Jack is flying his kite . He runs 100 feet away from his house and lets out 45 feet of string . The angle of elevation from the ground to the kite is 65 degrees . How far is the kite from the house ?

Respuesta :

Answer: 125.80 ft

Step-by-step explanation:

Asuming the described situation is as shown in the figure below, we need to find the distance [tex]h[/tex] between the kite and Jacks house, but first we need to find the [tex]x[/tex], [tex]y[/tex] and then [tex]d[/tex].

How?

We will use trigonometry, especifically the trigonometric functions sine and cosine:

For [tex]y[/tex]:

[tex]sin(65 \°)=\frac{y}{45 ft}[/tex] (1)

Where [tex]y[/tex] is the opposite side to the angle and [tex]45 ft[/tex] the hypotenuse.

Isolating [tex]y[/tex]:

[tex]y=40.78 ft[/tex] (2)

For [tex]x[/tex]:

[tex]cos(65 \°)=\frac{x}{45 ft}[/tex] (3)

Where [tex]x[/tex] is the adjacent side to the angle.

Isolating [tex]x[/tex]:

[tex]y=19.017 ft[/tex] (4)

Finding [tex]d[/tex]:

[tex]d=x+100 ft[/tex] (5)

[tex]d=19.017 ft +100 ft[/tex]

[tex]d=119.017 ft[/tex] (6)

Now that we have found these values, we have to work with a bigger triangle, where the hypotenuse is the distance between the kite and Jack's house [tex]h[/tex] and the sides are the values calculated in (4) and (6).

So, in this case we will use the Pithagorean theorem:

[tex]h^{2}=y^{2} +d^{2}[/tex] (7)

Isolating [tex]h[/tex] and writing with the known values:

[tex]h=\sqrt{y^{2} +d^{2}}[/tex] (8)

[tex]h=\sqrt{(19.017 ft)^{2} +(119.017 ft)^{2}}[/tex] (9)

[tex]h=125.80 ft[/tex] This is the distance between the kite and the house

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