18. Properties of Triangles’ Solutions Part 1#

  1. Let a=8a = 8 cm, b=10b = 10 cm and c=12c = 12 cm. So the smallest angle will be AA and greatest angle will be C.C. So from cosine rule,

    cosA=b2+c2a22bc=100+144642.10.12=34\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{100 + 144 - 64}{2.10.12} = \frac{3}{4}

    cosC=a2+b2c22ab=64+1001442.8.10=18\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{64 + 100 - 144}{2.8.10} = \frac{1}{8}

    cos2A=2cos2A1=2.9161=18=cosC\cos 2A = 2\cos^2A - 1 = 2.\frac{9}{16} - 1 = \frac{1}{8} = \cos C

    Thus, we see that greatest angle is double to that of smallest angle.

  2. Let b+c11=c+a12=a+b13,=k\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}, = k

    b+c=11k,c+a=12k,a+b=13ka+b+c=18k\therefore b + c = 11k, c + a = 12k, a + b = 13k \Rightarrow a + b + c = 18k

    a=7k,b=6k,c=5k\therefore a = 7k, b = 6k, c = 5k

    cosA7=b2+c2a22bc=36k2+25k249k22.6.7.7k2=135\frac{\cos A}{7} = \frac{b^2 + c^2 - a^2}{2bc} = \frac{36k^2 + 25k^2 - 49k^2}{2.6.7.7k^2} = \frac{1}{35}

    cosB19=c2+a2b22ca=25k2+49k236k22.5.7.19.k2=135\frac{\cos B}{19} = \frac{c^2 + a^2 - b^2}{2ca} = \frac{25k^2 + 49k^2 - 36k^2}{2.5.7.19.k^2} = \frac{1}{35}

    cosC25=a2+b2c22ab=49k2+36k225k22.7.6.25.k2=135\frac{\cos C}{25} = \frac{a^2 + b^2 - c^2}{2ab} = \frac{49k^2 + 36k^2 - 25k^2}{2.7.6.25.k^2} = \frac{1}{35}

  3. Given, Δ=a2(bc)2Δ=a2b2c2+2bc\Delta = a^2 - (b - c)^2 \Rightarrow \Delta = a^2 - b^2 - c^2 + 2bc

    b2+c2a2=2bcΔ2bccosA=2bc12bcsinA\Rightarrow b^2 + c ^2 - a^2 = 2bc - \Delta \Rightarrow 2bc\cos A = 2bc - \frac{1}{2}bc\sin A

    4cosA+sinA=44(12sin2A2)+2sinA2cosA2=4\Rightarrow 4\cos A + \sin A = 4 \Rightarrow 4\left(1 - 2\sin^2\frac{A}{2}\right) + 2\sin\frac{A}{2}\cos\frac{A}{2} = 4

    2sinA2(cosA24sinA2)=02\sin\frac{A}{2}\left(\cos\frac{A}{2} - 4\sin\frac{A}{2}\right) = 0

    \Rightarrow either sinA2=0\sin\frac{A}{2} = 0 or cosA24sinA2=0\cos\frac{A}{2} - 4\sin\frac{A}{2} = 0

    A=0A = 0 is not possible. tanA2=14\therefore \tan\frac{A}{2} = \frac{1}{4}

    tanA=2tanA21tan2A2=815.\tan A = \frac{2\tan\frac{A}{2}}{1 - \tan^2\frac{A}{2}} = \frac{8}{15}.

