CBSE Class 11 Mathematics Solution of Triangles MCQs Set 01

Answer: (d) Isosceles

Explanation:
1. 2 sin R cos P = sin Q = sin(P + R).
2. So sin(P + R) − 2 sin R cos P = 0, i.e. sin(R − P) = 0, hence R = P.
3. Two angles are equal, so the triangle is isosceles.

Teacher's Note:
1. Replace sin Q by sin(P + R) because Q = π − (P + R).
2. A product 2 sin R cos P suggests the sum-to-product identity.

Answer: (a) \( \sqrt{2} : 2 : (\sqrt{3}+1) \)

Explanation:
1. Angles are 30°, 45°, 105°. Sides are proportional to sin 30° : sin 45° : sin 105°.
2. \( = \dfrac{1}{2} : \dfrac{1}{\sqrt{2}} : \dfrac{\sqrt{3}+1}{2\sqrt{2}} \); multiplying by \( 2\sqrt{2} \) gives \( \sqrt{2} : 2 : (\sqrt{3}+1) \).

Teacher's Note:
1. Use the sine rule to turn an angle ratio into a side ratio.
2. sin 105° = sin(60° + 45°) = (√3 + 1)/(2√2).

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

Explanation:
1. The smallest angle is opposite the smallest side \( \sqrt{24} = 2\sqrt{6} \).
2. With \( a = 6 + 2\sqrt{3},\ b = 4\sqrt{3},\ c = 2\sqrt{6} \): \( \cos C = \dfrac{a^2 + b^2 - c^2}{2ab} = \dfrac{(48 + 24\sqrt{3}) + 48 - 24}{2(6 + 2\sqrt{3})(4\sqrt{3})} = \dfrac{\sqrt{3}}{2} \).
3. So the smallest angle is π/6.

Teacher's Note:
1. The smallest side lies opposite the smallest angle.
2. Apply the cosine rule to that angle.

Answer: (c) \( \dfrac{\pi}{3} \)

Explanation:
1. \( \cos B = \dfrac{c^2 + a^2 - b^2}{2ca} = \dfrac{ac}{2ac} = \dfrac{1}{2} \).
2. Hence B = π/3.

Teacher's Note:
1. Compare the given relation directly with the cosine rule for angle B.
2. cos B = 1/2 gives 60°.

Answer: (c) \( a + b + c \)

Explanation:
1. Regroup: \( (b\cos C + c\cos B) + (c\cos A + a\cos C) + (a\cos B + b\cos A) \).
2. By the projection formulae each bracket equals a, b, c respectively.
3. So the sum is a + b + c.

Teacher's Note:
1. The projection formulae say a = b cos C + c cos B and its cyclic forms.
2. Regrouping the six terms by pairs makes them appear.

Answer: (a) 0

Explanation:
1. Put a = k sin A etc. Then the sum is \( k[\sin A\sin(B - C) + \sin B\sin(C - A) + \sin C\sin(A - B)] \).
2. Since sin A = sin(B + C), sin A sin(B − C) = sin²B − sin²C. Cyclically the three terms add to 0.

Teacher's Note:
1. Use sin(B + C) sin(B − C) = sin²B − sin²C.
2. The three cyclic differences cancel in pairs.

Answer: (c) \( 3abc \)

Explanation:
1. Split into pairs: \( ab^2\cos A + a^2b\cos B = ab(b\cos A + a\cos B) = ab \cdot c \).
2. Likewise \( ac^2\cos A + a^2c\cos C = ac(c\cos A + a\cos C) = ac \cdot b \) and \( bc^2\cos B + b^2c\cos C = bc(c\cos B + b\cos C) = bc \cdot a \).
3. The six terms of the expression make exactly these three pairs, so the total is 3abc.

Teacher's Note:
1. Express each cosine through the cosine rule and put everything over abc.
2. A quick numerical check with a right triangle confirms the value.

Answer: (a) A.P.

