JEE Mathematics Properties and Solution of Triangles MCQs Set 03

Answer: (b) \(\frac rR\)

Explanation:
1. \( a\cos A+b\cos B+c\cos C=R(\sin2A+\sin2B+\sin2C)=4R\sin A\sin B\sin C \).
2. The denominator is \( 2R(\sin A+\sin B+\sin C)=8R\cos\frac A2\cos\frac B2\cos\frac C2 \).
3. The ratio is \( 4\sin\frac A2\sin\frac B2\sin\frac C2=\frac rR \).

Teacher's Note:
1. Write the numerator through sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.
2. Remember r = 4R sin(A/2) sin(B/2) sin(C/2).

Answer: (a) 18 and 22

Explanation:
1. Let PN = PM = n, QL = QN = n + 2, RM = RL = n + 4 (consecutive even integers when n is even). Then \( PQ=2n+2,\ PR=2n+4,\ QR=2n+6 \) and P (opposite the longest side QR) is the largest angle.
2. \( \cos P=\frac{(2n+2)^2+(2n+4)^2-(2n+6)^2}{2(2n+2)(2n+4)}=\frac{n-2}{2(n+1)}=\frac13 \), so \( 3n-6=2n+2 \) and n = 8.
3. The sides are 18, 20 and 22, so a possible pair is 18 and 22.

Teacher's Note:
1. Tangent lengths from one point to a circle are equal, so each side is the sum of two tangent lengths.
2. P is opposite the longest side QR.

Answer: (a) R, R, R

Explanation:
1. Angle BPC = 180° − A, so \( \sin\angle BPC=\sin A \).
2. Circumradius of PBC \( =\frac{BC}{2\sin\angle BPC}=\frac{a}{2\sin A}=R \). The other two are also R.

Teacher's Note:
1. The angle at the orthocentre is the supplement of the vertex angle.
2. Equal sines mean the same circumradius.

Answer: (d) all four of (i), (ii), (iii) and (iv)

Explanation:
1. (i) Perimeter of DEF \( =R(\sin2A+\sin2B+\sin2C)=4R\sin A\sin B\sin C \) and perimeter of ABC \( =2R(\sin A+\sin B+\sin C)=8R\cos\frac A2\cos\frac B2\cos\frac C2 \); the ratio simplifies to \( 4\sin\frac A2\sin\frac B2\sin\frac C2=\frac rR \).
2. (ii) Area \( =\frac12FD\cdot DE\sin(\pi-2A)=\frac12bc\cos B\cos C\sin2A=2\Delta\cos A\cos B\cos C \).
3. (iii) Area AEF \( =\frac12(c\cos A)(b\cos A)\sin A=\Delta\cos^2A \). (iv) \( R_{DEF}=\frac{FE\cdot ED\cdot DF}{4\Delta_{DEF}}=\frac{abc\cos A\cos B\cos C}{8\Delta\cos A\cos B\cos C}=\frac R2 \). All four are true.

Teacher's Note:
1. Use the side lengths a cos A, b cos B, c cos C of the pedal triangle.
2. The pedal triangle's circumradius is R/2 and its area is 2Δ cos A cos B cos C.

Answer: (b) 3

Explanation:
1. \( \frac12\cdot6\cdot10\sin C=15\sqrt3\Rightarrow\sin C=\frac{\sqrt3}2 \); C is obtuse, so C = 120°.
2. \( c^2=36+100+60=196 \), so c = 14.
3. \( r=\frac\Delta s=\frac{15\sqrt3}{15}=\sqrt3 \), so \( r^2=3 \).

Teacher's Note:
1. The obtuse condition picks C = 120° rather than 60°.
2. s = (6 + 10 + 14)/2 = 15.

