JEE Mathematics Properties and Solution of Triangles MCQs Set 04

Answer: (b) \(\sqrt[3]{7}\)

Explanation:
1. Since \(a=2R\sin A\), each ratio equals \((2R)^3\), so \((2R)^3=7\).
2. \(a=2R\sin A\le2R=\sqrt[3]{7}\), with equality for a right-angled triangle (\(A=90^\circ\)).

Teacher's Note:
1. Convert side lengths with \( a = 2R \sin A \) and use the standard identities for the triangle before trying algebra.
2. Test the final result on an equilateral or right-angled triangle as a quick check.

Answer: (b) the reciprocals of the altitudes are in A.P.

Explanation:
1. \(\sin A,\sin B,\sin C\) in \( A.P \). means \(a,b,c\) in \( A.P \).
2. Altitudes are \(h_a=\dfrac{2\Delta}{a}\) etc., so \(\dfrac1{h_a}=\dfrac{a}{2\Delta}\) is proportional to the sides.
3. Hence the reciprocals of the altitudes are in \( A.P \).

Teacher's Note:
1. Convert side lengths with \( a = 2R \sin A \) and use the standard identities for the triangle before trying algebra.
2. Test the final result on an equilateral or right-angled triangle as a quick check.

Answer: (b) \(P\in[3,4)\)

Explanation:
1. \((a+b+c)^2\ge3(ab+bc+ca)\) gives \(P\ge3\).
2. In a triangle \(a^2<a(b+c)\) etc., so \(a^2+b^2+c^2<2(ab+bc+ca)\) and \(P<4\).

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (a) \(a\ge2\lambda\sin\dfrac A2\)

Explanation:
1. \(a^2=b^2+c^2-2bc\cos A=(b-c)^2+2bc(1-\cos A)=(b-c)^2+4\lambda^2\sin^2\tfrac A2\).
2. So \(a\ge2\lambda\sin\tfrac A2\). The other two need \(ca=\lambda^2\) or \(ab=\lambda^2\), which are not given.

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (a) \(2ac\)

Explanation:
1. The two acute angles add to \(\pi/4\), so \(\tan(\theta_1+\theta_2)=1\).
2. \((c-a\tan\theta)^2=b^2(1+\tan^2\theta)\) gives \((a^2-b^2)t^2-2ac\,t+(c^2-b^2)=0\) with \(t=\tan\theta\).
3. \(\tan(\theta_1+\theta_2)=\dfrac{2ac/(a^2-b^2)}{1-(c^2-b^2)/(a^2-b^2)}=\dfrac{2ac}{a^2-c^2}=1\), so \(a^2-c^2=2ac\).

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (c) 120°

Explanation:
1. The largest side is \(\sqrt{1+\sin\alpha\cos\alpha}\).
2. \(\cos\theta=\dfrac{\sin^2\alpha+\cos^2\alpha-(1+\sin\alpha\cos\alpha)}{2\sin\alpha\cos\alpha}=\dfrac{-\sin\alpha\cos\alpha}{2\sin\alpha\cos\alpha}=-\dfrac12\).
3. So the greatest angle is 120 degrees.

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (d) \(\dfrac1{250}\)

Explanation:
1. \(\tan A+\tan B+\tan C=\tan A\tan B\tan C\), so the expression is \(\dfrac{2\tan B(\tan A+\tan C)}{\tan A\tan C}=2\tan B(\cot A+\cot C)\).
2. \(\cot A+\cot C=\dfrac{\sin B}{\sin A\sin C}\), so the value is \(\dfrac{2\sin^2B}{\cos B\sin A\sin C}=\dfrac{2b^2}{ac\cos B}\).
3. \(a^2+c^2=1001b^2\) gives \(2ac\cos B=1000b^2\), so \(ac\cos B=500b^2\) and the value is \(\dfrac{2}{500}=\dfrac1{250}\).

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (a) 60°

Explanation:
1. Altitude to the hypotenuse \(h=\dfrac{ab}{c}\) and \(c=4h\) give \(c^2=4ab\).
2. With \(a=c\sin A,\ b=c\cos A\): \(1=4\sin A\cos A=2\sin2A\), so \(\sin2A=\tfrac12\).
3. The acute angles differ by \(|90^\circ-2A|=60^\circ\) (since \(2A=30^\circ\) or \(150^\circ\)).

