CBSE Class 11 Mathematics Solution of Triangles MCQs Set 03

Answer: (a) No triangle is possible

Explanation:
1. From the sine rule, \( \sin C = \dfrac{c\sin B}{b} > 1 \) because \( b < c\sin B \).
2. This is impossible, so no triangle exists.

Teacher's Note:
1. Compare b with the height c sin B from C.
2. If the side b is shorter than the perpendicular, it cannot reach the base line.

Answer: (c) \( \dfrac{\pi}{8},\dfrac{3\pi}{8} \)

Explanation:
1. If the angles are θ and 90° − θ, then hypotenuse / perpendicular = \( \dfrac{2}{\sin 2\theta} \).
2. So \( \dfrac{2}{\sin 2\theta} = 2\sqrt{2} \) gives \( \sin 2\theta = \dfrac{1}{\sqrt{2}} \), i.e. 2θ = 45° or 135°.
3. The two acute angles are 22.5° = π/8 and 67.5° = 3π/8.

Teacher's Note:
1. Use h = c sin θ cos θ = (c/2) sin 2θ for the altitude on the hypotenuse.
2. Both 2θ = 45° and 135° describe the same pair of angles.

Answer: (a) \( c^2 - 3c - 7 = 0 \)

Explanation:
1. \( a^2 = b^2 + c^2 - 2bc\cos A \) gives \( 16 = 9 + c^2 - 2(3)c\left(\dfrac{1}{2}\right) \).
2. So \( c^2 - 3c - 7 = 0 \).

Teacher's Note:
1. Write the cosine rule for the given angle A.
2. Collect everything on one side.

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

Explanation:
1. The equation is sin 3x = k. Its solutions in (0, π/3) and (π/3, 2π/3) are x₁ = θ/3 and x₂ = (π − θ)/3 where sin θ = k.
2. So A + B = (θ + π − θ)/3 = π/3, hence C = π − π/3 = 2π/3.

Teacher's Note:
1. 3 sin x − 4 sin³x = sin 3x.
2. Two angles of a triangle must be the two solutions with sum π/3.

Answer: (b) \( \sqrt{\dfrac{9a^2 - c^2}{8c^2}} \)

Explanation:
1. b satisfies \( b^2 - (2c\cos A)b + (c^2 - a^2) = 0 \), so \( b_1 + b_2 = 2c\cos A \) and \( b_1b_2 = c^2 - a^2 \).
2. With b₂ = 2b₁: \( 3b_1 = 2c\cos A \) and \( 2b_1^2 = c^2 - a^2 \), so \( \dfrac{8c^2\cos^2 A}{9} = c^2 - a^2 \).
3. \( 8c^2(1 - \sin^2 A) = 9c^2 - 9a^2 \) gives \( \sin^2 A = \dfrac{9a^2 - c^2}{8c^2} \).

Teacher's Note:
1. Use the sum and product of the roots of the quadratic in the third side.
2. Replace cos²A by 1 − sin²A at the end.

Answer: (c) \( (a + b)^2 = c^2 + ab \) and \( \sqrt{2}(\sin A + \cos A) = \sqrt{3} \)

Explanation:
1. (a) tan A + tan B + tan C = tan A tan B tan C, which is 0 only if some angle is 0 or 180°: impossible.
2. (b) Sides in the ratio 2 : 3 : 7 violate the triangle inequality (2 + 3 < 7).
3. (c) \( c^2 = a^2 + b^2 + ab \) gives C = 120°, and \( \sin(A + 45°) = \dfrac{\sqrt{3}}{2} \) gives A = 15°, B = 45°. This triangle exists.
4. (d) The conditions would need cos A cos B = sin A sin B (so A + B = 90°) but then the stated sum of sines does not match, so no triangle exists.

Teacher's Note:
1. Test each statement against a basic triangle property.
2. Only (c) gives consistent angles adding to 180°.

Answer: (c) \( 2\sqrt{a^2 - b^2\sin^2 A} \)

Explanation:
1. c satisfies \( c^2 - (2b\cos A)c + (b^2 - a^2) = 0 \).
2. \( c_1 - c_2 = \sqrt{(c_1 + c_2)^2 - 4c_1c_2} = \sqrt{4b^2\cos^2 A - 4(b^2 - a^2)} = 2\sqrt{a^2 - b^2\sin^2 A} \).

Teacher's Note:
1. Use the sum and product of the two roots.
2. Replace cos²A by 1 − sin²A.

Answer: (b) 3 : 5 : 7

Explanation:
1. Let the sides be a − d, a, a + d, so s = 3a/2 and the equilateral triangle has side a.
2. \( \Delta^2 = \dfrac{3a}{2} \cdot \dfrac{a}{2}\left(\dfrac{a^2}{4} - d^2\right) \) and the condition is \( \Delta^2 = \dfrac{9}{25}\cdot\dfrac{3a^4}{16} \).
3. This gives \( d^2 = \dfrac{16a^2}{100} \), d = 0.4a, so the sides are 0.6a, a, 1.4a, i.e. 3 : 5 : 7.

Teacher's Note:
1. Take the sides as a − d, a, a + d for the A.P.
2. The equilateral triangle with the same perimeter has side equal to the middle term a.

Answer: (a) \( b\sin A = a,\ A < \dfrac{\pi}{2} \) or \( b\sin A < a,\ A < \dfrac{\pi}{2},\ b < a \)

Explanation:
1. If b sin A > a, the side a cannot reach the base line, so no triangle exists: options (b) and (c) fail.
2. For A acute, b sin A = a gives one right-angled triangle, and b sin A < a with b < a gives one triangle. So (a) describes triangles that exist.

Teacher's Note:
1. Compare a with the height b sin A.
2. Recall the case analysis of the ambiguous case.