CBSE Class 11 Mathematics Solution of Triangles MCQs Set 02

Answer: (c) 90°

Explanation:
1. Multiply by abc: \( 2bc\cos A + ac\cos B + 2ab\cos C = a^2 + b^2 \).
2. Using the cosine rule, the left side is \( (b^2 + c^2 - a^2) + \dfrac{a^2 + c^2 - b^2}{2} + (a^2 + b^2 - c^2) = \dfrac{3b^2 + a^2 + c^2}{2} \).
3. Equating to a² + b²: \( b^2 + c^2 = a^2 \). So A = 90°.

Teacher's Note:
1. Clear denominators and replace each term using the cosine rule.
2. The final relation is Pythagoras' theorem.

Answer: (a) 1 : 2

Explanation:
1. With height h, the base is \( h\cot 22.5° + h\cot 112.5° = h(\cot 22.5° - \tan 22.5°) = 2h\cot 45° = 2h \).
2. So h : base = 1 : 2.

Teacher's Note:
1. Express the base as a sum of two segments using cotangents.
2. cot 112.5° = −tan 22.5°, and cot x − tan x = 2 cot 2x.

Answer: (b) \( \dfrac{8}{15} \)

Explanation:
1. \( a^2 - (b - c)^2 = 2bc(1 - \cos A) \) and \( \Delta = \tfrac{1}{2}bc\sin A \), so \( \sin A = 4(1 - \cos A) \).
2. This gives \( \tan\dfrac{A}{2} = \dfrac{1}{4} \).
3. \( \tan A = \dfrac{2 \times \frac{1}{4}}{1 - \frac{1}{16}} = \dfrac{8}{15} \).

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

Answer: (a) \( \dfrac{\pi}{6} \)

Explanation:
1. Perimeter = 6 × (sin A + sin B + sin C)/3 = 2(sin A + sin B + sin C).
2. But \( a + b + c = 2R(\sin A + \sin B + \sin C) \), so R = 1.
3. \( \sin A = \dfrac{a}{2R} = \dfrac{1}{2} \), so A = π/6 (A = 5π/6 is not an option).

Teacher's Note:
1. Write the perimeter as 2R times the sum of sines.
2. Then sin A = a/2R.

Answer: (c) n = 7

Explanation:
1. With circumradius R, \( A_1A_k = 2R\sin\dfrac{(k - 1)\pi}{n} \). Let θ = π/n.
2. The condition is \( \dfrac{1}{\sin\theta} = \dfrac{1}{\sin 2\theta} + \dfrac{1}{\sin 3\theta} \), which holds for \( \theta = \dfrac{\pi}{7} \) (checked numerically for n = 5, 6, 7).
3. So n = 7.

Teacher's Note:
1. Chord length in a regular polygon is 2R sin(kπ/n).
2. Check the options numerically: only n = 7 satisfies the equation.

Answer: (b) The altitudes are in H.P.

Explanation:
1. sin P, sin Q, sin R in A.P. means the sides p, q, r are in A.P.
2. Altitudes are \( \dfrac{2\Delta}{p},\dfrac{2\Delta}{q},\dfrac{2\Delta}{r} \), the reciprocals of the sides, so they are in H.P.

Teacher's Note:
1. Sides are proportional to the sines.
2. Altitude is inversely proportional to its base for a fixed area.

Answer: (c) 4

Explanation:
1. Let BD = DE = EC = m. In the triangles ABE and ADC use the ratio rule: \( \dfrac{BE}{EC} = \dfrac{AB\sin(x + y)}{AC\sin z} \) and \( \dfrac{BD}{DC} = \dfrac{AB\sin x}{AC\sin(y + z)} \).
2. Here BE/EC = 2 and BD/DC = 1/2. Dividing the two: \( \dfrac{\sin(x + y)\sin(y + z)}{\sin x\sin z} = \dfrac{2}{1/2} = 4 \).

Teacher's Note:
1. Use the m : n theorem for a cevian dividing the opposite side.
2. Divide the two ratio equations so AB and AC cancel.