Answer: (d) \(3\sqrt5\)

Explanation:
1. Let BP = PQ = QC = x, so BC = 3x. Let b = AC and c = AB; then \( b^2+c^2=9x^2 \).
2. Apply Apollonius' theorem to the cevians: in triangle ABQ with P the mid-point of BQ, \( c^2+16=2(9+x^2) \); in triangle APC with Q the mid-point of PC, \( b^2+9=2(16+x^2) \).
3. Adding: \( b^2+c^2+25=50+4x^2\Rightarrow 9x^2=25+4x^2\Rightarrow x^2=5 \), so BC = 3x = \( 3\sqrt5 \).

Teacher's Note:
1. P is the mid-point of BQ and Q is the mid-point of PC, so the median formula applies twice.
2. Add the two relations and use b² + c² = BC² for the right angle at A.

Answer: (a) \(\frac14\)

Explanation:
1. Here \( f=R\cos A,\ g=R\cos B,\ h=R\cos C \).
2. \( \frac af+\frac bg+\frac ch=2(\tan A+\tan B+\tan C)=2\tan A\tan B\tan C \) and \( \frac{abc}{fgh}=\frac{8R^3\sin A\sin B\sin C}{R^3\cos A\cos B\cos C}=8\tan A\tan B\tan C \).
3. So \( 2=8\lambda \) and \( \lambda=\frac14 \).

Teacher's Note:
1. The distance from the circumcentre to side BC is R cos A.
2. In a triangle, tan A + tan B + tan C = tan A tan B tan C.

Answer: (c) 7 : 2

Explanation:
1. Take a = 3k, b = 7k, c = 8k, so s = 9k.
2. \( \Delta=\sqrt{9k\cdot6k\cdot2k\cdot k}=6\sqrt3\,k^2 \).
3. \( R=\frac{abc}{4\Delta}=\frac{168k^3}{24\sqrt3k^2}=\frac{7k}{\sqrt3} \) and \( r=\frac\Delta s=\frac{2k}{\sqrt3} \).
4. So R : r = 7 : 2.

Teacher's Note:
1. Use Heron's formula to get the area for the ratio-sides.
2. R = abc/(4Δ) and r = Δ/s.

Answer: (a) (A)-(s), (B)-(p), (C)-(r), (D)-(q)

Explanation:
1. (A) Perpendicular medians give \( a^2+b^2=5c^2 \), so \( c^2=5 \), \( \cos C=\frac{16+9-5}{24}=\frac56 \), \( \sin C=\frac{\sqrt{11}}6 \) and \( \Delta^2=\left(\frac12\cdot12\cdot\frac{\sqrt{11}}6\right)^2=11 \) (s).
2. (B) By A.M.-G.M.-H.M. on r1, r2, r3, \( (r_1r_2r_3)^{1/3}\ge\frac{3}{\frac1{r_1}+\frac1{r_2}+\frac1{r_3}}=3r \), so the minimum of \( \frac{r_1r_2r_3}{r^3} \) is 27 (p).
3. (C) \( \tan^2\frac C2=\frac{(s-a)(s-b)}{s(s-c)}=\frac{c^2-1}{81-c^2}=\frac79\Rightarrow c^2=36 \), c = 6 (r).
4. (D) \( (a-2b)^2+(a-c)^2=0 \) gives \( a=2b=c \), so \( \cos B=\frac{a^2+c^2-b^2}{2ac}=\frac78 \) and 8 cos B = 7 (q).

Teacher's Note:
1. For perpendicular medians from A and B, the standard condition is a² + b² = 5c².
2. In (D) the left side minus the right side is a sum of two squares.

Answer: (a) \(\left(\frac{3}{4\Delta}\right)^2\)

Explanation:
1. \( \frac{2\sin P-2\sin P\cos P}{2\sin P+2\sin P\cos P}=\frac{1-\cos P}{1+\cos P}=\tan^2\frac P2 \).
2. With \( s=4 \): \( \tan^2\frac P2=\frac{(s-b)(s-c)}{s(s-a)}=\frac{(s-b)^2(s-c)^2}{\Delta^2} \) (using \( \Delta^2=s(s-a)(s-b)(s-c) \)).
3. Since s − b = 1/2 and s − c = 3/2, the value is \( \left(\frac{3/4}{\Delta}\right)^2=\left(\frac{3}{4\Delta}\right)^2 \).

