They will rest after they hike `2` miles. Then they will hike the remaining `1.75` miles to the lookout. The trail the hikers wi
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There are several ways of going about this problem, but just know that it all boils down to triangles and the rules/laws of triangles.
So we start by making a large right triangle, because the hill is vertical with the horizontal ground, with 25° as the left angle (where John looks up to top), 90° is the right angle (where ground meets hill base). So we see that the top side of the triangle is the hypotenuse, and equals the line of sight from John to hilltop.
Now we've got additional information that if John walks 350ft towards the hill, his angle of elevation increases by 14. So that = 25+14 = 39. How does that possibly help us?? Well now we can make 2 triangles inside of the one we've already made. So that now we have the triangle base split between the left angle of 25° and where he stopped 350ft to the right of that.
Now the supplement of 39 is 141, and the remaining piece of that too left triangle = 180-25-141 = 14. What does that mean? Well now we have a triangle, where we know all 3 angles and 1 side --> we can find another side by the law of sines:
If a, b and c are the lengths of the legs of a triangle opposite to the angles A, B and C respectively; then the law of sines states:
a÷sinA = b÷sinB = c÷sinC
We really need the hypotenuse to then find our hill height, so we'll make hypotenuse = side b in attached image. That being so, then its opposite angle (B) = 141. And the top right angle (C) = 14 with its opposite side (c) = 350ft.
Now we only need b÷sinB = c÷sinC
So b/sin141 = 350/sin14 --> b = 350sin141/sin14 = 350×.63/
b = 220.3/.24 = 910.47 ft
Now that's our hypotenuse, so using our original large right triangle, we can use right triangular trig. to solve. Let's make the right side, our hill height, equal to x.
Sin ¥ = opp. side / hypotenuse -->
Sin ¥ = x / hypotenuse
Sin25 = x / 910.5
x = 910.5×sin25 = 910.5×.423
x = 384.8 ft
So we start by making a large right triangle, because the hill is vertical with the horizontal ground, with 25° as the left angle (where John looks up to top), 90° is the right angle (where ground meets hill base). So we see that the top side of the triangle is the hypotenuse, and equals the line of sight from John to hilltop.
Now we've got additional information that if John walks 350ft towards the hill, his angle of elevation increases by 14. So that = 25+14 = 39. How does that possibly help us?? Well now we can make 2 triangles inside of the one we've already made. So that now we have the triangle base split between the left angle of 25° and where he stopped 350ft to the right of that.
Now the supplement of 39 is 141, and the remaining piece of that too left triangle = 180-25-141 = 14. What does that mean? Well now we have a triangle, where we know all 3 angles and 1 side --> we can find another side by the law of sines:
If a, b and c are the lengths of the legs of a triangle opposite to the angles A, B and C respectively; then the law of sines states:
a÷sinA = b÷sinB = c÷sinC
We really need the hypotenuse to then find our hill height, so we'll make hypotenuse = side b in attached image. That being so, then its opposite angle (B) = 141. And the top right angle (C) = 14 with its opposite side (c) = 350ft.
Now we only need b÷sinB = c÷sinC
So b/sin141 = 350/sin14 --> b = 350sin141/sin14 = 350×.63/
b = 220.3/.24 = 910.47 ft
Now that's our hypotenuse, so using our original large right triangle, we can use right triangular trig. to solve. Let's make the right side, our hill height, equal to x.
Sin ¥ = opp. side / hypotenuse -->
Sin ¥ = x / hypotenuse
Sin25 = x / 910.5
x = 910.5×sin25 = 910.5×.423
x = 384.8 ft
Answer:
A.
Negative reciprocal gives you the perpendicular slope so negative reciprocal of 1/2 is -2.
Then apply point-slope form.
B. The answer is x = 6.
The midpoint of JK is
The line that goes through JK is just a horizontal line because the y-coordinate does not change. So its perpendicular bisector is the vertical line that goes through the x-coordinate of the midpoint, that is,
.