Explanation:
1. \( a\cdot\dfrac{1 + \cos C}{2} + c\cdot\dfrac{1 + \cos A}{2} = \dfrac{a + c}{2} + \dfrac{a\cos C + c\cos A}{2} = \dfrac{a + c + b}{2} \).
2. So \( \dfrac{a + b + c}{2} = \dfrac{3b}{2} \), giving a + c = 2b. The sides are in A.P.

Teacher's Note:
1. Use cos² x = (1 + cos 2x)/2 first.
2. Then a cos C + c cos A = b by the projection formula.

Answer: (b) \( \sqrt{\dfrac{(s - b)(s - c)}{s(s - a)}} \)

Explanation:
1. This is the half angle formula for tangent: tan(A/2) = sin(A/2)/cos(A/2).
2. With \( \sin\dfrac{A}{2} = \sqrt{\dfrac{(s - b)(s - c)}{bc}} \) and \( \cos\dfrac{A}{2} = \sqrt{\dfrac{s(s - a)}{bc}} \) the ratio is option (b).

Teacher's Note:
1. The factor s − a appears in the denominator for angle A.
2. Option (d) is cot(A/2).

Answer: (a) 0

Explanation:
1. Use \( \cot\dfrac{A}{2} = \dfrac{s - a}{r} \) and its cyclic forms.
2. The sum becomes \( \dfrac{1}{r}[(b - c)(s - a) + (c - a)(s - b) + (a - b)(s - c)] \).
3. The bracket equals s[(b − c) + (c − a) + (a − b)] − [a(b − c) + b(c − a) + c(a − b)] = 0 − 0 = 0.

Teacher's Note:
1. cot(A/2) = (s − a)/r is the key rewriting.
2. Cyclic sums such as (b − c) + (c − a) + (a − b) vanish.

Answer: (c) \( \cos^{-1}\left(\dfrac{7}{8}\right) \)

Explanation:
1. \( \cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{9 + 16 - 4}{24} = \dfrac{21}{24} = \dfrac{7}{8} \).
2. So A = cos⁻¹(7/8).

Teacher's Note:
1. Use the side opposite to A as the subtracted square.
2. Always simplify 21/24 before matching options.

Answer: (b) \( \dfrac{\pi}{3} \)

Explanation:
1. Convert sines to sides: \( (a + b + c)(a + b - c) = 3ab \).
2. So \( (a + b)^2 - c^2 = 3ab \), i.e. \( a^2 + b^2 - c^2 = ab \).
3. \( \cos C = \dfrac{ab}{2ab} = \dfrac{1}{2} \), so C = π/3.

Teacher's Note:
1. Divide the sine relation by the common 2R to get sides.
2. The difference of squares (a + b)² − c² is the key step.

Answer: (d) \( \dfrac{c^2}{ab} \)

Explanation:
1. With the right angle at C, tan A = a/b and tan B = b/a.
2. \( \tan A + \tan B = \dfrac{a^2 + b^2}{ab} = \dfrac{c^2}{ab} \) since \( a^2 + b^2 = c^2 \).

Teacher's Note:
1. Name the opposite and adjacent sides carefully for each angle.
2. Pythagoras turns a² + b² into c².

Answer: (b) 13

Explanation:
1. \( \cos A = \pm\dfrac{4}{5} \); the options only fit \( \cos A = \dfrac{4}{5} \) (A acute).
2. \( a^2 = b^2 + c^2 - 2bc\cos A = 400 + 441 - 2(20)(21)\dfrac{4}{5} = 841 - 672 = 169 \).
3. So a = 13. (The obtuse value gives a² = 1513, which is not a perfect square and matches no option.)

Teacher's Note:
1. Find cos A from sin A first.
2. Check which sign of cos A gives an option.

Answer: (a) \( \dfrac{3}{4} \)

Explanation:
1. s = (16 + 24 + 20)/2 = 30.
2. \( \cos^2\dfrac{B}{2} = \dfrac{s(s - b)}{ac} = \dfrac{30 \times 6}{16 \times 20} = \dfrac{9}{16} \).
3. So cos(B/2) = 3/4.

Teacher's Note:
1. Use the half-angle formula with the semi-perimeter.
2. B/2 is acute, so take the positive root.

