SAT SUBJECT TEST MATH LEVEL 2
REVIEW OF MAJOR TOPICS
1.3 Trigonometric Functions and Their Inverses
The final topic in trigonometry concerns the relationship between the angles and sides of a triangle that is not a right triangle. Depending on which of the sides and angles of the triangle are supplied, the following formulas can be used to find missing parts of a triangle. In ABC
Law of Sines:
used when the lengths of two sides and the value of the angle opposite one, or two angles and the length of one side are given.
Law of Cosines:
used when the lengths of two sides and the included angle, or the lengths of three sides, are given.
used when two sides and the included angle are given.
1. Find the number of degrees in the largest angle of a triangle whose sides are 3, 5, and 7.
The largest angle is opposite the longest side. Use the Law of Cosines:
Therefore, cos .
Since cos C < 0 and C is an angle of a triangle, 90° < C < 180°.
Therefore, C = 120°.
2. Find the number of degrees in the other two angles of ABC if , b = 150, and C = 30°.
Use the law of sines:
Therefore, B = 45° or 135°; A = 105° or 15° since there are 180° in the sum of the three angles of a triangle.
3. Find the area of ABC if a = 180 inches, b = 150 inches, and C = 30°.
If the lengths of two sides of a triangle and the angle opposite one of those sides are given, it is possible that two triangles, one triangle, or no triangle can be constructed with the data. This is called the ambiguous case. If the lengths of sides a and b and the value of A are given, the length of side b determines the number of triangles that can be constructed.
Case 1: If A > 90° and a b, no triangle can be formed because side a would not reach the base line. If a > b, one obtuse triangle can be drawn.
Let the length of the altitude from C to the base line be h. From the basic definition of sine, sin and thus, h = b sin A.
Case 2: If A < 90° and side a < b sin A, no triangle can be formed. If a = b sin A, one triangle can be formed. If a > b, there also will be only one triangle. If, on the other hand, b sin A < a < b, two triangles can be formed.
If a compass is opened the length of side a and a circle is drawn with center at C, the circle will cut the baseline at two points, B1 and B2. Thus, AB1C satisfies the conditions of the problem, as does AB2C.
1. How many triangles can be formed if a = 24, b = 31, and A = 30°?
Because A < 90°, b · sin A = 31 · sin 30° = 31 · = 15 · Since b · sin A < a < b, there are two triangles.
2. How many triangles can be formed if a = 24, b = 32, and A = 150°?
Since A > 90° and a < b, no triangle can be formed.
1. In ABC, A = 30°, b = 8, and a =4. Angle C could equal
2. In ABC, A = 30°, a = 6, and c = 8. Which of the following must be true?
(A) 0° < C < 90°
(B) 90° < C < 180°
(C) 45° < C < 135°
(D) 0° < C < 45° or 90° < C < 135°
(E) 0° < C < 45° or 135° < C < 180°
3. The angles of a triangle are in a ratio of 8 : 3 : 1. The ratio of the longest side of the triangle to the next longest side is
(B) 8 : 3
(D) 8 : 5
4. The sides of a triangle are in a ratio of 4 : 5 : 6. The smallest angle is
5. Find the length of the longer diagonal of a parallelogram if the sides are 6 inches and 8 inches and the smaller angle is 60°.
6. What are all values of side a in the figure below such that two triangles can be constructed?
(B) a > 8
7. In ABC, B = 30°, C = 105°, and b = 10. The length of side a equals
8. The area of ABC, = 24 , side a = 6, and side b = 16. The value of C is
(B) 30° or 150°
(D) 60° or 120°
(E) none of the above
9. The area of ABC= 12, side a = 6, and side b = 8. Side c =
(C) 2 or 2
(E) 10 or 12
10. Given the following data, which can form two triangles?
I. C = 30°, c = 8, b = 12
II. B = 45°, a = 12, b = 15
III. C = 60°, b = 12, c =5
(A) only I
(B) only II
(C) only III
(D) only I and II
(E) only I and III