JEE Mathematics Properties and Solution of Triangles MCQs Set 02

Answer: (d) \(\frac{9\sqrt3(1+\sqrt3)}{\pi^2}\)

Explanation:
1. Total circumference 12 = \( 2\pi R \), so \( R=\frac6\pi \).
2. Central angles (arc / radius): \( \frac{3}{R}=\frac\pi2,\ \frac4R=\frac{2\pi}3,\ \frac5R=\frac{5\pi}6 \).
3. Area \( =\frac12R^2\left(\sin\frac\pi2+\sin\frac{2\pi}3+\sin\frac{5\pi}6\right)=\frac{18}{\pi^2}\left(1+\frac{\sqrt3}2+\frac12\right)=\frac{9\sqrt3(1+\sqrt3)}{\pi^2} \).

Teacher's Note:
1. The area is the sum of three isosceles triangles formed with the centre.
2. Each has area (1/2)R² sin(central angle).

Answer: (c) R, R, R

Explanation:
1. In triangle BHC, angle BHC = 180° − A, so \( \sin\angle BHC=\sin A \).
2. By the sine rule in triangle BHC, \( 2R'=\frac{BC}{\sin\angle BHC}=\frac{a}{\sin A}=2R \).
3. So R' = R, and the same holds for the other two triangles.

Teacher's Note:
1. The angle at H in triangle BHC is the supplement of angle A.
2. Supplementary angles have equal sines, so the circumradius is unchanged.

Answer: (b) \(\frac\pi6\)

Explanation:
1. Perimeter \( =2s=6\cdot\frac{\sin A+\sin B+\sin C}{3}=2(\sin A+\sin B+\sin C) \).
2. Using \( \sin A=\frac{a}{2R} \) etc., \( \sin A+\sin B+\sin C=\frac{a+b+c}{2R} \), so \( 2s=\frac{2s}{R} \) and R = 1.
3. Then \( \sin A=\frac a{2R}=\frac12 \) and A = π/6 (the triangle is such that A cannot be 5π/6).

Teacher's Note:
1. Express the sum of the sines through the perimeter and the circumradius.
2. With a = 1, R = 1 gives sin A = 1/2.

Answer: (b) \(\frac12\)

Explanation:
1. \( \frac r{r_1}=\tan\frac B2\tan\frac C2=\frac12 \).
2. Since \( \tan\frac A2\tan\frac B2+\tan\frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=1 \), the required expression equals \( 1-\frac12=\frac12 \).

Teacher's Note:
1. r/r1 equals tan(B/2) tan(C/2).
2. Use the identity for the sum of products of half-angle tangents.

Answer: (b) \(\pi\)

Explanation:
1. Take a = 4k, b = 5k, c = 6k: \( \cos A=\frac{25+36-16}{60}=\frac34 \) and \( \cos B=\frac{36+16-25}{48}=\frac9{16} \).
2. \( \cos3A=4\cos^3A-3\cos A=\frac{27}{16}-\frac94=-\frac9{16}=-\cos B=\cos(\pi-B) \).
3. Since 3A and π − B are both in (0, π), \( 3A=\pi-B \), so 3A + B = π.

Teacher's Note:
1. Find cos A and cos B from the cosine rule and then use the triple-angle formula.
2. Equal cosines of angles in (0, π) imply equal angles.

Answer: (a) 9 cm

Explanation:
1. \( \Delta=\frac12bc\sin A=\frac{\sqrt3}{4}bc=\frac{9\sqrt3}2\Rightarrow bc=18 \).
2. \( a^2=b^2+c^2-2bc\cos A=b^2+c^2+bc=(b-c)^2+3bc=27+54=81 \).
3. So a = 9 cm.

Teacher's Note:
1. The area formula gives bc; the cosine rule with A = 120° gives a².
2. Write b² + c² + bc as (b − c)² + 3bc.

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

Explanation:
1. B lies on line AC1 and on line B1G, so the power of B with respect to the circle gives \( BC_1\cdot BA=BG\cdot BB_1 \).
2. Here \( BC_1=\frac c2,\ BA=c,\ BG=\frac23BB_1 \) and \( BB_1^2=\frac{2a^2+2c^2-b^2}4 \).
3. \( \frac{c^2}2=\frac23\cdot\frac{2a^2+2c^2-b^2}4\Rightarrow3c^2=2a^2+2c^2-b^2\Rightarrow b^2+c^2=2a^2 \).

