In the JEE (Joint Entrance Examination) syllabus, the topic of "Solution of Triangles" refers to the study of finding unknown side lengths and angles of a triangle using various trigonometric ratios and laws. The main concepts covered in the Solution of Triangles include.
Law of Sines: This law relates the ratios of the lengths of the sides of a triangle to the sines of their opposite angles. It is given by the equation: a/sin(A) = b/sin(B) = c/sin(C) where a, b, and c are the side lengths, and A, B, and C are the corresponding angles of the triangle.
Law of Cosines: This law relates the lengths of the sides of a triangle to the cosine of one of its angles. It is given by the equation:
c^2 = a^2 + b^2 - 2ab*cos(C)
where c is the side opposite to angle C, and a and b are the lengths of the other two sides.
Use of Trigonometric Ratios: The Solution of Triangles involves the application of trigonometric ratios such as sine, cosine, and tangent to find unknown side lengths and angles in a triangle. This includes using the sine rule, cosine rule, and tangent rule in various scenarios.
Solving for Unknowns: The Solution of Triangles also involves solving systems of equations, often using the above trigonometric laws, to find the unknown side lengths or angles of a triangle. This may include using the concept of congruence and similarity of triangles.
In JEE, questions related to the Solution of Triangles can range from finding missing side lengths or angles in a given triangle to solving complex geometric problems involving multiple triangles and their properties.
It is important to have a strong understanding of trigonometric concepts, such as trigonometric ratios, identities, and basic trigonometric functions, to effectively solve problems related to the Solution of Triangles in the JEE examination.
Here are 10 basic JEE-level questions on the solution of triangles along with their solutions.
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.