Respuesta :
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 [tex]\big{(} \frac{\text{opposite}}{\text{adjacent}} \big{)}[/tex] to create an equation with the two angles.
- [tex]\displaystyle \text{tan(45)} = \frac{h}{x}[/tex]
- [tex]\displaystyle \text{tan(30)} = \frac{h}{x+200} }[/tex]
Let's take these two equations and solve for x in both of them.
[tex]\textbf{Equation I}[/tex]
- [tex]\displaystyle \text{tan(45)} = \frac{h}{x}[/tex]
tan(45) = 1, so we can rewrite this equation.
- [tex]\displaystyle 1=\frac{h}{x}[/tex]
Multiply x to both sides of the equation.
- [tex]\displaystyle x = h[/tex]
[tex]\textbf{Equation II}[/tex]
- [tex]\displaystyle \text{tan(30)} = \frac{h}{x+200} }[/tex]
Multiply x + 200 to both sides and divide h by tan(30).
- [tex]\displaystyle \text{x + 200} = \frac{h}{\text{tan (30)}}[/tex]
Subtract 200 from both sides of the equation.
- [tex]\displaystyle \text{x} = \frac{h}{\text{tan (30)}} - 200[/tex]
Simplify h/tan(30).
- [tex]x=\sqrt{3}h - 200[/tex]
[tex]\textbf{Equation I = Equation II}[/tex]
Take Equation I and Equation II and set them equal to each other.
- [tex]h=\sqrt{3}h-200[/tex]
Subtract √3 h from both sides of the equation.
- [tex]h-\sqrt{3}h=-200[/tex]
Factor h from the left side of the equation.
- [tex]h(1-\sqrt{3}) =-200[/tex]
Divide both sides of the equation by 1 - √3.
- [tex]\displaystyle h=\frac{-200}{1-\sqrt{3} }[/tex]
Rationalize the denominator by multiplying the numerator and denominator by the conjugate.
- [tex]\displaystyle h=\frac{-200}{1-\sqrt{3} } \big{(} \frac{1+\sqrt{3} }{1+\sqrt{3}} \big{)}[/tex]
- [tex]\displaystyle h=\frac{-200+200\sqrt{3} }{1-3}[/tex]
- [tex]\displaystyle h =\frac{-200+200\sqrt{3} }{-2}[/tex]
Simplify this equation.
- [tex]\displaystyle h=100+100\sqrt{3}[/tex]
- [tex]h=273.20508075[/tex]
The height of the lighthouse is about 273 meters.
![Ver imagen Supernova](https://us-static.z-dn.net/files/df4/871f2fc03de094548ce1477741d5ebd6.jpeg)
Answer:
[tex]h\approx 273.21 \text{ meters}[/tex]
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:
[tex]h=x[/tex]
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:
[tex]\text{Shorter leg}=h[/tex]
And the longer leg is given by:
[tex]\text{Longer leg}=x+200[/tex]
So, the relationship between the shorter leg and longer leg is:
[tex]\sqrt3h = (x+200)[/tex]
And since we know that h is equivalent to x, we can write:
[tex]\sqrt3h=(h+200)[/tex]
Now, we just have to solve for h. We can subtract h from both sides:
[tex]\sqrt3h-h=200[/tex]
Factoring out the h yields:
[tex]h(\sqrt3-1)=200[/tex]
Therefore:
[tex]\displaystyle h=\frac{200}{\sqrt3-1}[/tex]
Approximate. So, the height of the lighthouse is approximately:
[tex]h \approx 273.2050 \text{ meters}[/tex]
![Ver imagen xKelvin](https://us-static.z-dn.net/files/dc6/345c43de87a886d0abca6d14b61feb14.jpg)