Teacher's Note:
1. The expression is tan²(P/2) after cancelling 2 sin P.
2. Multiply numerator and denominator by (s − a)... via Heron's formula to write it using Δ.

Answer: (b) orthocentre

Explanation:
1. The internal bisector of angle A passes through I and I1, and I2I3 is the external bisector at A, so AI1 is perpendicular to I2I3. The same holds at B and C.
2. So the internal bisectors AI1, BI2, CI3 are the altitudes of triangle I1I2I3, and they meet at I.
3. Hence I is the orthocentre of the excentral triangle.

Teacher's Note:
1. Internal and external bisectors at a vertex are perpendicular.
2. Lines through the vertices of the excentral triangle perpendicular to opposite sides are its altitudes.

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

Explanation:
1. For a right angle at B, \( r=s-b=\frac{AB+BC-AC}{2} \), so (i) is true and (ii), (iii) are false.
2. The circumradius is half the hypotenuse, \( R=\frac{AC}{2} \). Also \( s-r=b \), so \( R=\frac{s-r}2 \) and (iv) is true.

Teacher's Note:
1. In a right triangle the inradius is s minus the hypotenuse.
2. R is half the hypotenuse and s − r equals the hypotenuse.

Answer: (c) \(\frac c{\sqrt3}\)

Explanation:
1. From \( (a+b)^2-c^2=ab \) we get \( a^2+b^2-c^2=-ab \), so \( \cos C=-\frac12 \) and C = 120°.
2. Then \( R=\frac{c}{2\sin C}=\frac{c}{2\sin120^\circ}=\frac c{\sqrt3} \).

Teacher's Note:
1. The given relation is the cosine rule in disguise.
2. The sine rule gives R once the angle opposite c is known.

Answer: (d) \(4Rr^2\)

Explanation:
1. The distance of I from A is \( r\,\mathrm{cosec}\frac A2 \), and similarly for B and C.
2. Product \( =\frac{r^3}{\sin\frac A2\sin\frac B2\sin\frac C2}=\frac{r^3}{r/(4R)}=4Rr^2 \).

Teacher's Note:
1. In the right triangle formed by I, the touch point and A, IA = r / sin(A/2).
2. Use r = 4R sin(A/2) sin(B/2) sin(C/2).

Answer: (a) A.P.

Explanation:
1. Use \( \cos^2\frac A2=\frac{s(s-a)}{bc} \) and \( \cos^2\frac B2=\frac{s(s-b)}{ac} \).
2. The condition becomes \( \frac sc[(s-a)+(s-b)]=\frac{3c}2\Rightarrow\frac sc\cdot c=\frac{3c}2 \), so \( s=\frac{3c}{2} \).
3. Thus a + b + c = 3c, i.e. a + b = 2c, so a, c, b are in A.P.

Teacher's Note:
1. Half-angle formulas for cos² turn the condition into one in s and c.
2. (s − a) + (s − b) = c, which simplifies the algebra.

Answer: (d) 2

Explanation:
1. \( \cot\frac B2\cot\frac C2=\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}=\frac{s}{s-a} \).
2. With b + c = 3a: \( s=2a,\ s-a=a \), so the value is 2.

Teacher's Note:
1. The product of the two half-angle cotangents collapses to s/(s − a).
2. b + c = 3a gives s = 2a.