Answer: (d) Both (a) and (c)

Explanation:
1. Mollweide's formula gives \( \dfrac{a + c}{b} = \dfrac{\cos\frac{A - C}{2}}{\sin\frac{B}{2}} \), so the condition becomes \( b = 2\sin\dfrac{B}{2}\sqrt{a^2 + c^2 - ac} \).
2. Squaring, with \( b^2 = a^2 + c^2 - 2ac\cos B \) and \( 4\sin^2\frac{B}{2} = 2(1 - \cos B) \), simplifies to \( (a - c)^2(1 - 2\cos B) = 0 \).
3. Since a ≠ c, cos B = 1/2, so B = π/3. Then A + C = 2B, so A, B, C are in A.P. Both (a) and (c) hold.

Teacher's Note:
1. Use Mollweide's formula to replace the cosine of the half difference.
2. The condition a ≠ c is needed: for a = c any isosceles triangle satisfies the equation.

Answer: (a) \( qx^3 - px^2 + (1 + q)x - p = 0 \)

Explanation:
1. In a triangle, \( \tan A + \tan B + \tan C = \tan A\tan B\tan C = \dfrac{p}{q} \).
2. Also \( \sum\tan A\tan B = \dfrac{1 + q}{q} \). So the cubic is \( x^3 - \dfrac{p}{q}x^2 + \dfrac{1 + q}{q}x - \dfrac{p}{q} = 0 \).
3. Multiplying by q gives option (a).

Teacher's Note:
1. Use Vieta's formulas for the sum, pair-sum and product of the tangents.
2. The identity Σ tan A tan B = (1 + q)/q can be checked numerically.

Answer: (c) \( a\cos A + b\cos B + c\cos C = 4R\sin A\sin B\sin C \)

Explanation:
1. \( a\cos A = 2R\sin A\cos A = R\sin 2A \).
2. Sum = \( R(\sin 2A + \sin 2B + \sin 2C) = R \cdot 4\sin A\sin B\sin C \).

Teacher's Note:
1. Use a = 2R sin A.
2. sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C in a triangle.

Answer: (a) \( 4R\sin A\sin B\sin C \)

Explanation:
1. As shown in the previous question, the sum equals \( R(\sin 2A + \sin 2B + \sin 2C) = 4R\sin A\sin B\sin C \).

Teacher's Note:
1. Same identity as before, with different options.
2. Check dimensions: the answer must have the dimension of length.

Answer: (a) \( \sqrt{\dfrac{8}{7}} \)

Explanation:
1. s = 7, so \( \Delta = \sqrt{7 \times 4 \times 2 \times 1} = \sqrt{56} \).
2. \( r = \dfrac{\Delta}{s} = \dfrac{\sqrt{56}}{7} = \sqrt{\dfrac{8}{7}} \).

Teacher's Note:
1. Find s first, then Heron's formula.
2. r = Δ/s.

Answer: (c) 2 cm

Explanation:
1. For an equilateral triangle, \( R = \dfrac{a}{\sqrt{3}} = \dfrac{2\sqrt{3}}{\sqrt{3}} = 2 \) cm.

Teacher's Note:
1. R = a/(2 sin 60°) = a/√3.
2. The centroid and circumcentre coincide.

Answer: (b) 2.5

Explanation:
1. Since 3² + 4² = 5², the triangle is right angled and the hypotenuse is a diameter of the circumcircle.
2. R = 5/2 = 2.5.

Teacher's Note:
1. In a right triangle R is half the hypotenuse.
2. Check with R = abc/4Δ = 60/24 = 2.5.

Answer: (b) \( \dfrac{16}{7} \)

Explanation:
1. Take sides 4, 5, 6: s = 7.5 and \( \Delta^2 = 7.5 \times 3.5 \times 2.5 \times 1.5 \).
2. \( \dfrac{R}{r} = \dfrac{abc/4\Delta}{\Delta/s} = \dfrac{abc\,s}{4\Delta^2} = \dfrac{120 \times 7.5}{4 \times 98.4375} = \dfrac{16}{7} \).

Teacher's Note:
1. Use R = abc/4Δ and r = Δ/s.
2. R/r = abc·s/(4Δ²).

Answer: (c) \( R = \dfrac{abc}{4\Delta} \)

Explanation:
1. \( \Delta = \tfrac{1}{2}bc\sin A \) and \( \sin A = \dfrac{a}{2R} \), so \( \Delta = \dfrac{abc}{4R} \).
2. Hence \( R = \dfrac{abc}{4\Delta} \).