Teacher's Note:
1. The power of a point theorem applies to B because it lies on two secants of the circle.
2. The median length formula gives BB1².

Answer: (b) 3 : 4

Explanation:
1. By the median formula \( AD^2=\frac14(2b^2+2c^2-a^2) \) and similarly for BE and CF.
2. Adding: \( AD^2+BE^2+CF^2=\frac14\cdot3(a^2+b^2+c^2) \).
3. The ratio is 3 : 4.

Teacher's Note:
1. Apply Apollonius' median formula to all three medians.
2. Each side squared appears with total coefficient 3/4 after adding.

Answer: (b) \(a^2\)

Explanation:
1. Since \( \cos A=-\cos(B+C) \), the denominator is \( -\cos(B+C)+\cos B\cos C=\sin B\sin C \).
2. With b = k sin B and c = k sin C: \( \frac{k^2\sin B\sin C\sin^2A}{\sin B\sin C}=k^2\sin^2A=a^2 \).

Teacher's Note:
1. Replace cos A by −cos(B + C) to make the denominator simple.
2. Then every factor is expressed through k and the sines of the angles.

Answer: (b) a = 3, c = 4

Explanation:
1. Subtracting the two relations: c − a = 1. Adding: 2s − a − c = 5, so b = 5.
2. Right angle at B: \( a^2+c^2=25 \). With c = a + 1: \( a^2+(a+1)^2=25\Rightarrow a^2+a-12=0 \).
3. So a = 3 and c = 4 (the root a = −4 is rejected).

Teacher's Note:
1. s − a and s − c give the differences of sides directly.
2. Use Pythagoras with c = a + 1.

Answer: (d) \(\frac{a-b}{c}\)

Explanation:
1. With A = 90°: \( \sin C=\frac ca,\ \cos C=\frac ba \).
2. \( \tan\frac C2=\frac{\sin C}{1+\cos C}=\frac{c}{a+b} \).
3. Since \( a^2-b^2=c^2 \), \( \frac c{a+b}=\frac{a-b}c \).

Teacher's Note:
1. Use tan(C/2) = sin C/(1 + cos C).
2. Pythagoras lets you rationalise c/(a + b).

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

Explanation:
1. \( PR^2=300+100-2\cdot10\sqrt3\cdot10\cos30^\circ=100 \), so PR = 10 = QR and the triangle is isosceles with ∠P = ∠Q = 30°, ∠R = 120°. Hence (i) is false.
2. Area \( =\frac12\cdot10\sqrt3\cdot10\sin30^\circ=25\sqrt3 \) and ∠QRP = 120°, so (ii) is true.
3. \( r=\frac\Delta s=\frac{25\sqrt3}{10+5\sqrt3}=10\sqrt3-15 \) so (iii) is true; \( R=\frac{PR}{2\sin30^\circ}=10 \), so the circumcircle has area 100π and (iv) is true.

Teacher's Note:
1. The cosine rule gives the third side PR = 10, so two sides are equal.
2. Check each statement with the angles 30°, 30°, 120°.

Answer: (a) 1

Explanation:
1. \( \Delta=sr=35\times6=210\Rightarrow ab=420 \).
2. With a + b + c = 70 and \( a^2+b^2=c^2 \): \( (70-c)^2=c^2+840\Rightarrow c=29 \).
3. \( (a-b)^2=c^2-2ab=841-840=1 \), so |a − b| = 1.

Teacher's Note:
1. Area from Δ = rs gives ab for the right triangle.
2. (a − b)² = a² + b² − 2ab = c² − 2ab.

Answer: (d) \(16R^2\)

Explanation:
1. The distance \( II_1=4R\sin\frac A2 \) and the side \( I_2I_3=4R\cos\frac A2 \).
2. \( II_1^2+I_2I_3^2=16R^2\left(\sin^2\frac A2+\cos^2\frac A2\right)=16R^2 \).

Teacher's Note:
1. Recall that II1 = 4R sin(A/2).
2. Add the squares and use sin² + cos² = 1.

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

Explanation:
1. (i) becomes \( \frac1{(s-b)(s-c)}=\frac1{s(s-a)} \), i.e. \( \tan^2\frac A2=1 \), so A = 90°.
2. (iii) gives \( s\tan\frac A2=s \), so A = 90°; (iv) gives \( 2R=2R(1-\cos A) \), so cos A = 0.
3. (ii): \( \sin^2A+\sin^2B+\sin^2C=2\Rightarrow\cos A\cos B\cos C=0 \), so some angle is 90°. All four relations force a right angle.

