JEE Mathematics Properties of Triangles MCQs Set 03

Answer: (c) \(1\)

Explanation:
1. The bracket is \(\sin(B+C)=\sin(\pi-A)=\sin A\).
2. So the product is \(\operatorname{cosec}A\cdot\sin A=1\).

Teacher's Note:
1. Use \( A + B + C = \pi \) to convert \( \sin(B + C) \) to \( \sin A \).
2. The product of a function and its reciprocal is 1.

Answer: (d) \(\left(\frac{c^2-a^2}{2ca}\right)^2\)

Explanation:
1. \(\frac{\sin3B}{\sin B}=3-4\sin^2B=4\cos^2B-1\).
2. Here \(\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{a^2+c^2}{4ac}\), so \(4\cos^2B-1=\frac{(a^2+c^2)^2-4a^2c^2}{4a^2c^2}=\left(\frac{c^2-a^2}{2ac}\right)^2\).

Teacher's Note:
1. Use \( \sin 3B = 3 \sin B - 4 \sin^3 B \).
2. Substitute \( \cos B \) from the cosine rule using the given relation.

Answer: (a) AP

Explanation:
1. Cross-multiplying: \(\sin A\sin(B-C)=\sin C\sin(A-B)\).
2. Using \(\sin(P-Q)\sin(P+Q)=\sin^2P-\sin^2Q\) this becomes \(\sin^2B-\sin^2C=\sin^2A-\sin^2B\).
3. So \(b^2-c^2=a^2-b^2\), i.e. \(2b^2=a^2+c^2\): an AP.

Teacher's Note:
1. Reduce the sine ratios to squares of sines using \( \sin(A+B) = \sin C \).
2. Equal differences of squares give an AP.

Answer: (a) \(bc\)

Explanation:
1. The half-angle formula gives \(\sin^2\frac A2=\frac{(s-b)(s-c)}{bc}\).
2. Hence \(x=bc\).

Teacher's Note:
1. Memorise \( \sin^2(A/2) \) = (s-b)(s-c)/bc.
2. Compare coefficients directly.

Answer: (a) \(\sqrt3:(2+\sqrt3)\)

Explanation:
1. The angles are \(120^\circ,30^\circ,30^\circ\).
2. Sides are proportional to sines: \(\frac{\sqrt3}2:\frac12:\frac12\).
3. Ratio \(=\frac{\sqrt3/2}{\sqrt3/2+1}=\frac{\sqrt3}{\sqrt3+2}\).

Teacher's Note:
1. Sides are proportional to the sines of opposite angles.
2. The ratio of angles 4:1:1 sums to 6 parts of 30 degrees.

Answer: (c) \(\frac3{a+b+c}\)

Explanation:
1. With \(\angle C=60^\circ\) the cosine rule gives \(c^2=a^2+b^2-ab\).
2. The sum is \(\frac{a+b+2c}{a^2+b^2+c(a+b)}\), since \((a+c)(b+c)=ab+c(a+b)+c^2=a^2+b^2+c(a+b)\).
3. Also \((a+b+2c)(a+b+c)=(a+b)^2+3c(a+b)+2c^2=3[a^2+b^2+c(a+b)]\), so the sum is \(\frac3{a+b+c}\).

Teacher's Note:
1. Use the cosine rule at \( C = 60 \) degrees to replace \( c^2 \).
2. Check with an equilateral triangle \( a = b = c = 1 \): \( 1 = 3/3 \).

Answer: (b) \(\frac12\)

Explanation:
1. Since \(2b=a+c\), \(s=\frac{3b}2\) and \(s-b=\frac b2\).
2. Using \(\sin\frac A2\sin\frac C2=\sqrt{\frac{(s-b)(s-c)}{bc}\cdot\frac{(s-a)(s-b)}{ab}}=\frac{s-b}{b}\sqrt{\frac{(s-a)(s-c)}{ac}}\) and \(\sin\frac B2=\sqrt{\frac{(s-a)(s-c)}{ac}}\), the ratio is \(\frac{s-b}{b}=\frac12\).

Teacher's Note:
1. Express the half-angle sines in terms of \( s - a, s - b, s - c \).
2. Test with the 4, 5, 6 triangle: the value is 0.5.

Answer: (c) \(5-\sqrt6\text{ or }5+\sqrt6\)

Explanation:
1. Angles in AP give \(B=60^\circ\), the angle opposite the middle side 9.
2. The cosine rule: \(81=100+x^2-10x\), i.e. \(x^2-10x+19=0\), so \(x=5\pm\sqrt6\).
3. Both values (2.55 and 7.45) are less than 9, so both are valid.