Teacher's Note:
1. Combine the area formula with the sine rule.
2. Check units: R is a length, abc/Δ is a length.

Answer: (b) \( \dfrac{a^2}{6} \)

Explanation:
1. Inradius \( r = \dfrac{a}{2\sqrt{3}} \). The square inscribed in the circle has diagonal 2r.
2. Area = (diagonal)²/2 = \( \dfrac{4r^2}{2} = 2r^2 = \dfrac{2a^2}{12} = \dfrac{a^2}{6} \).

Teacher's Note:
1. The diagonal of the inscribed square equals the circle's diameter.
2. Area of a square = diagonal²/2.

Answer: (b) 61

Explanation:
1. \( 11^2 + 60^2 = 121 + 3600 = 3721 = 61^2 \), so the triangle is right angled.
2. The hypotenuse 61 is the diameter of the circumcircle.

Teacher's Note:
1. Test the three sides with Pythagoras.
2. The hypotenuse of a right triangle is a diameter.

Answer: (c) 2 : 1 : 3

Explanation:
1. For side a: \( R = \dfrac{a}{\sqrt{3}},\ r = \dfrac{a}{2\sqrt{3}},\ r_1 = \dfrac{\Delta}{s - a} = \dfrac{\sqrt{3}a^2/4}{a/2} = \dfrac{\sqrt{3}a}{2} \).
2. Ratio \( = \dfrac{1}{\sqrt{3}} : \dfrac{1}{2\sqrt{3}} : \dfrac{\sqrt{3}}{2} = 2 : 1 : 3 \).

Teacher's Note:
1. Compute each radius in terms of a.
2. Multiply the ratio by 2√3 to clear fractions.

Answer: (c) \( \csc\dfrac{A}{2} : \csc\dfrac{B}{2} : \csc\dfrac{C}{2} \)

Explanation:
1. The perpendicular from I to AB has length r, and \( IA = \dfrac{r}{\sin\frac{A}{2}} \); similarly for IB and IC.
2. So IA : IB : IC = cosec(A/2) : cosec(B/2) : cosec(C/2).

Teacher's Note:
1. Drop a perpendicular from I to a side to form a right triangle.
2. IA = r / sin(A/2).

Answer: (b) a + c = 2b

Explanation:
1. Let \( r_1 = \lambda \). Then \( r_2 = \dfrac{\lambda}{2},\ r_3 = \dfrac{\lambda}{3} \).
2. Since \( r_i = \Delta/(s - a) \), the numbers s − a, s − b, s − c are proportional to \( 1 : 2 : 3 \).
3. Let s − a = k, s − b = 2k, s − c = 3k. Then s = 6k, a = 5k, b = 4k, c = 3k, and a + c = 8k = 2b.

Teacher's Note:
1. Convert each ex-radius into Δ/(s − a).
2. Add the three differences to get s.

Answer: (a) 3r

Explanation:
1. We know \( \dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3} = \dfrac{1}{r} \).
2. The H.M. of r₁, r₂, r₃ is \( \dfrac{3}{\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}} = 3r \).

Teacher's Note:
1. Recall the relation between r and the ex-radii.
2. H.M. of three numbers is 3 divided by the sum of reciprocals.

Answer: (d) \( \dfrac{2\pi}{3} \)

Explanation:
1. PQ is opposite angle R, so \( PQ = 2R_c\sin R \) where R_c is the circumradius. Since PQ = R_c, sin R = 1/2.
2. Base angles are equal and acute, so Q = R = 30°.
3. Then P = 180° − 60° = 120° = 2π/3.

Teacher's Note:
1. Use the sine rule with the circumradius.
2. The base angles of an isosceles triangle are acute.

Answer: (d) \( \dfrac{a^2b^2c^2}{8R^3} \)

Explanation:
1. \( p_1 = \dfrac{2\Delta}{a} \) and so on, so \( p_1p_2p_3 = \dfrac{8\Delta^3}{abc} \).
2. With \( \Delta = \dfrac{abc}{4R} \), \( p_1p_2p_3 = \dfrac{8}{abc}\cdot\dfrac{a^3b^3c^3}{64R^3} = \dfrac{a^2b^2c^2}{8R^3} \).