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (c) \(\dfrac1{\sqrt6}\)

Explanation:
1. By the ratio theorem, \(\dfrac{BD}{DC}=\dfrac{AB\sin\angle BAD}{AC\sin\angle CAD}\).
2. \(\dfrac{\sin\angle BAD}{\sin\angle CAD}=\dfrac13\cdot\dfrac{AC}{AB}=\dfrac13\cdot\dfrac{\sin B}{\sin C}=\dfrac13\cdot\dfrac{\sqrt3/2}{\sqrt2/2}=\dfrac1{\sqrt6}\).

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (b) 1

Explanation:
1. The angles are \(\pi/7,2\pi/7,4\pi/7\) and \(\lambda=\dfrac{\prod(\sin^2A-\sin^2B)}{\prod\sin^2A}\).
2. Evaluating (for example numerically, or with \( \sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B) \)) gives 1.

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (d) the angles of the triangle are in A.P.

Explanation:
1. If \( B \) were 90 degrees the left side would be \( -1 \), so \( B \) is not the right angle. \( \tan A \) real means \( A \) is not 90 degrees either, so \( C = 90 \) degrees.
2. Then \( \cos B = \sin A \) and \( 4 \sin^2 A = 1 \), so \( A = 30 \) degrees, \( B = 60 \) degrees, \( C = 90 \) degrees.
3. 30, 60, 90 are in \( A.P \)., while the sides \( 1, sqrt3, 2 \) are not.

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (a) 1.5 min

Explanation:
1. Let the height be \(h\): distances are \(h\cot30^\circ=h\sqrt3\) and \(h\cot60^\circ=\dfrac h{\sqrt3}\).
2. Distance covered in \( 3 \min = h\sqrt3-\dfrac h{\sqrt3}=\dfrac{2h}{\sqrt3} \).
3. Remaining distance \(\dfrac h{\sqrt3}\) is half of it, so it takes \( 1.5 \min \).

Teacher's Note:
1. For height and distance problems draw the right triangles and write each tangent ratio as height over horizontal distance.
2. Equate the unknown horizontal distances from the two observations.

Answer: (d) \(60+15\sqrt3\)

Explanation:
1. Put the first foot at 0. The 30 degree staff has height \(h_1\): \(x_A=h_1\sqrt3,\ x_B=h_1/\sqrt3\), and \(x_A-x_B=30\) gives \(h_1=15\sqrt3\), \(x_B=15,\ x_A=45\).
2. For the second staff of height \(h_2\): its foot is at \(45+h_2/\sqrt3\) (from \( A, 60 \) degrees) and at \(15+h_2\) (from \( B, 45 \) degrees).
3. So \(h_2(1-1/\sqrt3)=30\), \(h_2=45+15\sqrt3\) and the distance between the feet is \(15+h_2=60+15\sqrt3\).

Teacher's Note:
1. For height and distance problems draw the right triangles and write each tangent ratio as height over horizontal distance.
2. Equate the unknown horizontal distances from the two observations.

Answer: (b) 7

Explanation:
1. Distance = \(100(\cot30^\circ-\cot60^\circ)=100\left(\sqrt3-\dfrac1{\sqrt3}\right)=\dfrac{200\sqrt3}{3}\).
2. So \( a = 200 \) and \( b = 3 \), giving \( 200/20 - 3 = 7 \).

Teacher's Note:
1. For height and distance problems draw the right triangles and write each tangent ratio as height over horizontal distance.
2. Equate the unknown horizontal distances from the two observations.

Answer: (c) 4

Explanation:
1. By the ratio theorem on \(D\): \(\dfrac{BD}{DC}=\dfrac{1}{2}=\dfrac{AB\sin x}{AC\sin(y+z)}\), and on \(E\): \(\dfrac{BE}{EC}=2=\dfrac{AB\sin(x+y)}{AC\sin z}\).
2. Dividing the second by the first: \( \sin(x+y) \sin(y+z) / (\sin x \sin z) = 2 / (1/2) = 4 \). \( A \) numerical check on a sample triangle also gives 4.

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (a) 2

Explanation:
1. Let \(BC=3x\) with midpoint \(M\); \(AM=\dfrac{3x}2\) (right angle at \( A \)) and \(PM=QM=\dfrac x2\).
2. Apollonius in triangle \(APQ\): \(AP^2+AQ^2=2AM^2+2PM^2\): \(25=\dfrac{9x^2}2+\dfrac{x^2}2=5x^2\).
3. \(x^2=5\), \(BC=3\sqrt5\), so \( a = 3, b = 5 \) and \( |a - b| = 2 \).

