Trigonometry Problems and Questions with Solutions - Grade 11

Grade 11 trigonometry problems and questions with answers and solutions are presented.



  1. A ferris wheel with a radius of 25 meters makes one rotation every 36 seconds. At the bottom of the ride, the passenger is 1 meter above the ground.

    a) Let h be the height, above ground, of a passenger. Determine h as a function of time if h = 51 meter at t = 0.

    b) Find the height h after 45 seconds.

  2. Linda measures the angle of elevation from a point on the ground to the top of the tree and find it to be 35o. She then walks 20 meters towards the tree and finds the angle of elevation from this new point to the top of the tree to be 45o. Find the height of the tree. (Round answer to three significant digits)

  3. From the top of a cliff 200 meters high, the angles of depression of two fishing boats in the same line of sight on the water are 13 degrees and 15 degrees. How far apart are the boats? (Round your answer to 4 significant digits)

  4. Prove that [ cos(x) - sin(x) ][ cos(2x) - sin(2x) ] = cos(x) - sin(3x)

  5. The graph of function f is the graph of function g(x) = a sin(x - pi/3) translated vertically by 2. Also f(pi/2) = 1. Find a formula in terms of x for function f.

  6. Find sin(x) and tan(x) if cos(pi/2 - x) = - 3/5 and sin(x + pi/2) = 4/5?

  7. Find the exact value of [ tan (25o)+ tan (50o ] / [ 1 - tan( 25o) tan(50o) ]

  8. What is the angle B of triangle ABC, given that A = 46o, b = 4 and c = 8?(Note: side a faces angle A, side b faces angle B and side c faces angle C).

  9. Find the exact value of tan (s + t) given that sin s = 1/4, with s in quadrant 2, and sin t = -1/2, with t in quadrant 4.

  10. Find all angles of a triangle with sides 9, 12 and 15.

  11. Write an equation for a sine function with an amplitude of 5/3 , a period of pi/2, and a vertical shift of 4 units up.

  12. Find the exact values of cos (13pi/12).

  13. Two gears are interconnected. The smaller gear has a radius of 4 inches, and the larger gear has a radius of 10 inches. The smaller gear rotates 890 degrees in 4 seconds. What is the angular speed, in degrees per minute, of the larger rotate?

  14. A ladder of length 20 meters is resting against the wall. The base of the ladder is x meters away from the base of the wall and the angle made by the wall and the ladder is t.

    a) Find x in terms of t.

    b) Starting from t = 0 (the ladder against the wall) and then gradually increase angle t; for what size of angle t will x be the quarter of the length of the ladder?

Solutions to the Above Problems

  1. a) Let P be the position of the passenger (see figure below)

    ferris wheel problem solution.



    The height h of the passenger is given by

    h = 1 + 25 + y = y + 26

    y depends on the angle of rotation A.

    sin(pi/2 - A) = y/25 which gives y = 25 cos(A)

    Angle A depends on the angular speed w as follows

    A = w t where t is the time.

    The angular speed w is given by

    w = 2pi / 36 = Pi / 18 (radians/second)

    We now substitute to find h as follows h(t) = 25 cos( (pi/18) t) + 26 , where t is in seconds and y in meters.

    b) h(45) = 25 cos( (pi/18) 45) + 26 = 25 cos(3pi/2) + 26 = 26 meters.

  2. Using the figure below we write the following equations:

    tan(35o) = h / x and tan(45o) = h / (x - 20) , where h is the height of the tree.

    Solve both equations for x to find

    x = h / tan(35o) and x = h / tan(45o) + 20

    Which gives h / tan(35o) = h / tan(45o) + 20

    Solve for h; h = [ 20 tan(35o) tan(45o) ] / [ tan(45o) - tan(35o) ] = 46.7 meters (3 significant digits)

    tree problem solution.


  3. Using the figure below we write the following equations:

    tan(75o) = y / 200 and tan(87o) = (y + x) / 200

    Eliminate y from the two equations and solve for x: x = 200 [ tan(87o) - tan(75o) ] = 3070 meters (rounded to 4 significant figures)

    boat problem solution.


