JEE Mathematics Properties of Triangles and Heights and Distances MCQs Set 01

Answer: (b) 2

Explanation:
1. For A + B + C = π, cos²A + cos²B + cos²C + 2cos A cos B cos C = 1.
2. So sin²A + sin²B + sin²C = 3 − (cos²A + cos²B + cos²C) = 2 + 2cos A cos B cos C.
3. Hence the expression equals 2.

Teacher's Note:
1. Memorise the identity for cos² terms in a triangle.
2. Test with an equilateral triangle: 3 × 3/4 − 2 × 1/8 = 2.

Answer: (d) 120°

Explanation:
1. With R = AB = AC = c, the sine rule gives c = 2R sin C, so sin C = 1/2.
2. B = C and both are acute, so B = C = 30°.
3. A = 180° − 60° = 120°.

Teacher's Note:
1. Use c = 2R sin C.
2. The two equal base angles must be acute.

Answer: (d) \( \sqrt3 \)

Explanation:
1. cos A/sin A = cos B/sin B = cos C/sin C (since a : b : c = sin A : sin B : sin C), so cot A = cot B = cot C.
2. All three angles are equal, so the triangle is equilateral with side 2.
3. Area = (√3/4) × 2² = √3.

Teacher's Note:
1. Replace the sides by sines using the sine rule.
2. Equal cotangents mean equal angles.

Answer: (a) 13

Explanation:
1. cot A + cot B = sin C/(sin A sin B) and cot C = cos C/sin C, so the ratio = sin²C/(sin A sin B cos C) = c²/(ab cos C).
2. ab cos C = (a² + b² − c²)/2, and a² + b² = 17c²/9 so a² + b² − c² = 8c²/9.
3. The ratio = c² / (4c²/9) = 9/4, so m + n = 13.

Teacher's Note:
1. Convert sines and cosines to sides with the sine and cosine rules.
2. Use the given relation to express a² + b² in terms of c².

Answer: (a) 30°

Explanation:
1. By the sine rule, sin C = c sin A / a = 2 × (√2/2) / (2√2) = 1/2.
2. So C = 30° or 150°. Since A = 45°, C = 150° would make A + C > 180°, so C = 30°.

Teacher's Note:
1. Reject any value that makes the angle sum exceed 180°.

Answer: (c) Both 1 and 2

Explanation:
1. sin A + sin B = sin C would mean a + b = c, but in a triangle a + b > c, so no such triangle exists: statement 1 is correct.
2. Angles in ratio 1 : 2 : 3 are 30°, 60°, 90°, and the sides are proportional to sin 30° : sin 60° : sin 90° = 1/2 : √3/2 : 1 = 1 : √3 : 2: statement 2 is correct.

Teacher's Note:
1. Sides are proportional to the sines of the opposite angles.
2. The triangle inequality rules out a + b = c.

Answer: (c) sin²A

Explanation:
1. The numerator is 16Δ² (Heron's formula: 16s(s − a)(s − b)(s − c) = 16Δ²).
2. So the expression = 4Δ²/(b²c²), and with Δ = ½bc sin A this equals sin²A.

Teacher's Note:
1. Recognise Heron's formula in the numerator.
2. Then substitute Δ = ½bc sin A.

Answer: (b) 4 : 1 : 3

Explanation:
1. BC = a, so DC = BD = AD = AC = a/2 = b.
2. In triangle ABD, ∠ADB = 180° − 60° = 120°, so c² = (a/2)² + (a/2)² − 2(a/2)² cos 120° = 3a²/4.
3. So a² : b² : c² = a² : a²/4 : 3a²/4 = 4 : 1 : 3.

Teacher's Note:
1. Find all lengths in terms of a.
2. Use the cosine rule in triangle ABD with the 120° angle.

Answer: (b) 2a² = 3h²

Explanation:
1. Let P be the top of the tower and A, B, C the vertices. Each triangle PAB has apex angle 60° and PA = PB, so it is equilateral and PA = a.
2. The distance from the centre to a vertex is a/√3.
3. h² = PA² − a²/3 = 2a²/3, so 2a² = 3h².