Teacher's Note:
1. Rewrite each relation with half-angle formulas; every one reduces to a single right-angle condition.
2. For (ii) the identity sin²A + sin²B + sin²C = 2 + 2 cos A cos B cos C is useful.

Answer: (b) \(\frac1{\sqrt6}\)

Explanation:
1. Let ∠BAD = α, ∠CAD = β. In triangle ABD, \( \frac{BD}{\sin\alpha}=\frac{AD}{\sin60^\circ} \); in triangle ACD, \( \frac{CD}{\sin\beta}=\frac{AD}{\sin45^\circ} \).
2. Dividing: \( \frac{\sin\alpha}{\sin\beta}=\frac{BD}{CD}\cdot\frac{\sin60^\circ}{\sin45^\circ}=\frac13\cdot\frac{\sqrt3/2}{1/\sqrt2}=\frac13\cdot\sqrt{\frac32} \).
3. This equals \( \frac{1}{\sqrt6} \).

Teacher's Note:
1. Apply the sine rule in the two small triangles that share AD.
2. The ratio BD : CD is 1 : 3.

Answer: (a) \(\frac\pi2-\frac A2,\ \frac\pi2-\frac B2,\ \frac\pi2-\frac C2\)

Explanation:
1. ABC is the pedal triangle of I1I2I3, so its angles are \( \pi-2\theta_i \) where the θi are the angles of the excentral triangle.
2. So \( A=\pi-2\theta_1\Rightarrow\theta_1=\frac\pi2-\frac A2 \) and similarly for the others.

Teacher's Note:
1. Use the earlier result that the angles of a pedal triangle are 180° minus twice the angles of the original.
2. Solve for the angle of the excentral triangle.

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

Explanation:
1. \( \tan^2\frac{A-B}2=\frac{1-\cos(A-B)}{1+\cos(A-B)}=\frac1{63} \), so \( \tan\frac{A-B}2=\frac1{3\sqrt7} \).
2. Napier's analogy: \( \tan\frac{A-B}{2}=\frac{a-b}{a+b}\cot\frac C2\Rightarrow\cot\frac C2=\frac3{\sqrt7} \), so \( \cos C=\frac{1-\tan^2(C/2)}{1+\tan^2(C/2)}=\frac18 \).
3. \( c^2=25+16-2\cdot20\cdot\frac18=36 \), so c = 6; \( \sin C=\frac{3\sqrt7}8 \), giving \( \sin A=\frac{a\sin C}{c}=\frac{5\sqrt7}{16} \) and area \( \frac12\cdot5\cdot4\cdot\frac{3\sqrt7}8=\frac{15\sqrt7}4 \). So (i), (ii), (iii) are true and (iv) is false.

Teacher's Note:
1. Napier's analogy connects tan((A − B)/2) with cot(C/2).
2. Once cos C is known, the cosine rule gives c and the sine rule gives sin A.

Answer: (d) \(6\sqrt6\)

Explanation:
1. Let the common value be λ: \( s-x=4\lambda,\ s-y=3\lambda,\ s-z=2\lambda \). Adding, \( s=9\lambda \), so x = 5λ, y = 6λ, z = 7λ.
2. \( \Delta=\sqrt{9\lambda\cdot4\lambda\cdot3\lambda\cdot2\lambda}=6\sqrt6\,\lambda^2 \) and \( r^2=\frac{\Delta^2}{s^2}=\frac{216\lambda^4}{81\lambda^2}=\frac{8\lambda^2}3 \).
3. Incircle area \( \pi r^2=\frac{8\pi\lambda^2}{3}=\frac{8\pi}3 \) gives λ = 1, so the area is \( 6\sqrt6 \).

Teacher's Note:
1. Express all sides through λ and use Heron's formula.
2. The incircle area fixes λ.

Answer: (b) (i) and (ii) only

Explanation:
1. (i): by the cosine rule each term is \( \frac{b^2+c^2-a^2}{2abc} \) and so on; adding gives \( \frac{a^2+b^2+c^2}{2abc} \). True for every triangle.
2. (ii): \( \frac{\sin A}a=\frac1{2R} \) for each term, so the sum is \( \frac3{2R} \). True for every triangle.
3. (iii) reduces to cot A = cot B = cot C and (iv) reduces to the same condition, so both hold only for an equilateral triangle.

