Question 1:
In triangle ABC, if angle A = 60 degrees, angle B = 45 degrees, and side AC = 8 units, find the length of side BC.
Solution:
Using the Law of Sines:
BC / sin(A) = AC / sin(B)
BC / sin(60°) = 8 / sin(45°)
BC / (√3/2) = 8 / (√2/2)
BC = (8 * √3) / √2
BC = 4√6 units
Question 2:
Triangle DEF has angle D = 30 degrees, angle E = 60 degrees, and side DE = 5 units. Determine the length of side EF.
Solution:
Using the Law of Sines:
EF / sin(D) = DE / sin(E)
EF / sin(30°) = 5 / sin(60°)
EF / (1/2) = 5 / (√3/2)
EF = (5 * 2) / √3
EF = (10√3) / 3 units
Question 3:
In triangle PQR, if side PQ = 10 units, angle P = 30 degrees, and angle Q = 45 degrees, calculate the length of side QR.
Solution:
Using the Law of Sines:
QR / sin(P) = PQ / sin(Q)
QR / sin(30°) = 10 / sin(45°)
QR / (1/2) = 10 / (√2/2)
QR = (10 * 2) / √2
QR = 10√2 units
Question 4:
Triangle XYZ has angle X = 45 degrees, angle Y = 60 degrees, and side XY = 6 units. Find the length of side YZ.
Solution:
Using the Law of Sines:
YZ / sin(X) = XY / sin(Y)
YZ / sin(45°) = 6 / sin(60°)
YZ / (√2/2) = 6 / (√3/2)
YZ = (6 * √2) / √3
YZ = (6√6) / 3 units
YZ = 2√6 units
Question 5:
In triangle ABC, if side AB = 8 units, side BC = 6 units, and side AC = 10 units, calculate the measure of angle A.
Solution:
Using the Law of Cosines:
AC^2 = AB^2 + BC^2 - 2 * AB * BC * cos(A)
10^2 = 8^2 + 6^2 - 2 * 8 * 6 * cos(A)
100 = 64 + 36 - 96 * cos(A)
cos(A) = (100 - 100) / 96
cos(A) = 0
A = 90 degrees
Question 6:
Triangle DEF has side DE = 7 units, side EF = 9 units, and side FD = 10 units. Determine the measure of angle E.
Solution:
Using the Law of Cosines:
EF^2 = DE^2 + FD^2 - 2 * DE * FD * cos(E)
9^2 = 7^2 + 10^2 - 2 * 7 * 10 * cos(E)
81 = 49 + 100 - 140 * cos(E)
cos(E) = (100 - 81 - 49) / (-140)
cos(E) = -30/(-140)
cos(E) = 3
/14
E ≈ 67.4 degrees
Question 7:
In triangle PQR, if side PQ = 5 units, angle P = 30 degrees, and angle Q = 60 degrees, find the length of side PR.
Solution:
Using the Law of Sines:
PR / sin(P) = PQ / sin(Q)
PR / sin(30°) = 5 / sin(60°)
PR / (1/2) = 5 / (√3/2)
PR = (5 * 2) / √3
PR = (10√3) / 3 units
Question 8:
Triangle STU has angle S = 45 degrees, angle T = 60 degrees, and side ST = 9 units. Determine the length of side SU.
Solution:
Using the Law of Sines:
SU / sin(S) = ST / sin(T)
SU / sin(45°) = 9 / sin(60°)
SU / (√2/2) = 9 / (√3/2)
SU = (9 * √2) / √3
SU = (9√6) / 3 units
SU = 3√6 units
Question 9:
In triangle XYZ, if angle X = 60 degrees, angle Y = 45 degrees, and side XY = 5 units, find the length of side YZ.
Solution:
Using the Law of Sines:
YZ / sin(X) = XY / sin(Y)
YZ / sin(60°) = 5 / sin(45°)
YZ / (√3/2) = 5 / (√2/2)
YZ = (5 * √3) / √2
YZ = (5√6) / 2 units
Question 10:
Triangle ABC has side AB = 12 units, side BC = 9 units, and side AC = 15 units. Calculate the measure of angle A.
Solution:
Using the Law of Cosines:
AC^2 = AB^2 + BC^2 - 2 * AB * BC * cos(A)
15^2 = 12^2 + 9^2 - 2 * 12 * 9 * cos(A)
225 = 144 + 81 - 216 * cos(A)
cos(A) = (225 - 144 - 81) / (-216)
cos(A) = 0
A = 90 degrees
These 10 questions cover various concepts related to the solution of triangles and their solutions demonstrate the application of the Law of Sines and the Law of Cosines to find unknown side lengths and angles in triangles.
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