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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.

**Eager**

A California-based travel writer, lover of food, oceans, and nature.

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