =-200$

Divide both sides of the equation by 1 - √3.

• $\displaystyle h=\frac{-200}{1-\sqrt{3} }$

Rationalize the denominator by multiplying the numerator and denominator by the conjugate.

• $\displaystyle h=\frac{-200}{1-\sqrt{3} } \big{(} \frac{1+\sqrt{3} }{1+\sqrt{3}} \big{)}$
• $\displaystyle h=\frac{-200+200\sqrt{3} }{1-3}$
• $\displaystyle h =\frac{-200+200\sqrt{3} }{-2}$

Simplify this equation.

• $\displaystyle h=100+100\sqrt{3}$
• $h=273.20508075$

The height of the lighthouse is about 273 meters.