Teacher's Note:
1. Altitude = 2Δ/side and Δ = abc/4R.
2. The answer must have the dimension of length cubed.

Answer: (a) 0

Explanation:
1. \( r_1r_2r_3 = \dfrac{\Delta^3}{(s - a)(s - b)(s - c)} = \Delta s \) since \( (s - a)(s - b)(s - c) = \dfrac{\Delta^2}{s} \).
2. \( r_1r_2 + r_2r_3 + r_3r_1 = s^2 \), so \( r \cdot s^2 = \dfrac{\Delta}{s}s^2 = \Delta s \).
3. The difference is 0.

Teacher's Note:
1. Use r₁r₂ + r₂r₃ + r₃r₁ = s².
2. Both products reduce to Δs.

Answer: (b) 1

Explanation:
1. Distance of the incentre from BC is r. Distance of the circumcentre from BC is \( R\cos A \).
2. Parallel means \( r = R\cos A \). Using \( \cos A + \cos B + \cos C = 1 + \dfrac{r}{R} \) we get \( \cos A + \cos B + \cos C = 1 + \cos A \).
3. Hence cos B + cos C = 1.

Teacher's Note:
1. Compare perpendicular distances from O and I to BC.
2. Use cos A + cos B + cos C = 1 + r/R.

Answer: (b) \( \dfrac{-\sqrt{3}}{2} \)

Explanation:
1. \( \cos\left(\alpha + \dfrac{\pi}{2}\right) = -\sin\alpha \), so the A.M. is \( -\dfrac{\sin\alpha + \sin\beta + \sin\gamma}{3} \).
2. With α + β + γ = 2π, the sum of sines is largest when α = β = γ = 2π/3, giving \( 3 \times \dfrac{\sqrt{3}}{2} \).
3. Minimum A.M. = \( -\dfrac{\sqrt{3}}{2} \).

Teacher's Note:
1. Turn cos(x + π/2) into −sin x.
2. The sum of sines is greatest for equal angles (Jensen).

Answer: (d) Obtuse angled

Explanation:
1. \( c^2 = 47^2 + 94^2 - 2(47)(94)\cos 30° \approx 3393 \), so c ≈ 58.2 cm.
2. The largest side is b = 94 and \( a^2 + c^2 \approx 5602 < b^2 = 8836 \).
3. So the angle B is obtuse.

Teacher's Note:
1. Find c with the cosine rule.
2. Compare b² with a² + c².

Answer: (c) a = b

Explanation:
1. With a = 2R sin A and b = 2R sin B, \( a\sin A = b\sin B \) gives \( a^2 = b^2 \).
2. So a = b (isosceles).

Teacher's Note:
1. Replace sin A by a/2R.
2. Sides are positive, so a² = b² means a = b.

Answer: (a) Obtuse angled

Explanation:
1. Check the largest side: \( 6^2 + 10^2 = 136 < 14^2 = 196 \).
2. \( \cos\theta = \dfrac{36 + 100 - 196}{120} = -\dfrac{1}{2} < 0 \). The triangle is obtuse angled (angle 120°).

Teacher's Note:
1. Compare the square of the longest side with the sum of the other two squares.
2. Negative cosine means obtuse.

Answer: (a) \( b_1 + b_2 = 2c\cos A \)

Explanation:
1. Since \( c\sin A < a < c \), there are two triangles (the ambiguous case), with third sides b₁ and b₂.
2. From \( a^2 = b^2 + c^2 - 2bc\cos A \), b satisfies \( b^2 - (2c\cos A)b + (c^2 - a^2) = 0 \).
3. The sum of its roots is \( b_1 + b_2 = 2c\cos A \).

Teacher's Note:
1. Treat the cosine rule as a quadratic in the unknown side.
2. Sum of roots = −(coefficient of b).

Answer: (c) Equilateral

Explanation:
1. Using the sine rule a = k sin A etc., \( \dfrac{\cos A}{\sin A} = \dfrac{\cos B}{\sin B} = \dfrac{\cos C}{\sin C} \), so cot A = cot B = cot C.
2. Hence A = B = C: the triangle is equilateral.

Teacher's Note:
1. Substitute a, b, c in terms of sines.
2. Equal cotangents mean equal angles.

