The Law of Sines, Section 7.1 #4, 6, 22, 36.

4. C = 135^o, c = 45, B = 10^o
   A = 180^o - B - C = 35^o
   c / sin C = a / sin A, a = sin A (c / sin C) = sin 35^o( 45 / sin 135^o) = 36.50,
   b = sin B (c / sin C) = sin 10^o( 45 / sin 135^o) = 11.05.

6. A = 60^o, a = 9, c = 10
   Recall that if the side opposite an acute angle has length less than an adjacent side such that
   h < a < c, we have the ambiquous case. To test this, h = c sin A = 5sqrt(3) = 8.6603, h < a < c so two solutions.
   Case 1, C is acute
   C = arcsin(10/9*sin 60^o) = 74.21^o,
   B = 180^o - A - C = 45.79^o, b = sin B(9 / sin A) = 7.449.
   Case 2, C is obtuse, we have C₂ the supplement of C₁,
   C₂ = 180^o - C₁ = 105.79^o, B₂ = 180^o - A - C₂ = 14.207^o, b = sin B(9 / sin A) = 2.551.

22. A = 22 5/6 ^o, B = 96^o, c = 30 meters, height = a.
    C = 180^o - A - B = 61 1/6^o, a = sin 22 5/6^o(30 / sin 61 1/6^o) = 13.29 meters.

36. Find the area of the triangle given B = 72.5^o, a = 105, c = 64.
    Recall the area of an oblique triangle is given by 1/2 the product of two sides times the sine of their included angle, so
    area = 1/2 (a * c * sin B)
      = 1/2 * 105 * 64 * sin 72.5^o = 3204.5 square units.

The Law of Cosines, Section 7.2 #8, 20, 29, 38.

8. a = 1.42, b = 0.75, c = 1.25.
   cos C = (a² + b² - c²) / (2ab) = 1.42² + 0.75² - 1.25²) / (2 * 1.42 * 0.75) = 0.47718,
   C = 61.5^o,
   B = arcsin(0.75(sin 61.5^o / 1.25) = 31.8^o,
   A = arcsin(1.42(sin 61.5^o / 1.25) = 86.7^o, check A + B + C = 180^o.

20. Determine the angle q given the side opposite q is 4.5, and the sides adjacent q are 3 and 2 units in length.
    q = arccos( (3² + 2² - (4 1/2)²) / (2 * 3 * 2) ) = 127.17^o.

29. An engine has a 7 inch connecting rod fastened to a crank which rotates on a 1.5 inch radius.
    a.) Use the law of cosines to write an equation giving the relationship between x and q, where x is the distance from the center of the connecting rod to the center of the crank. See figure on p. 548.
    cos q = (1.5² + x² - 7²) / (2 * 1.5 * x).
    
    b.) Write x as a function of q.
    Writing the above equation in quadratic form
    x² - 3x cos q - 46.75 = 0.
    Applying the positive term from the quadratic equation, where a = 1, b = -3 cos q, and c = -46.75 gives
    x = (3 cos q + sqrt( (3 cos q)² -4 * (-46.75) ) / 2.
    
    c.)Use a graphing utility to graph the function in part (b).
    y₁ = (3 cos x + sqrt( 9 cos x cos x + 187) ) / 2, x_min = 0, x_max = 2p, y_min = 0, y_max = 10, mode in radians.
    
    d.) Use the graph in part (c) to determine the maximum distance the piston moves in one cycle.
    The maximum value of x is 8.5 inches and the minimum value of x is 5.5 inches which gives a total travel of 6 inches in one cycle.

38. Heron's area formula is area = sqrt(s(s - a)(s - b)(s - c)) / 2, where s = (a + b + c) / 2.
    a = 75.4, b = 52, c = 52, so area = 1350.2 square units.

Demoivre's Theorem, Section 7.5 #47, 72, 84.

47. Let z₁ = 3(cos p/3 + i sin p/3), let z₂ = 4(cos p/6 + i sin p/6).
    To find z₁z₂, recall that,
    z₁z₂ = r₁r₂[cos(q₁ + q₂) + i sin(q₁ + q₂)]
      = 12[cos(p/3 + p/6) + i sin(p/3 + p/6)]
      = 12[cos(p/2) + i sin(p/2)].

72. Use Demoivre's Theorem, zⁿ = [r(cos q + i sin q)]ⁿ = rⁿ(cos nq + i sin nq),
    to find (2 + 2i)⁶.
    r = sqrt(2² + 2²) = sqrt(8) is the modulus of z,
    q = arctan(2/2) = p/4 is the argument of z.
    We have
    z⁶ = 8^6/2[cos( 6p/4) + i sin(6p/4)]
      = 8³[0 + i(-1)]
      = -512i.

84. Find (sqrt(5) - 4i)³.
    r = sqrt(5 + (-4)²) = sqrt(21) is the modulus of z,
    q = arctan -4/sqrt(5) = -1.06106 is the argument of z.
    We have
    z³ = 21^3/2[cos( 3q) + i sin(3q)]
      = -96.15 + 4i.

Essential Trigonometry, Math Links

by Jack Tompkins