  4. Start with right hand side: cos(x) - sin(3x) = cos(x) - sin( x + 2x)

    = cos(x) - sin(x)cos(2x) - cos(x)sin(2x)

    We now expand the left hand side: [ cos(x) - sin(x) ][ cos(2x) - sin(2x) ]

    = cos(x) cos(2x) - cos(x) sin(2x) - sin(x) cos(2x) + sin(x) sin(2x)

    Use the identities cos(2x) = 1 - 2 sin2 and sin(2x) = 2 sin(x) cos(x) to transform the first two terms (only) in the above expression.

    = cos(x)(1 - 2 sin2) + sin(x) 2 sin(x) cos(x) - cos(x) sin(2x) - sin(x) cos(2x)

    = cos(x) - 2 cos(x) sin2 + 2 cos(x) sin2 - cos(x) sin(2x) - sin(x) cos(2x)

    = cos(x) - cos(x) sin(2x) - sin(x) cos(2x)

    The left hand side has been transformed so that it is equal to the right hand side.

  5. f has the following form f(x) = a sin(x - pi/3) + 2 : shifting graph of g 2 units up.

    f(pi/2) = a sin(pi/2 - pi/3) + 2 = 1

    Solve for a to find a = -2

    f(x) = -2 sin(x - pi/3) + 2

  6. Expand and simplify: cos(pi/2 - x) = cos(pi/2)cos(x) + sin(pi/2)sin(x) = sin(x) = -3/5

    Expand and simplify: sin(x + pi/2) = sin(x) cos(pi/2) + cos(x) sin(pi/2) = cos(x) = 4/5

    tan(x) = sin(x) / cos(x) = (-3/5) / (4/5) = -3/4

  7. The addition formula for the tangent may be used to write

    [ tan (25o)+ tan (50o) ] / [ 1 - tan( 25o) tan(50o) ] = tan(25o + 50o)

    = tan(75o)

    = tan(45o + 30o)

    = [ tan(45o) + tan(30o) ] / [1 - tan(45o)tan(30o) ]

    = [ 1 + sqrt(3) / 3 ] / [ 1 - 1*sqrt(3) / 3 ]

    = sqrt(3) + 2

  8. Use cosine rule to find side a

    a = sqrt(16 + 64 - 2*4*8*cos(46o))

    We now use the sine rule to find angle B as follows

    sin(B) / 4 = sin(A) / a

    B = arcsin (4 sin(A) / a) = 29 degrees (rounded to the nearest unit)

  9. Given sin(s) = 1/4 and sin(t) = -1/2 and their quadrants, find cos(s) and cos(t).

    cos(s) = - sqrt(15) / 4 and cos(t) = sqrt(3) / 2

    We now expand:

    tan (s + t) = sin(s + t) / cos(s + t)

    = [ sin(s)cos(t) + cos(s)sin(t) ] / [ cos(s)cos(t) - sin(s)sin(t) ]

    Substitute

    = - [ 4 sqrt(3) + sqrt(15) ] / 11

  10. Note that 152 = 122 + 92 which means that the triangle in question is a right triangle.

    Let A be the angle facing side with length 9; hence sin(A) = 9/15

    A = 37o (rounded to the nearest degree)

    The third angle = 90o - 37o = 53o

  11. y = (5/3) sin(B x) + 4 , B > 0

    2 pi / B = pi/2 , solve for B: B = 4

    y = (5/3) sin(4 x) + 4

  12. cos(13 pi/12) = cos(pi/12 + pi) = - cos(pi/12)

    = - cos( (1/2)(pi/6) ) = - sqrt [ ( (1/2)(1 + cos(pi/6)) ] half angle formula

    = - sqrt [ 1/2 + srqt(3) / 4 ]



  13. Let R1 and R2 be the radii of gear 1 and 2. Let S1 and S2 be the arcs of rotation of gears 1 and 2. The interconnected gears have equal tangential velocity (measured in inches/second), therefore arcs S1 and S2 are equal in length.

    R1 * t1 = R2 * t2

    t1 and t2 are the angles of rotation of the larger and smaller gears respectively.

    10 * t1 = 4 * 890o

    t1 = 356o

    Angular speed = 356o / 4 second = 89o / second

    = 89o * 60 / (1 second * 60) = 5340o / minute

    gear problem solution.


  14. The ladder, the wall and the ground make a right a right triangle. Hence

    a) tan(t) = x / 20 or x = 20 tan(t)

    b) x = (1/4) 20 = 20 tan(t) Solve for t: t = arctan(1/4) = 14o (rounded to 2 significant digits)


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Updated: 2 April 2013

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