Respuesta :
Answer:
1) sin([tex]-\frac{\pi }{12}[/tex]) = [tex]-\frac{\sqrt{2-\sqrt{3} } }{2}[/tex] (Decimal: -0.258. . .)
2) π/24, 5π/24, 13π/24, 17π/24
3) The replacement ramp forms an angle of 14.1 ° with the horizontal
Step-by-step explanation:
1)
sin([tex]-\frac{\pi }{12}[/tex])
use the following property: sin(-x) = -sin(x)
sin([tex]-\frac{\pi }{12}[/tex]) = -sin([tex]\frac{\pi }{12}[/tex])
sin([tex]\frac{\pi }{12}[/tex]) = [tex]\frac{\sqrt{2-\sqrt{3} } }{2}[/tex]
= [tex]-\frac{\sqrt{2-\sqrt{3} } }{2}[/tex]
2)
2sin(4x) = 1
sin(4x) = 1/2
4x = arcsin(1/2) = π/6, 5π/6, plus those +2π, 4π, etc
π/6+2π = π/6+12π/6 = 13π/6
5π/6+2π = 5π/6+12π/6 = 17π/6
π/6+4π = π/6+24π/6 = 25π/6
5π/6+4π = 5π/6+24π/6 = 29π/6
x = above divided by 4, so solutions are
(1/4)(π/6, 5π/6, 13π/6, 17π/6)
or
π/24, 5π/24, 13π/24, 17π/24
3)
To carry out this problem we have to think of the ramp as a right triangle. The length of the ramp will be the hypotenuse and the height will be the opposite leg to the angle they give us.
To calculate the degree of the new ramp we have to first calculate the height of our ramp that would be the same on the 2 ramps
well to start we have to know the relationship between angles, legs and the hypotenuse in a right triangle
α = 24°
a: adjacent
o: opposite = ?
h: hypotenuse = 30
sin α = o/h
cos α = a/h
tan α = o/a
we see that it has (angle, hypotenuse, opposite)
is the sine sin α = o/h
sin 24° = o/30
sin 24° * 30 = o
12.2 = o
The height of the ramp is 12.2ft
Now we have the length of the hypotenuse and the opposite leg and we can calculate the angle
sin α = o/h
sin α = 12.2/50
α = sin^-1(12.2/50)
α = 14.12°
if we round to the nearest tenth
α = 14.12° = 14.1°
(1) The exact value be "[tex]\frac{\sqrt{3}-2 }{2\sqrt{2} }[/tex]".
(2) The solution will be "[tex]\\\frac{\pi}{24}, \frac{5\pi}{24}, \frac{13\pi}{24}, \frac{17\pi}{24}[/tex]".
(3) The angle will be "[tex]14.1^{\circ}[/tex]".
Trigonometry:
(1)
[tex]\to \sin (-\frac{pi}{12})\\\\\to \sin (-\frac{180}{12})\\\\\to \sin (- 15)\\\\ \to \sin 15^{\circ} = \sin (45-30)^{\circ}[/tex]
Using formula:
[tex]\to \bold{\sin (A-B) = \sin A \cos B -\cos A \sin B}[/tex]
[tex]\to \sin (45-30) = \sin 45^{\circ} \cos 30^{\circ} -\cos 45^{\circ} \sin 30^{\circ}[/tex]
[tex]= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{{2}} -\frac{1}{\sqrt{2}} \times \frac{1}{2}\\\\= \frac{\sqrt{3}}{2\sqrt{2}} -\frac{1}{2\sqrt{2}} \\\\= \frac{\sqrt{3}-1}{2\sqrt{2}} \\\\[/tex]
(2)
[tex]\to 2 \sin (4x)=1\\\\\to \sin (4x)= \frac{1}{2}\\\\\to \sin (4x)= \sin 30^{\circ}\\\\\to (4x)= 30^{\circ}\\\\\therefore\\\\\to (4x)= \frac{\pi}{6} \ \ Or\ \ \frac{5\pi}{6} \\\\[/tex]
adding value [tex]2\pi, 4\pi, etc[/tex] in the above-given value.
[tex]\frac{\pi}{6}+2\pi = \frac{\pi+ 12 \pi}{6} = \frac{13\pi}{6}\\\\\frac{5\pi}{6}+2\pi = \frac{5\pi+ 12 \pi}{6} = \frac{17\pi}{6}\\\\\frac{\pi}{6}+4\pi = \frac{\pi+ 24 \pi}{6} = \frac{25\pi}{6}\\\\\frac{5\pi}{6}+4\pi = \frac{5\pi+ 24 \pi}{6} = \frac{29\pi}{6}\\\\[/tex]
To calculate x we divide the value by 4 then:
[tex](\frac{1}{4}) (\frac{\pi}{6} , \frac{5\pi}{6} , \frac{13\pi}{6} , \frac{17\pi}{6})\\or \\\frac{\pi}{24}, \frac{5\pi}{24}, \frac{13\pi}{24}, \frac{17\pi}{24}[/tex]
(3)
- To solve this problem, we must consider the ramp to be a right triangle. The ramp's length will be the hypotenuse, and its elevation is the opposite leg to the angle they provide for us.
- When calculating the degree of the new ramp, we must first determine the height of our ramp, which is indeed the same on both ramps.
- To begin, we must understand the relationship between angles, legs, and the hypotenuse in a right-angled triangle.
[tex]\alpha = 24^{\circ}\\\\a: adjacent \\\\o: opposite = ?\\\\h: hypotenuse = 30\\\\\sin \alpha = \frac{O}{h}\\\\\cos \alpha = \frac{a}{h}\\\\\tan \alpha = \frac{O}{a}\\\\[/tex]
Given: (angle, hypotenuse, opposite)
[tex]\sin \alpha = \frac{O}{h}\\\\\sin 24^{\circ} = \frac{O}{30}\\\\\sin 24^{\circ} \times 30 = 0\\\\\to 12.2 = 0\\\\[/tex]
We know,
Height of ramp= 12.2ft
Now we have the length of the hypotenuse and the opposite leg and we can calculate the angle
[tex]\sin \alpha = \frac{O}{h}\\\\\sin \alpha = \frac{12.2}{50}\\\\\alpha = sin^{-1}(\frac{12.2}{50}) = 14.12^{\circ}\\\\\text{rounding the value}\\\\\alpha = 14.12^{\circ} = 14.1^{\circ} \\\\[/tex]
Thus the above approach is correct.
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