Respuesta :
In solving problems like this, it's best to work on a diagram (see attached image). First, given the angle formed by the ramp and the length of the ramp, we can apply the properties of right triangles and recall that
[tex] sin \theta = \frac{opposite}{hypotenuse} [/tex]
Hence, we have
[tex] \sin 30^{0} = \frac{h}{10} [/tex]
[tex] \frac{1}{2} = \frac{h}{10} [/tex]
[tex] h = 5 [/tex]
Now that we have the window's height, we can compute for the angle, θ, if the ramp has a length of 15.
[tex] \sin \theta = \frac{5}{15} [/tex]
[tex] \theta = \sin^{-1}(\frac{1}{3}) [/tex]
[tex] \theta = 19.47 [/tex]
Therefore, for the 15-foot ramp to reach the window, the ramp must form an angle of 19.47°.
Answer: 19.47°
[tex] sin \theta = \frac{opposite}{hypotenuse} [/tex]
Hence, we have
[tex] \sin 30^{0} = \frac{h}{10} [/tex]
[tex] \frac{1}{2} = \frac{h}{10} [/tex]
[tex] h = 5 [/tex]
Now that we have the window's height, we can compute for the angle, θ, if the ramp has a length of 15.
[tex] \sin \theta = \frac{5}{15} [/tex]
[tex] \theta = \sin^{-1}(\frac{1}{3}) [/tex]
[tex] \theta = 19.47 [/tex]
Therefore, for the 15-foot ramp to reach the window, the ramp must form an angle of 19.47°.
Answer: 19.47°