Answer: (a) \( \dfrac{5\pi}{9} \)

Explanation:
1. Let the angles be x, 3x, 5x. Then 9x = π, so x = π/9.
2. The greatest angle is 5x = 5π/9.

Teacher's Note:
1. The three angles must add up to π.
2. An angle of 11π/9 is impossible in a triangle.

Answer: (c) \( c = a \)

Explanation:
1. 2 cos A sin C = sin B = sin(A + C).
2. So sin(A + C) − 2 cos A sin C = 0, i.e. sin(C − A) = 0, hence C = A.
3. Equal angles mean equal opposite sides: c = a.

Teacher's Note:
1. Write cosec C as 1/sin C and clear the denominator.
2. sin(C − A) = 0 forces C = A.

Answer: (d) 75°

Explanation:
1. Angles in A.P. give B = 60°.
2. \( \sin C = \dfrac{c\sin B}{b} = \dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2} = \dfrac{1}{\sqrt{2}} \), so C = 45° (C < B because c < b).
3. A = 180° − 60° − 45° = 75°.

Teacher's Note:
1. Three angles in A.P. always include 60° as the middle one.
2. The smaller side c has the smaller angle, so C = 45°, not 135°.

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

Explanation:
1. \( \sin(A - B)\sin(A + B) = \sin^2 A - \sin^2 B \), so \( \dfrac{\sin(A - B)}{\sin(A + B)} = \dfrac{\sin^2 A - \sin^2 B}{\sin^2(A + B)} \).
2. Since sin(A + B) = sin C, the sine rule turns this into \( \dfrac{a^2 - b^2}{c^2} \).

Teacher's Note:
1. Use sin²A − sin²B = sin(A + B) sin(A − B).
2. sin(A + B) = sin C.

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

Explanation:
1. The largest angle is opposite the side 7.
2. \( \cos\theta = \dfrac{9 + 25 - 49}{2 \times 3 \times 5} = -\dfrac{1}{2} \).
3. So θ = 2π/3.

Teacher's Note:
1. The longest side faces the largest angle.
2. A negative cosine means an obtuse angle.

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

Explanation:
1. Since \( 3\alpha^2 + \beta^2 - (\alpha + \beta)^2 = 2\alpha(\alpha - \beta) > 0 \), the third side is the largest.
2. \( \cos\theta = \dfrac{(\alpha - \beta)^2 + (\alpha + \beta)^2 - (3\alpha^2 + \beta^2)}{2(\alpha - \beta)(\alpha + \beta)} = \dfrac{\beta^2 - \alpha^2}{2(\alpha^2 - \beta^2)} = -\dfrac{1}{2} \).
3. So θ = 2π/3.

Teacher's Note:
1. First decide which side is the longest.
2. Apply the cosine rule for the angle opposite it.

Answer: (c) 1

Explanation:
1. The bracket is sin(B + C) = sin(180° − A) = sin A.
2. So the expression is (1/sin A) × sin A = 1.

Teacher's Note:
1. Recognise the compound-angle expansion of sin(B + C).
2. sin(180° − A) = sin A.

Answer: (d) x = 7

Explanation:
1. \( x^2 = a^2 + b^2 - 2ab\cos C = 81 + 64 - 2(9)(8)\left(\dfrac{2}{3}\right) = 145 - 96 = 49 \).
2. So x = 7.

Teacher's Note:
1. Substitute cos C = 2/3 directly into the cosine rule.
2. The side c lies opposite angle C.

Answer: (b) \( \dfrac{\sqrt{3} + 1}{2} \) cm²

Explanation:
1. A = 105°, and \( \Delta = \dfrac{a^2\sin B\sin C}{2\sin A} \).
2. \( a^2 = 4 + 2\sqrt{3},\ \sin B\sin C = \dfrac{1}{2\sqrt{2}},\ \sin 105° = \dfrac{\sqrt{3} + 1}{2\sqrt{2}} \).
3. \( \Delta = \dfrac{(4 + 2\sqrt{3})/(2\sqrt{2})}{(\sqrt{3} + 1)/\sqrt{2}} = \dfrac{\sqrt{3} + 1}{2} \) cm².