Teacher's Note:
1. In an AP of angles the middle angle is 60 degrees.
2. Check that each root keeps the stated order of the sides.

Answer: (b) \(xy\sqrt3\)

Explanation:
1. Area \(=\frac12ab\sin C=\frac12(2x)(2y)\sin120^\circ=2xy\cdot\frac{\sqrt3}2=\sqrt3\,xy\).

Teacher's Note:
1. Area = (1/2) ab \( \sin C \).
2. \( \sin 120 \) degrees = \( sqrt3/2 \).

Answer: (b) \(\cos A\)

Explanation:
1. The first term is \(\cos^2\frac A2\) and the second is \(\sin^2\frac A2\).
2. Their difference is \(\cos A\).

Teacher's Note:
1. Recall \( \cos^2(A/2) \) = s(s-a)/bc and \( \sin^2(A/2) \) = (s-b)(s-c)/bc.
2. \( \cos^2 - \sin^2 \) of the half-angle is \( \cos \) of the angle.

Answer: (a) \(\frac{2\Delta}{s}\)

Explanation:
1. \(b+c-a=2(s-a)\) and \(\tan\frac A2=\frac{\Delta}{s(s-a)}\).
2. Product \(=\frac{2\Delta}{s}\).

Teacher's Note:
1. \( \tan(A/2) = \Delta / (s(s-a)) \).
2. \( 2(s - a) \) cancels the factor.

Answer: (d) None of these

Explanation:
1. Apollonius: \(4AD^2=2b^2+2c^2-a^2\).
2. With \(A=60^\circ\), \(a^2=b^2+c^2-bc\), so \(4AD^2=b^2+c^2+bc\), which matches none of the options.

Teacher's Note:
1. Use the median length formula and the cosine rule.
2. Be careful with the sign of the bc term.

Answer: (b) right angled

Explanation:
1. Rewrite as \(2-\sin^2A-\sin^2C=\sin^2B\), i.e. \(\sin^2A+\sin^2B+\sin^2C=2\).
2. This holds exactly when one angle is a right angle.

Teacher's Note:
1. The identity \( \sin^{2A} + \sin^{2B} + \sin^{2C} = 2 \) + 2cosA cosB cosC.
2. So the product of cosines is zero.

Answer: (b) 0

Explanation:
1. By the sine rule \(\sin B=\frac{b\sin A}{a}=\frac{7\cdot3/4}{5}=\frac{21}{20}>1\).
2. This is impossible, so no triangle exists.

Teacher's Note:
1. Compute \( \sin B \) from the sine rule and check it is at most 1.
2. A value above 1 means no triangle.

Answer: (b) \(\frac{a^2+b^2+c^2}{4\Delta^2}\)

Explanation:
1. Altitude \(p_1=\frac{2\Delta}a\), so \(p_1^{-2}=\frac{a^2}{4\Delta^2}\).
2. The sum is \(\frac{a^2+b^2+c^2}{4\Delta^2}\).

Teacher's Note:
1. Altitude = \( 2 \Delta \) divided by the base.
2. Square the reciprocals.

Answer: (c) \(\frac{2ab}{a+b}\cos\frac C2\)

Explanation:
1. Area of \(\triangle ABC\) = sum of areas of the two parts: \(\frac12ab\sin C=\frac12a\cdot CD\sin\frac C2+\frac12b\cdot CD\sin\frac C2\).
2. Using \(\sin C=2\sin\frac C2\cos\frac C2\): \(CD=\frac{2ab\cos\frac C2}{a+b}\).

Teacher's Note:
1. Split the area along the bisector.
2. Use the double angle formula for \( \sin C \).

Answer: (c) \(c^2\)

Explanation:
1. Expand: \(a^2+b^2-2ab(\cos^2\frac C2-\sin^2\frac C2)=a^2+b^2-2ab\cos C\).
2. By the cosine rule this is \(c^2\).

Teacher's Note:
1. Group the terms in \( a^2 + b^2 \) and 2ab.
2. \( \cos^2 - \sin^2 \) of half-angle is \( \cos C \).

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

Explanation:
1. Take \(a=4,b=5,c=6\): \(s=7.5,\ \Delta=\frac{15\sqrt7}4\).
2. Then \(R=\frac{abc}{4\Delta}=\frac{8}{\sqrt7}\) and \(r=\frac\Delta s=\frac{\sqrt7}{2}\).
3. Ratio \(R:r=\frac{16}7\).

Teacher's Note:
1. Use \( R \) = abc/4Delta and \( r = \Delta/s \).
2. The ratio does not depend on the scale factor.