Teacher's Note:
1. An isosceles triangle with a 60° apex is equilateral.
2. The circumradius of an equilateral triangle of side a is a/√3.

Answer: (a) 0.5

Explanation:
1. sin C = sin(A + B) = sin A cos B + cos A sin B, so sin A cos B = 0 and cos B = 0: B = 90°.
2. Then b is the hypotenuse and c/b = 3/5, so cos A = 3/5 and sin A = 4/5.
3. tan(A/2) = sin A/(1 + cos A) = (4/5)/(8/5) = 0.5.

Teacher's Note:
1. Use sin C = sin(A + B) to show the triangle is right angled.
2. tan(A/2) = sin A/(1 + cos A).

Answer: (b) \( \frac43 \)

Explanation:
1. a² − (b − c)² = 2bc − (b² + c² − a²) = 2bc − 2bc cos A = 2bc(1 − cos A).
2. Also 2Δ = bc sin A, so sin A = 2(1 − cos A). Since sin A/(1 − cos A) = cot(A/2), we get cot(A/2) = 2 and tan(A/2) = 1/2.
3. tan A = 2(1/2)/(1 − 1/4) = 4/3.

Teacher's Note:
1. Use the cosine rule to rewrite a² − (b − c)².
2. Then apply the double-angle formula for tan A.

Answer: (c) they are in A.P.

Explanation:
1. sin²A − sin²B = sin(A + B) sin(A − B) = sin C sin(A − B) and sin²B − sin²C = sin A sin(B − C).
2. Hence sin(A − B)/sin(B − C) = [(sin²A − sin²B)/sin C] / [(sin²B − sin²C)/sin A] = (sin A/sin C)(sin²A − sin²B)/(sin²B − sin²C).
3. Equating with sin A/sin C gives sin²A − sin²B = sin²B − sin²C, i.e. a² + c² = 2b², so a², b², c² are in A.P.

Teacher's Note:
1. Use sin²P − sin²Q = sin(P + Q) sin(P − Q).
2. Then switch to sides with the sine rule.

Answer: (c) \( \frac{\sqrt3}{\sqrt3+2} \)

Explanation:
1. The angles are 120°, 30°, 30°. Sides are proportional to sin 120° : sin 30° : sin 30° = √3/2 : 1/2 : 1/2.
2. Greatest side / perimeter = (√3/2)/(√3/2 + 1) = √3/(√3 + 2).

Teacher's Note:
1. Divide the angle sum 180° in the given ratio.
2. Sides are proportional to the sines of the opposite angles.

Answer: (d) \( \frac{191}{60} \)

Explanation:
1. Using r₁ = Δ/(s − a) etc., s − b = 2(s − a) and s − c = 3(s − a). Let s − a = k, so s − b = 2k and s − c = 3k.
2. s = (s − a) + (s − b) + (s − c) = 6k, so a = 5k, b = 4k and c = 3k.
3. a/b + b/c + c/a = 5/4 + 4/3 + 3/5 = (75 + 80 + 36)/60 = 191/60.

Teacher's Note:
1. Write each exradius as Δ divided by (s minus the side).
2. Add the three quantities s − a, s − b, s − c to get s.

Answer: (a) 1

Explanation:
1. Since A/2 + B/2 = π/2 − C/2, tan(A/2 + B/2) = cot(C/2).
2. (tan(A/2) + tan(B/2))/(1 − tan(A/2)tan(B/2)) = 1/tan(C/2), which gives the sum of pair products = 1.

Teacher's Note:
1. Use tan(x + y) with x + y = π/2 − C/2.

Answer: (d) 7/8

Explanation:
1. Rewrite the condition as (a − 2b)² + (a − c)² = 0.
2. So a = 2b and a = c. Take b = 1, a = c = 2.
3. cos B = (a² + c² − b²)/(2ac) = (4 + 4 − 1)/8 = 7/8.

