Test 6:  Law of Sines, Cosines, and area formulas

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MQ6 TEST 6A  (answers below)                                                                                      Mr. Gary Jaye

INSTRUCTIONS: 1. Express all answers in simplest form , SHOW ALL WORK, and box answers.
                                        2. NO graphing calculators     3. Use pen (except for graphs)

1. In a-j, fill in all the answers. [4, 3, 3 off for first 3 wrong answers]
In #a-c, fill in the three Pythagorean identities.
a. b. c.
In d-e, state the two standard equations of the unit circle, center at origin.
d. e.
f. The definition of radian measure in
    terms of s, r, and q  is:		 g. equals: h. The change of base formula is:
i. The transformation that maps the graph of y = bx onto     the graph of is: j. In terms of logbA, logb(1/A) equals:

 

 

 

 

 

 

 

 

 

 

 

2. Fill in the answer. [3, 3, 3 for 1st three wrong]
a. The formula for the distance between points (x1 , y1) and (x2, y2) is:
b. In terms of q , sin(-q) equals: c. In terms of q, cos(-q) equals:
d. Two formulas for the area of parallelogram ABCD are: (fill in below)
1. (if altitude is known): 2. (if m<A is known):

 

 

 

 

 

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In 3-4, write formula and legend, set up equation, and solve for unknown.
3. In triangle PQR, p = 67, q = 59, r = 93. Find m<R
to the nearest hundredth. [6]

 

 

 

 4. In triangle PQR, p = 77, q = 67, r = 84. Find area to the nearest tenth. [6]

 

 

 

 

 

 

 

 

 

 

5. Problems a-e refer to triangle PQR. Sketch the given data, draw and determine altitude, and use
    sketch to determine and state the number of distinct triangles that can be determined: 0, 1, 1 right,
    or 2 triangles.    [4 each]
a. P = 30° , r = 14, p = 15.

 

 

 c. P = 30° , r = 14, p = 9.

 

 
b. P = 30° , r = 14, p = 5.

 

 

 d. P = 30° , r = 14, p = 7.

 

 
e. P = 150° , r = 14, p = 11

 

 

 

 

 

 

 

 

 

 

 

 

 

6. In triangle PQR, <P = 26° , p = 45, and q = 92. Find all the possible measures for <Q and <R to nearest tenth      degree.  [15]

 

 

 

 



 

 

 

 

 

7. In triangle ABC, solve for angle C (nearest integer) given area = 72, a = 12, and b = . [11]

 

 

 

 

 






 

 

 

 

 

8. The heading and air speed of an airplane are 90° and 300 MPH, respectively. The wind is from 210° at 75 MPH.       [22]
a. Use the parallelogram of forces to represent the given data and sketch the resultant vector. Remember that north      is 0° , east is 90° , etc...

 

 

 

 
b. Find the plane’s ground speed (nearest MPH).

 

 

 

 

 

 c. Find the plane’s course (nearest degree)

 

 

 

 

 


 

 

 

 

 

 

 

 

 

 

 

 

 

ANSWERS:  1a-1c. sin2q + cos2q = 1, tan2q + 1 = sec2q, cot2q + 1 = csc2q,  1d-1e.  sin2q + cos2q = 1,
x2 + y2 = 1    1f. | q | = s/r    1g. N    1h. logCA = logbA/logbC    1i. ry = x    1j. -logbA   
2a. d = Ö[(x2 - x1)2 + (y2 - y1)2]     2b. -sinq     2c. cosq    2d(1). k = bh    2d(2). k = AB(AD)sinA
3. m<R = 94.93    4. k = 2438.7    5a. opp > adj, therefore 1 triangle    5b. opp < hyp, therefore 0 triangles
5c. hyp < opp < adj, therefore 2 triangles    5d. opp = hyp, therefore 1 right triangle    5e. Not ambiguous!! Since
m<P ³ 90°, length of opposite side must be greatest. Therefore, 0 triangles   6. (Q = 63.7° and R = 90.3°) or
(Q = 180° - 63.7° = 116.3° and R = 37.7°)    7. C = 60° or C = 180° - 60° = 120°
8b. ground speed = 343.693... = 344 MPH     8c. Use unrounded result from 8b: course is 79°

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