Answer: (c) \(4Rr^2\)

Explanation:
1. \( r_1-r=\frac\Delta{s-a}-\frac\Delta s=\frac{\Delta a}{s(s-a)}=a\tan\frac A2 \).
2. Using \( \tan\frac A2=\frac r{s-a} \): \( \prod\tan\frac A2=\frac{r^3}{(s-a)(s-b)(s-c)}=\frac rs \).
3. So the product is \( abc\cdot\frac rs=\frac{4R\Delta r}{s}=4Rr^2 \) (since \( \Delta=rs \)).

Teacher's Note:
1. Show that r1 − r equals a tan(A/2) first.
2. Then abc = 4RΔ and Δ = rs finish the job.

Answer: (a) 4

Explanation:
1. \( r_1+r_2=\frac{\Delta c}{(s-a)(s-b)} \), so the product is \( \frac{\Delta^3abc}{(s-a)^2(s-b)^2(s-c)^2}=\frac{abc\,s^2}{\Delta} \) (using \( \Delta^2=s(s-a)(s-b)(s-c) \)).
2. Since \( abc=4R\Delta \), the product equals \( 4Rs^2 \) and the ratio is 4.

Teacher's Note:
1. r1 + r2 simplifies to Δc/((s − a)(s − b)).
2. Then abc = 4RΔ turns the product into 4Rs².

Answer: (a) \(\frac{\sqrt7}2\)

Explanation:
1. \( \sin P=\frac p{2R}=\frac{\sqrt3}2 \) and \( \sin Q=\frac q{2R}=\frac12 \). So P = 60° or 120° and Q = 30° or 150°. Since the triangle is not right angled and P + Q < 180°, P = 120° and Q = 30° (P = 60° would give R = 90°).
2. Then R = 30° and \( r=2R_c\sin30^\circ=1 \) (r is the side PQ).
3. Median formula: \( RS=\frac12\sqrt{2p^2+2q^2-r^2}=\frac12\sqrt{6+2-1}=\frac{\sqrt7}2 \).

Teacher's Note:
1. The sine rule with circumradius 1 gives the sines of P and Q directly.
2. The non-right-angled condition removes the possibility P = 60°.

Answer: (b) both right angled and isosceles

Explanation:
1. Altitude from A is \( c\sin B\ge a \) and altitude from B is \( a\sin C\ge b \).
2. Using the sine rule: \( \sin C\sin B\ge\sin A \) and \( \sin A\sin C\ge\sin B \). Multiplying gives \( \sin^2C\ge1 \), so \( C=90^\circ \).
3. Then the altitude from A is b and that from B is a, so \( b\ge a \) and \( a\ge b \), i.e. a = b. The triangle is right angled and isosceles.

Teacher's Note:
1. Write each altitude through the sine of an angle and combine the two inequalities.
2. Since sin C cannot exceed 1, equality must hold throughout.

Answer: (b) both isosceles and right angled

Explanation:
1. \( \sin C=\frac{1-\cos A\cos B}{\sin A\sin B}\le1\Rightarrow\cos(A-B)\ge1 \), so \( \cos(A-B)=1 \) and A = B.
2. Then \( \sin C=\frac{1-\cos^2A}{\sin^2A}=1 \), so C = 90°.
3. The triangle is right angled and isosceles.

Teacher's Note:
1. Since sin C is at most 1, the numerator must be at most the denominator.
2. This gives cos(A − B) = 1, which forces A = B.

Answer: (a) \(1+\sqrt3\)

Explanation:
1. Cosine rule at C: \( a^2+b^2-c^2=2ab\cos30^\circ=\sqrt3\,ab \).
2. Substituting and factorising gives \( (x^2-1)\left[(2-\sqrt3)(x^2+x)-(1+\sqrt3)\right]=0 \). Since b = x² − 1 > 0, \( x^2+x=\frac{1+\sqrt3}{2-\sqrt3}=5+3\sqrt3 \).
3. Solving \( x^2+x-(5+3\sqrt3)=0 \): \( x=\frac{-1\pm(3+2\sqrt3)}{2} \), i.e. \( x=1+\sqrt3 \) or \( x=-(2+\sqrt3) \). The negative root makes c = 2x + 1 negative, so x = 1 + √3.