Teacher's Note:
1. Look for a sum of two squares equal to zero.
2. Then apply the cosine rule.

Answer: (a) \( \frac{\sqrt3+1}{2} \)

Explanation:
1. The third angle is 105°, and the included side c is opposite it.
2. Area = c² sin A sin B/(2 sin C) = (√3 + 1)² (1/2)(√2/2)/(2 sin 105°), with sin 105° = (√6 + √2)/4.
3. This simplifies to (√3 + 1)/2.

Teacher's Note:
1. The side between two given angles is opposite the third angle.
2. Use Area = c² sin A sin B / (2 sin C).

Answer: (c) \( \left(\frac{3}{4\Delta}\right)^2 \)

Explanation:
1. (2 sin P − sin 2P)/(2 sin P + sin 2P) = (1 − cos P)/(1 + cos P) = tan²(P/2).
2. s = 4, so tan²(P/2) = (s − b)(s − c)/(s(s − a)) = (1/2)(3/2)/(4 × 2) = 3/32.
3. Heron: Δ² = 4 × 2 × (1/2) × (3/2) = 6, and (3/(4Δ))² = 9/96 = 3/32, which matches option (c).

Teacher's Note:
1. Cancel 2 sin P from numerator and denominator.
2. tan²(P/2) = (s − b)(s − c)/(s(s − a)).

Answer: (b) sin(A − B)

Explanation:
1. a² + b² = c², so (a² − b²)/(a² + b²) = sin²A − sin²B (sine rule, c = 2R).
2. sin²A − sin²B = sin(A + B) sin(A − B) = sin 90° sin(A − B) = sin(A − B).

Teacher's Note:
1. Use a = 2R sin A and c = 2R.
2. sin²A − sin²B = sin(A + B) sin(A − B).

Answer: (b) \( \frac{abc}{R^3} \)

Explanation:
1. ∠BOC = 2A, so R₁ = a/(2 sin 2A) = 2R sin A/(4 sin A cos A) = R/(2 cos A).
2. a/R₁ = 2R sin A · 2 cos A/R = 2 sin 2A, so the sum is 2(sin 2A + sin 2B + sin 2C) = 8 sin A sin B sin C.
3. 8 sin A sin B sin C = 8 · abc/(8R³) = abc/R³.

Teacher's Note:
1. Use the sine rule in triangle OBC.
2. sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C for a triangle.

Answer: (d) 8

Explanation:
1. c/a = 1/sin A and c/b = 1/cos A, so the expression is ((sin A + cos A)/(sin A cos A))².
2. Let t = sin A + cos A ∈ (1, √2]; sin A cos A = (t² − 1)/2, so the bracket is 2t/(t² − 1), which decreases as t increases.
3. Minimum at t = √2: 2√2, so the minimum of the square is 8.

Teacher's Note:
1. Express in terms of t = sin A + cos A.
2. The minimum occurs when A = 45°.

Answer: (d) 4

Explanation:
1. d₁ = R cos A, d₂ = R cos B, d₃ = R cos C, so a/d₁ = 2 tan A etc.
2. The left side is λ · 2(tan A + tan B + tan C) = 2λ tan A tan B tan C.
3. abc/(d₁d₂d₃) = 8R³ sin A sin B sin C/(R³ cos A cos B cos C) = 8 tan A tan B tan C. So λ = 4.

Teacher's Note:
1. Express d₁, d₂, d₃ using R and the cosines of the angles.
2. tan A + tan B + tan C = tan A tan B tan C in a triangle.

Answer: (b) circumcentre

Explanation:
1. Equal elevation from A, B and C means equal horizontal distances from the foot of the pole to A, B and C (the height is the same).
2. The point equidistant from the three vertices is the circumcentre.

Teacher's Note:
1. Same elevation and same height imply same horizontal distance.
2. The point equidistant from the vertices is the circumcentre.