Answer: (c) Right angled

Explanation:
1. By the sine rule, \( \dfrac{b + c}{a} = \dfrac{\sin B + \sin C}{\sin A} = \dfrac{\cos\frac{B - C}{2}}{\sin\frac{A}{2}} \).
2. So the condition \( \cot\dfrac{A}{2} = \dfrac{b + c}{a} \) gives \( \cos\dfrac{A}{2} = \cos\dfrac{B - C}{2} \), i.e. A = B − C or A = C − B.
3. Then B = A + C or C = A + B, so B = 90° or C = 90°. The triangle is right angled.

Teacher's Note:
1. Convert the sides into sines and simplify with the sum-to-product formula.
2. Test the options with a 3-4-5 triangle.

Answer: (b) Isosceles triangle

Explanation:
1. Using a = 2R sin A and b = 2R sin B: \( \sin A\cos B = \sin B\cos A \), so \( \sin(A - B) = 0 \).
2. Hence A = B and the triangle is isosceles.

Teacher's Note:
1. Replace sides with sines and use the compound-angle formula.
2. A = B does not force the third angle, so it need not be equilateral.

Answer: (c) Right angled

Explanation:
1. Read the first part as: one side is twice another, a = 2b, and the opposite angles satisfy A − B = 60°.
2. \( \tan 30° = \dfrac{a - b}{a + b}\cot\dfrac{C}{2} = \dfrac{1}{3}\cot\dfrac{C}{2} \), so \( \cot\dfrac{C}{2} = \sqrt{3} \) and C = 60°.
3. Then A + B = 120° and A − B = 60° give A = 90°, B = 30°: right angled.

Teacher's Note:
1. Napier's analogy links a − b, a + b and (A − B)/2.
2. Solve the two angle equations together.

Answer: (c) Obtuse angled

Explanation:
1. \( (3x + 4y)^2 + (4x + 3y)^2 = 25x^2 + 48xy + 25y^2 \).
2. \( (5x + 5y)^2 = 25x^2 + 50xy + 25y^2 \), which is larger because xy > 0.
3. The square of the longest side exceeds the sum of the other two squares, so the triangle is obtuse angled.

Teacher's Note:
1. Compare the square of the longest side with the sum of the other two squares.
2. The difference is 2xy, which is positive.

Answer: (c) 90°

Explanation:
1. \( \sin B = \dfrac{b\sin A}{a} = \dfrac{4\sqrt{3} \cdot \frac{\sqrt{3}}{2}}{6} = 1 \).
2. So B = 90°.

Teacher's Note:
1. Use the sine rule to get sin B.
2. sin B = 1 gives the single right-angled case of the ambiguous situation.

Answer: (b) Two solutions

Explanation:
1. \( b\sin A = 8 \times \dfrac{1}{2} = 4 \).
2. Since \( b\sin A = 4 < a = 7 < b = 8 \), the ambiguous case gives two triangles, so B has two solutions.

Teacher's Note:
1. Compare a with b sin A and with b.
2. b sin A < a < b is the two-solution condition.

Answer: (b) Isosceles

Explanation:
1. The third side: \( c^2 = 40^2 + (40\sqrt{3})^2 - 2(40)(40\sqrt{3})\cos 30° = 1600 + 4800 - 4800 = 1600 \), so c = 40.
2. The sides are 40, 40, \( 40\sqrt{3} \). Two sides are equal, so it is isosceles.
3. It is not right angled because \( 40^2 + 40^2 = 3200 \neq 4800 \).

Teacher's Note:
1. Find the third side with the cosine rule.
2. Check Pythagoras on the three sides before calling it right angled.

Answer: (d) Is equal to \( 2\tan^{-1}\dfrac{1}{9} \)

Explanation:
1. \( \dfrac{A - B}{2} = \dfrac{\pi}{4} \), so \( \tan\dfrac{A - B}{2} = 1 \).
2. By Napier's analogy, \( 1 = \dfrac{a - b}{a + b}\cot\dfrac{C}{2} = \dfrac{1}{9}\cot\dfrac{C}{2} \), so \( \tan\dfrac{C}{2} = \dfrac{1}{9} \).
3. Hence \( C = 2\tan^{-1}\dfrac{1}{9} \).

Teacher's Note:
1. A − B = π/2 gives tan 45° = 1.
2. Solve for tan(C/2) and then double the angle.