trig

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A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 4 mi apart, to be x = 27° and y = 53°, as shown in the figure. (Round your answers to two decimal places.)

(a) Find the distance of the plane from point A.
mi

(b) Find the elevation of the plane.
mi

  • trig -

    let me start you off

    draw a straight line representing 4miles apart..... now at the ends of the line make angles of 27 and 53


    now figure it out

  • trig -

    i don't get it

  • trig -

    I'm strugglin so hard on this

  • trig -

    I THINK I GOT IT
    okay so draw a triangle (not a right triangle)
    the bottom number is 4
    draw an imaginary line straight down from the tip of the triangle to the bottom
    half a circle is 180. so the total middle area of the triangle is 180-27-53 which is 100.
    keep in mind that 27 and 53 are the ANGLES OF DEPRESSION. so draw a straight line across the top of your triangle and the area between this straight top line and the side of your triangle is the angle of depression.
    now look at your imaginary line.
    we know that 1/4 of a triangle is 90 deg
    90-27=63.
    so your top angle in THIS HALF of the triangle is 63degrees
    basically treat each half like 2 different triangles.
    we can determine the 3rd angle (i have this marked as A) by 180-90-63 (the 90 is where our straight line down is)
    so we know that A is 27
    now we have all 3 angles to one side of our triangle
    now repeat this process to the other half of the triangle
    (keep in mind we can NOT divide the 4 miles)
    so we should have 90, 37 and 53 for triangle half #2.
    NOW look at the triangle as one big triangle
    the top is 180, the sides are 27 and 53
    the bottom length is 4
    look at this as a ASA triangle
    27deg, 4, 53deg
    now use law of sines
    a/sinA= c/sinC
    plug in numbers
    a/sin27= 4/sin180
    now a= (4x sin27)/ sin180
    plug that into your calculator and you have side a
    repeat this to find side b
    the answer to your first question will be the length of side b

    for the second question you need to take these same numbers
    divide your triangle in half again
    solve for the new "side a" aka the line you drew down the middle

    tada you have your answers!

    yay math

  • trig -

    HOLD UP HOLD UP
    okay sorry y'all one minor error.... ignore my last post here is the new and improved and correct explanation



    I THINK I GOT IT
    okay so draw a triangle (not a right triangle)
    the bottom number is 4
    draw an imaginary line straight down from the tip of the triangle to the bottom
    half a circle is 180. so the total middle area of the triangle is 180-27-53 which is 100.
    keep in mind that 27 and 53 are the ANGLES OF DEPRESSION. so draw a straight line across the top of your triangle and the area between this straight top line and the side of your triangle is the angle of depression.
    now look at your imaginary line.
    we know that 1/4 of a triangle is 90 deg
    90-27=63.
    so your top angle in THIS HALF of the triangle is 63degrees
    basically treat each half like 2 different triangles.
    we can determine the 3rd angle (i have this marked as A) by 180-90-63 (the 90 is where our straight line down is)
    so we know that A is 27
    now we have all 3 angles to one side of our triangle
    now repeat this process to the other half of the triangle
    (keep in mind we can NOT divide the 4 miles)
    so we should have 90, 37 and 53 for triangle half #2.
    NOW look at the triangle as one big triangle
    the top is 100, the sides are 27 and 53
    the bottom length is 4
    look at this as a ASA triangle
    27deg, 4, 53deg
    now use law of sines
    a/sinA= c/sinC
    plug in numbers
    a/sin27= 4/sin100
    now a= (4x sin27)/ sin100
    plug that into your calculator and you have side a
    repeat this to find side b
    the answer to your first question will be the length of side b

    for the second question you need to take these same numbers
    divide your triangle in half again
    solve for the new "side a" aka the line you drew down the middle

    tada you have your answers!

    yay math

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