Math 5  Trigonometry  Chapter 5 Test Solutions  fall ’06  

 

1.        What is the length of the arc subtended by an angle  in a circle of radius 14?
SOLN:  

2.        Find the central angle on a circle of radius 4 that subtends and arc of length 6.   Give both the radian measure and the degree measure of the angle.  
SOLN: 

3.        What is the radius of a circle where a sector with central angle of 36˚ has area = 9?
SOLN:  36˚ = , so we solve  

4.        Three circles with radii 3, 5 and 12 are externally tangent to one another, as shown in the figure at right. a.        Show that . SOLN:  so, by the converse of Pythagorus’ theorem, it’s a right triangle. b.       Approximate to the nearest hundredth of a degree measures for  and , interior to triangle ABC. SOLN:    c.        Find the area of the shaded region between the three circles.

     SOLN:  Area of triangle minus areas of sectors  

5.        A tricycle with little wheels of diameter 10cm and a big wheel with diameter 40cm is rolling along so that the big wheel is rolling at 35 rotations per minute. 

a.        What is the angular speed of the little wheels?
SOLN:  The angular speed of the little wheel is

b.       What is the linear speed of the tricycle?

6.        The elongation α for Mercury is the angle formed by the planet, Earth and Sun, as shown in the diagram at right.  Assume the distance from Mercury to the sun is 0.387 AU (38.7% of the distance from Earth to Sun)  and that α = 18˚.   Find the possible distances from Earth to Mercury. SOLN: By the law of sines, the acute angle at Mercury can be found like so:  so the obtuse angle is 180˚  52.99˚ = 127.01˚, thus the angle at the sun is either 34.99˚ or 109.01˚.

This means the sun to earth distance can be either   or

7.        Find the area of the shaded region in the figure at right. SOLN:  108˚ = .  Area of sector minus area of triangle is  then 8.        By the law of cosines, the largest angle is

By the law of sines, the smallest angle is   
That means the third angle is 180˚  41.4˚  82.8˚ = 57.8˚

9.        By the law of cosines,

10.     The vectors are perpendicular and have equal lengths, so the sum and difference have the same value:  

11.     (This is a problem from the text in the section on law of sines)
 In the picture, we can see that, since , in triangle ABC, by the law of sines,  and in triangle ADC, .    Also, since  in the big triangle BCD,  we have .  Thus  
(1.1)                 

Applying the law of sines again in triangle ABC,   Substituting this into (1.1) we get  , from which the result follows: .