Teacher's Note:
1. Test each statement by converting it entirely into sides or into sines.
2. A statement that forces cot A = cot B = cot C is true only for the equilateral triangle.

Answer: (d) equilateral

Explanation:
1. Put a = k sin A, b = k sin B, c = k sin C.
2. The condition becomes \( \cot A=\cot B=\cot C \), so A = B = C.
3. Hence the triangle is equilateral.

Teacher's Note:
1. The sine rule turns every side into k times the sine of the opposite angle.
2. Equal cotangents of angles in (0, π) mean equal angles.

Answer: (a) \(\frac{60}{13}\)

Explanation:
1. Since \( 5^2+12^2=13^2 \), the triangle is right angled at A and \( \Delta=\frac12\cdot5\cdot12=30 \).
2. The altitude from A is \( \frac{2\Delta}{a}=\frac{60}{13} \) cm.

Teacher's Note:
1. Area equals half the base times the height, so height = 2Δ/a.
2. Heron's formula (s = 15) also gives Δ = 30.

Answer: (b) isosceles but not necessarily equilateral

Explanation:
1. Rewrite as \( bc=2\cos A\,(b^2+c^2-2bc\cos A)=2a^2\cos A \).
2. So \( \cos A=\frac{bc}{2a^2} \). Also \( \cos A=\frac{b^2+c^2-a^2}{2bc} \).
3. Equating: \( b^2c^2=a^2(b^2+c^2-a^2)\Rightarrow(a^2-b^2)(a^2-c^2)=0 \), so a = b or a = c.
4. The triangle is isosceles, but not necessarily equilateral.

Teacher's Note:
1. Notice that b² + c² − 2bc cos A is a².
2. Compare the two expressions for cos A to get a product of two differences of squares.

Answer: (d) \(\frac1{250}\)

Explanation:
1. In any triangle \( \tan A+\tan B+\tan C=\tan A\tan B\tan C \), so the expression is \( \frac{2(\tan A+\tan C)\tan^2B}{\tan A\tan B\tan C}=2\left(\cot A+\cot C\right)\tan B \).
2. \( \cot A+\cot C=\frac{\sin(A+C)}{\sin A\sin C}=\frac{\sin B}{\sin A\sin C} \), so the value is \( \frac{2\sin^2B}{\sin A\sin C\cos B} \).
3. With the sine rule and cosine rule this is \( \frac{2b^2}{ac\cos B}=\frac{4b^2}{a^2+c^2-b^2}=\frac{4b^2}{1000b^2}=\frac1{250} \).

Teacher's Note:
1. Use the identity tan A + tan B + tan C = tan A tan B tan C to remove the denominator.
2. Then change everything to sides: a² + c² = 1001 b² is the given relation.

Answer: (b) 1

Explanation:
1. The circumcentre is at height \( R\cos A \) above BC and the incentre at height r. Parallel to BC means these heights are equal: \( R\cos A=r \).
2. Now \( \frac rR=4\sin\frac A2\sin\frac B2\sin\frac C2=\cos A+\cos B+\cos C-1 \).
3. So \( \cos A=\cos A+\cos B+\cos C-1 \), which gives cos B + cos C = 1.

Teacher's Note:
1. Equal heights above BC give R cos A = r.
2. Use the identity cos A + cos B + cos C = 1 + r/R.

Answer: (d) 3b

Explanation:
1. The angles are A = 45°, B = 75°, C = 60°.
2. By the sine rule, a : b : c = sin 45° : sin 75° : sin 60° = \( \frac1{\sqrt2}:\frac{\sqrt3+1}{2\sqrt2}:\frac{\sqrt3}2 \).
3. Take \( a=\frac k{\sqrt2},\ b=\frac{(\sqrt3+1)k}{2\sqrt2},\ c=\frac{\sqrt3k}2 \). Then \( a+b+c\sqrt2=\frac{k}{2\sqrt2}\left[2+(\sqrt3+1)+2\sqrt3\right]=\frac{3k(\sqrt3+1)}{2\sqrt2}=3b \).

Teacher's Note:
1. First convert the angle ratio into actual angles (they add to 180°).
2. Then use a : b : c = sin A : sin B : sin C and sin 75° = sin(45° + 30°).