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (b) 2

Explanation:
1. \(AD=\dfrac{2bc\cos(A/2)}{b+c}\) and \(AI=\dfrac{r}{\sin(A/2)}\), so \(ID=AD-AI\).
2. Hence the sum of the three terms equals 2abc (the correction terms cancel when summed over \( A, B \) and \( C \); checked numerically on a sample triangle), so \( k = 2 \).
3. The three terms sum to 2abc, so \( k = 2 \).

Teacher's Note:
1. Convert side lengths with \( a = 2R \sin A \) and use the standard identities for the triangle before trying algebra.
2. Test the final result on an equilateral or right-angled triangle as a quick check.

Answer: (c) \(\sin^3\theta\)

Explanation:
1. \(\sin(A-\theta)=\sin\theta\sin A(\cot\theta-\cot A)\), similarly for \( B \) and \( C \).
2. The product is \(\sin^3\theta\sin A\sin B\sin C\,(\cot\theta-\cot A)(\cot\theta-\cot B)(\cot\theta-\cot C)\). With \(\cot\theta=\sum\cot A\), \(\cot\theta-\cot A=\cot B+\cot C=\dfrac{\sin A}{\sin B\sin C}\).
3. The three factors multiply to \(\dfrac1{\sin A\sin B\sin C}\), leaving \(\sin^3\theta\).

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (d) 3/2

Explanation:
1. \(\cos A+\cos B+\cos C=1+4\sin\tfrac A2\sin\tfrac B2\sin\tfrac C2=1+\dfrac rR\).
2. Since \(R\ge2r\), the maximum is \(1+\dfrac12=\dfrac32\) for the equilateral triangle.

Teacher's Note:
1. Convert side lengths with \( a = 2R \sin A \) and use the standard identities for the triangle before trying algebra.
2. Test the final result on an equilateral or right-angled triangle as a quick check.

Answer: (a) \(\dfrac{41}{9}\)

Explanation:
1. Horizontal distances from the foot: \(\dfrac h{0.6}=\dfrac{5h}3\) (south) and \(\dfrac h{0.75}=\dfrac{4h}3\) (east).
2. The two directions are perpendicular, so \(AB^2=\dfrac{25h^2}9+\dfrac{16h^2}9=\dfrac{41h^2}9\).

Teacher's Note:
1. For height and distance problems draw the right triangles and write each tangent ratio as height over horizontal distance.
2. Equate the unknown horizontal distances from the two observations.

Answer: (a) (i) and (iii) only

Explanation:
1. \(b\cos^2\tfrac A2+a\cos^2\tfrac B2=\dfrac{a+b}2+\dfrac{b\cos A+a\cos B}2=\dfrac{a+b+c}2\) (projection rule).
2. So \( a + b + c = 3c \), i.e. \( c = (a + b)/2 \).
3. \( (i) c = (a+b)/2 \) >= \( \sqrt{ab} \): true. (iii) AM-GM on three positive terms with product 1: true.
4. (ii) \( 2c - a = b \) and \( 2c - b = a \), so the sum is \( (a+c)/b + (b+c)/a = 3a/(2b) + 3b/(2a) + 1 \) >= 4, with equality only for \( a = b \): not always true.

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (a) (i), (ii) and (iv) only

Explanation:
1. Triangle inequality \(a+b>c\): \(1+r>r^2\), so \(r<\tfrac{1+\sqrt5}2\): \( (i) \) true.
2. Sides increase, so \(A<B<C\), hence \(C>\pi/3\): (iv) true.
3. \(b^2=ac\) gives \(\cos B=\dfrac{a^2+c^2-ac}{2ac}\ge\dfrac12\), so \(B<\tfrac\pi3\) (\( r \) > 1): (ii) true and (iii) false.

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (a) 6

Explanation:
1. \(h_1=\dfrac{2\Delta}a,\ r=\dfrac\Delta s\) give \(\dfrac{h_1+r}{h_1-r}=\dfrac{2/a+1/s}{2/a-1/s}=\dfrac{2s+a}{2s-a}=\dfrac{2a+b+c}{b+c}\).
2. \(\dfrac{2a+b+c}{b+c}=1+\dfrac{2a}{b+c}\), so the sum is \(3+2\sum\dfrac a{b+c}\ge3+2\cdot\dfrac32=6\) by Nesbitt's inequality.
3. Equality holds for an equilateral triangle.

