Basics of Trigonometry (Level - 3)
Q1–10 of 30Let ABCD be a parallelogram. The lengths of side AD and diagonal AC are 10 cm and 20 cm, respectively. If ∠ADC = 30°, then the area of the parallelogram, in sq cm, is:
In a triangle ABC, the lengths of sides AB and AC are 4 cm and 6 cm respectively. If the area of △ABC is 6√3 sq cm, what are the possible values for ∠A?
Hint: Area formula ½·AB·AC·sinA = 6√3 ⟹ ½(4)(6)sinA = 6√3 ⟹ sinA = √3/2. Both 60° and 120° are valid acute and obtuse angles in a triangle.
Find the number of real solutions of the equation 2cos(x(x+1)) = 2x + 2−x.
Hint: Apply AM-GM on RHS: 2x + 2−x ≥ 2. On LHS, since cosθ ≤ 1, 2cos(x(x+1)) ≤ 21 = 2. Both sides equal 2 simultaneously only when x = 0.
A chord of length 5 cm subtends an angle of 60° at the center of a circle. The length, in cm, of a chord that subtends an angle of 120° at the center of the same circle is:
Hint: A 60° chord forms an equilateral triangle with radii, so R = 5 cm. A chord subtending 120° has length 2Rsin60° = 2(5)(√3/2) = 5√3 cm.
In △ABC, AB = 1 cm, BC = 1 cm, and ∠ABC = 30°. Find the length of side AC.
In △ABC, AB = 4, BC = 5, and AC = 6. Point D is on BC such that AD ⊥ BC. Find the length of segment BD.
Hint: Use Cosine Rule on ∠B: cosB = (4²+5²−6²)/(2·4·5) = 5/40 = 1/8. In right triangle △ABD, BD = AB·cosB = 4×1/8 = 0.5.
A vertical line segment OP of height h stands at the center O of a square ABCD of side b. Suppose ∠APB = 60°, then the relationship between h and b is:
Hint: Isosceles △APB with vertex 60° is equilateral, so AP = b. Half-diagonal AO = b/√2. Apply Pythagoras in △AOP: b² = (b/√2)² + h² ⟹ 2h² = b².
A ladder leans against a vertical wall. The top of the ladder is 8 m above the ground. When the bottom is moved 2 m farther away, the top slides down and rests against the foot of the wall. Find the ladder's length.
Hint: Let initial base distance be x. Ladder length L = x+2. Pythagoras: 8²+x² = (x+2)² ⟹ 64+x² = x²+4x+4 ⟹ x=15 ⟹ L=17 m.
In a quadrilateral ABCD, ∠B = 90°, ∠D = 90°, AB = 3, BC = 4, and CD = 1. Find the value of sin(∠BAD).
In an isosceles trapezium ABCD with AB ∥ CD, AB = 10 cm, CD = 4 cm, and non-parallel sides AD = BC = 5 cm. Find cos(∠A).
Hint: Drop perpendiculars from C and D to AB. Base projection on each side is (10−4)/2 = 3 cm. In the right triangle formed at corner A, cos(∠A) = base/hypotenuse = 3/5.