Teacher's Note:
1. Use the area formula that needs one side and two angles.
2. Note that 4 + 2√3 = (√3 + 1)².

Answer: (a) \( a^2 + b^2 = c^2 \)

Explanation:
1. \( 2\cos^2\dfrac{A}{2} = 1 + \cos A = \dfrac{b + c}{c} \), so \( \cos A = \dfrac{b}{c} \).
2. Using the cosine rule, \( \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{b}{c} \), i.e. \( c^2 = a^2 + b^2 \).

Teacher's Note:
1. Use 2cos²(A/2) = 1 + cos A.
2. The result shows the right angle is at C.

Answer: (b) \( \dfrac{b}{c} \)

Explanation:
1. sin(A + B) = sin(180° − C) = sin C.
2. So the ratio is sin B / sin C = b / c by the sine rule.

Teacher's Note:
1. A + B and C are supplementary.
2. Do not cancel A or add sides in the sine rule.

Answer: (c) \( (c - a)^2 + ca = b^2 \)

Explanation:
1. \( b^2 = a^2 + c^2 - 2ac\cos 60° = a^2 + c^2 - ac = (c - a)^2 + ac \).
2. Option (d) would give a² + c² = b² + ac, which is a different relation.

Teacher's Note:
1. Write the cosine rule for the side opposite 60°.
2. Then complete the square: a² + c² − ac = (c − a)² + ac.

Answer: (c) 0

Explanation:
1. \( \cot A = \dfrac{b^2 + c^2 - a^2}{4\Delta} \) and cyclically.
2. The expression equals \( \dfrac{3a^2 - b^2 - c^2}{4\Delta} \).
3. With \( b^2 + c^2 = 3a^2 \) it is 0.

Teacher's Note:
1. Each cotangent equals (sum of two squares − third square)/4Δ.
2. Add the three numerators carefully with their signs.

Answer: (c) \( \dfrac{2}{3} \)

Explanation:
1. \( \sin B = \dfrac{b\sin A}{a} = \dfrac{8 \times \frac{1}{2}}{6} = \dfrac{2}{3} \).
2. So x = 2/3.

Teacher's Note:
1. Use the sine rule to find sin B.
2. Since b > a there are two angles B, but x is the same for both.

Answer: (a) \( \dfrac{2c}{a + b + c} \)

Explanation:
1. \( \tan\dfrac{A}{2}\tan\dfrac{B}{2} = \dfrac{s - c}{s} \).
2. So the expression is \( 1 - \dfrac{s - c}{s} = \dfrac{c}{s} = \dfrac{2c}{a + b + c} \).

Teacher's Note:
1. Memorise tan(A/2) tan(B/2) = (s − c)/s.
2. Use 2s = a + b + c at the end.

Answer: (b) Right angled

Explanation:
1. \( \sin^2 B = 1 - \cos^2 B \), so \( \cos^2 A + \cos^2 B + \cos^2 C = 1 \).
2. In any triangle this sum is \( 1 - 2\cos A\cos B\cos C \), so one cosine is 0.
3. Hence one angle is 90°: the triangle is right angled.

Teacher's Note:
1. Use the identity cos²A + cos²B + cos²C = 1 − 2 cos A cos B cos C.
2. A zero cosine means a right angle.

Answer: (c) \( \dfrac{3}{a + b + c} \)

Explanation:
1. With C = 60°, \( c^2 = a^2 + b^2 - ab \).
2. \( \dfrac{1}{a + c} + \dfrac{1}{b + c} = \dfrac{a + b + 2c}{(a + c)(b + c)} \) and \( (a + c)(b + c) = a^2 + b^2 + c(a + b) \).
3. Also \( (a + b + 2c)(a + b + c) = 3a^2 + 3b^2 + 3c(a + b) \) (using the relation for c²), so \( (a + c)(b + c) = \dfrac{(a + b + 2c)(a + b + c)}{3} \).
4. Hence the sum is \( \dfrac{3}{a + b + c} \).