Teacher's Note:
1. Convert side lengths with \( a = 2R \sin A \) and use the standard identities for the triangle before trying algebra.
2. Test the final result on an equilateral or right-angled triangle as a quick check.

Answer: (c) 3

Explanation:
1. \(\cot\tfrac A2,\cot\tfrac B2,\cot\tfrac C2\) are in \( A.P \)., so \(\cot\tfrac A2+\cot\tfrac C2=2k\) with \(k=\cot\tfrac B2\).
2. Also \(\cot\tfrac A2+\cot\tfrac B2+\cot\tfrac C2=\cot\tfrac A2\cot\tfrac B2\cot\tfrac C2\), so \(3k=k\cot\tfrac A2\cot\tfrac C2\) and the product of the outer two is 3.
3. AM-GM: \(2k\ge2\sqrt3\), so \(k^2\ge3\).

Teacher's Note:
1. Convert side lengths with \( a = 2R \sin A \) and use the standard identities for the triangle before trying algebra.
2. Test the final result on an equilateral or right-angled triangle as a quick check.

Answer: (b) 5

Explanation:
1. The triangle formed by the altitudes has sides proportional to \( 1/a, 1/b, 1/c \), and \(\Delta_1/\Delta=\Delta_2/\Delta_1=k\) (the same constant \( k \) at each step; verified numerically on several triangles).
2. \(k=\dfrac{20}{30}=\dfrac23\), so \(\Delta=\dfrac{30}{2/3}=45\).
3. \(\dfrac\Delta9=5\).

Teacher's Note:
1. Convert side lengths with \( a = 2R \sin A \) and use the standard identities for the triangle before trying algebra.
2. Test the final result on an equilateral or right-angled triangle as a quick check.

Answer: (d) 6

Explanation:
1. The circle is the incircle with \(r=\dfrac23\).
2. \(p_1+p_2+p_3=2\Delta\sum\dfrac1a=2rs\sum\dfrac1a=2r\left(s\sum\dfrac1a\right)\ge2r\cdot\dfrac92=9r=6\) since \(s\sum\dfrac1a\ge\dfrac92\) by AM-HM.

Teacher's Note:
1. Convert side lengths with \( a = 2R \sin A \) and use the standard identities for the triangle before trying algebra.
2. Test the final result on an equilateral or right-angled triangle as a quick check.

Answer: (c) \(x^3-(4R+r)x^2+s^2x-rs^2=0\)

Explanation:
1. \(r_1+r_2+r_3=4R+r\), \(r_1r_2+r_2r_3+r_3r_1=s^2\) and \(r_1r_2r_3=rs^2\).
2. Form the cubic \( x^3 \) - (sum) \( x^2 \) + (pair sum) \( x \) - (product) = 0.

Teacher's Note:
1. Convert side lengths with \( a = 2R \sin A \) and use the standard identities for the triangle before trying algebra.
2. Test the final result on an equilateral or right-angled triangle as a quick check.

Answer: (a) \(2s^2(s+r+2R)\)

Explanation:
1. Expand with the symmetric functions: \(s^3+s^2(4R+r)+s\cdot s^2+rs^2\).
2. \(=2s^3+s^2(4R+2r)=2s^2(s+r+2R)\).

Teacher's Note:
1. Convert side lengths with \( a = 2R \sin A \) and use the standard identities for the triangle before trying algebra.
2. Test the final result on an equilateral or right-angled triangle as a quick check.

Answer: (a) a > b sin A and a < b

Explanation:
1. The discriminant \(4(a^2-b^2\sin^2A)>0\) needs \(a>b\sin A\).
2. Product of the roots \(b^2-a^2>0\) and sum \(2b\cos A>0\) require both roots positive, so \(a<b\).

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.

Answer: (a) only one triangle is possible

Explanation:
1. With \(a=b\) the equation becomes \(c^2-2bc\cos A=0\), so \(c=0\) or \(c=2b\cos A\).
2. \( c = 0 \) is not a triangle, so exactly one triangle is possible.

Teacher's Note:
1. Use the cosine rule and the projection rule to switch between sides and angles.
2. Check any identity numerically on a convenient triangle before trusting it.