Teacher's Note:
1. All sides must be positive, which rules out the negative root.
2. The quadratic in x comes out as x² + x = 5 + 3√3.

Answer: (c) (i), (iii) and (iv) only

Explanation:
1. Take BD = DE = EC = 1. Since the angles at D and E are equal, triangle ADE is isosceles and the foot of the perpendicular from A is the mid-point of DE. Let its height be h.
2. Then \( \tan\theta=\frac{h}{1/2}=2h \), while \( \tan B=\frac h{3/2} \) and \( \tan C=\frac h{3/2} \). So \( \tan\theta=3\tan B=3\tan C \) and B = C.
3. Then \( A=\pi-2B\Rightarrow\tan A=-\tan2B=-\frac{2\tan B}{1-\tan^2B}=\frac{6\tan\theta}{\tan^2\theta-9} \). So (i), (iii), (iv) are true and (ii) is false.

Teacher's Note:
1. Triangle ADE is isosceles, so the foot of the perpendicular from A bisects DE.
2. Equal tangents of B and C force B = C.

Answer: (d) \(\frac{\sqrt5+1}{2}\)

Explanation:
1. Let the sides be vectors a and b; the diagonals are a + b and a − b. Their cross product has magnitude \( 2ab\sin30^\circ=ab \) and their dot product is \( a^2-b^2 \).
2. The angle between them is 45°, so \( \tan45^\circ=\frac{ab}{a^2-b^2}=1\Rightarrow a^2-b^2=ab \).
3. With \( t=\frac ab \): \( t^2-t-1=0\Rightarrow t=\frac{1+\sqrt5}2 \) (t > 1).

Teacher's Note:
1. Treat the sides as vectors; the diagonals are their sum and difference.
2. The tangent of the angle between two vectors is |cross product| / dot product.

Answer: (a) \(\frac{(p^2+q^2)\sin\theta}{p\cos\theta+q\sin\theta}\)

Explanation:
1. In the right triangle BCD, \( BD=\sqrt{p^2+q^2} \), \( \sin\alpha=\frac p{\sqrt{p^2+q^2}} \) and \( \cos\alpha=\frac{q}{\sqrt{p^2+q^2}} \) where α = ∠BDC. Since AB ∥ CD, ∠ABD = α.
2. In triangle ABD the angles are θ at D, α at B, so the angle at A is \( \pi-(\theta+\alpha) \). Sine rule: \( \frac{AB}{\sin\theta}=\frac{BD}{\sin(\theta+\alpha)} \).
3. \( AB=\frac{\sqrt{p^2+q^2}\sin\theta}{\sin\theta\cos\alpha+\cos\theta\sin\alpha}=\frac{(p^2+q^2)\sin\theta}{q\sin\theta+p\cos\theta} \).

Teacher's Note:
1. Alternate angles in the parallel lines AB and CD give ∠ABD = ∠BDC.
2. Apply the sine rule in triangle ABD.

Answer: (c) \(\frac12Rr^2\)

Explanation:
1. In triangle MIN, IM = IN = r and angle MIN = 180° − A, so \( MN=2r\cos\frac A2 \) and \( x=\frac{MN}{2\sin(180^\circ-A)}=\frac{r}{2\sin\frac A2} \).
2. Similarly \( y=\frac r{2\sin\frac B2},\ z=\frac r{2\sin\frac C2} \).
3. \( xyz=\frac{r^3}{8\sin\frac A2\sin\frac B2\sin\frac C2}=\frac{r^3}{8\cdot\frac r{4R}}=\frac12Rr^2 \).

Teacher's Note:
1. The quadrilateral AMIN is cyclic, so angle MIN is the supplement of A.
2. Use r = 4R sin(A/2) sin(B/2) sin(C/2) at the end.