Teacher's Note:
1. Use the cosine rule with cos 60° = 1/2.
2. Test with an equilateral triangle: 1/(2a) + 1/(2a) = 1/a = 3/(3a).

Answer: (c) \( 24/25 \)

Explanation:
1. Since 3² + 4² = 5², the right angle is at C.
2. \( \sin B = \dfrac{4}{5},\ \cos B = \dfrac{3}{5} \), so \( \sin 2B = 2 \times \dfrac{4}{5} \times \dfrac{3}{5} = \dfrac{24}{25} \).

Teacher's Note:
1. Check for a right triangle first.
2. Use sin 2B = 2 sin B cos B.

Answer: (a) 30°

Explanation:
1. \( \tan\dfrac{B - C}{2} = \dfrac{b - c}{b + c}\cot\dfrac{A}{2} \) gives \( 1 = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}\cot\dfrac{A}{2} \).
2. \( \cot\dfrac{A}{2} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3} = \cot 15° \).
3. So A/2 = 15° and A = 30°.

Teacher's Note:
1. Napier's analogy connects B − C with the sides b − c and b + c.
2. cot 15° = 2 + √3.

Answer: (b) Equilateral

Explanation:
1. \( \cos A + \cos B + \cos C = 1 + 4\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2} \), whose maximum value is 3/2.
2. The maximum occurs only when A = B = C = 60°.
3. So the triangle is equilateral.

Teacher's Note:
1. The sum of the three cosines never exceeds 3/2.
2. Equality holds only for the equilateral triangle.

Answer: (c) cot A, cot B, cot C

Explanation:
1. \( \cot A = \dfrac{b^2 + c^2 - a^2}{4\Delta} \) and cyclically.
2. \( \cot A + \cot C - 2\cot B = \dfrac{4b^2 - 2a^2 - 2c^2}{4\Delta} \), which is 0 exactly when \( a^2 + c^2 = 2b^2 \).
3. So cot A, cot B, cot C are in A.P. when a², b², c² are in A.P.

Teacher's Note:
1. Write each cotangent as (sum of squares − third square)/4Δ.
2. a², b², c² in A.P. means a² + c² = 2b².

Answer: (b) \( \dfrac{a - b}{a + b} \)

Explanation:
1. \( \cot\dfrac{A + B}{2} = \tan\dfrac{C}{2} \).
2. By Napier's analogy, \( \tan\dfrac{A - B}{2} = \dfrac{a - b}{a + b}\cot\dfrac{C}{2} \).
3. The product is \( \dfrac{a - b}{a + b} \).

Teacher's Note:
1. (A + B)/2 = 90° − C/2.
2. The tan(C/2) and cot(C/2) cancel.

Answer: (c) A.P.

Explanation:
1. \( \cot\dfrac{A}{2} = \dfrac{s - a}{r},\ \cot\dfrac{B}{2} = \dfrac{s - b}{r},\ \cot\dfrac{C}{2} = \dfrac{s - c}{r} \).
2. If a, b, c are in A.P., so are s − a, s − b, s − c. Hence the cotangents are in A.P.

Teacher's Note:
1. Use cot(A/2) = (s − a)/r.
2. Subtracting each term from s keeps an A.P.

Answer: (c) A.P.

Explanation:
1. From question 36, \( a^2, b^2, c^2 \) in A.P. is equivalent to cot A, cot B, cot C in A.P.
2. So the squares of the sides are in A.P.

Teacher's Note:
1. Each cotangent is (sum of two squares − third square)/4Δ.
2. The A.P. condition on cotangents becomes a² + c² = 2b².

Answer: (c) \( \dfrac{\pi}{4} \)

Explanation:
1. \( \tan\angle BAD = \dfrac{BD}{AD} = \dfrac{1}{3},\ \tan\angle CAD = \dfrac{CD}{AD} = \dfrac{1}{2} \).
2. \( \tan A = \dfrac{\frac{1}{3} + \frac{1}{2}}{1 - \frac{1}{6}} = 1 \).
3. So A = π/4.

Teacher's Note:
1. Split angle A into two right triangles with common height AD.
2. Use the tan(x + y) formula.