  4. Since A,B,CA, B, C are in A.P. 2B=A+C\therefore 2B = A + C

    A+B+C=180B=60\because A + B + C = 180^\circ \Rightarrow B = 60^\circ

    cosB=c2+a2b22ca12=c2+a2b22aca2ac+c2=b2\cos B = \frac{c^2 + a^2 - b^2}{2ca} \Rightarrow \frac{1}{2} = \frac{c^2 + a^2 - b^2}{2ac} \Rightarrow a^2 - ac + c^2 = b^2

    a+ca2ac+a2=a+cb\Rightarrow \frac{a + c}{\sqrt{a^2 - ac + a^2}} = \frac{a + c}{b}

    =k(sinA+sinC)ksinB=2sinA+C2cosAC2sinB= \frac{k(\sin A + \sin C)}{k\sin B} = \frac{2\sin\frac{A + C}{2}\cos\frac{A - C}{2}}{\sin B}

    =2sinBcosAC2sinB=2cosAC2= \frac{2\sin B\cos \frac{A - C}{2}}{\sin B} = 2\cos\frac{A - C}{2}

  5. Δ=12.a.p11p1=aΔ\Delta = \frac{1}{2}.a.p_1 \therefore \frac{1}{p_1} = \frac{a}{\Delta}

    Similarly, 1p2=bΔ,1p3=cΔ\frac{1}{p_2} = \frac{b}{\Delta}, \frac{1}{p_3} = \frac{c}{\Delta}

    L.H.S. =a+bc2Δ=(a+b)2c22Δ(a+b+c)= \frac{a + b - c}{2\Delta} = \frac{(a + b)^2 - c^2}{2\Delta(a + b + c)}

    =2ab+a2+b2c22Δ(a+b+c)=2ab+2abcosC2Δ(a+b+c)= \frac{2ab + a^2 + b^2 - c^2}{2\Delta(a + b + c)} = \frac{2ab + 2ab\cos C}{2\Delta(a + b + c)}

    =ab(1+cosC)Δ(a+b+c)=2abcos2C2Δ(a+b+c)= \frac{ab(1 + \cos C)}{\Delta(a + b + c)} = \frac{2ab\cos^2\frac{C}{2}}{\Delta(a + b + c)}

  6. :tanθ=2ababsinC2\because :\tan\theta = \frac{2\sqrt{ab}}{a - b}\sin\frac{C}{2}

    (ab)2tan2θ=4absin2C2(ab)2(sec2θ1)=4absin2C2\Rightarrow (a - b)^2\tan^2\theta = 4ab\sin^2\frac{C}{2} \Rightarrow (a - b)^2(\sec^2\theta - 1) = 4ab\sin^2\frac{C}{2}

    (ab)2sec2θ=a2b22ab(12sin2C2)(a - b)^2\sec^2\theta = a^2b^2 -2ab\left(1 - 2\sin^2\frac{C}{2}\right)

    (ab)2sec2θ=a2+b22abcosC=c2[cosC=a2+b2c22ab](a - b)^2\sec^2\theta = a^2 + b^2 - 2ab\cos C = c^2[\because \cos C = \frac{a^2 + b^2 - c^2}{2ab}]

    c=(ab)secθ.\Rightarrow c = (a - b)\sec\theta.

  7. ΔABC=12bcsinA=126.3.sinC=9sinC\Delta ABC = \frac{1}{2}bc\sin A = \frac{1}{2}6.3.\sin C = 9\sin C

    tan2AB2=1cos(AB)1+cos(AB)=19\tan^2\frac{A - B}{2} = \frac{1 - \cos(A - B)}{1 + \cos(A - B)} = \frac{1}{9}

    tanAB2=±13\tan\frac{A - B}{2} = \pm\frac{1}{3}

    0<AB2<90tanAB2=13\because 0 < \frac{A - B}{2} < 90^\circ \therefore \tan\frac{A - B}{2} = \frac{1}{3}

    tanAB2=aba+bcotC2cotC2=1\tan\frac{A - B}{2} = \frac{a - b}{a + b}\cot\frac{C}{2} \Rightarrow \cot\frac{C}{2} = 1

    C=90\Rightarrow C = 90^\circ

    Thus, required area =8sin90=9= 8\sin90^\circ = 9

  8. The diagram is given below:

    Problem 8

    From question, A=75,C=60B=45\angle A = 75^\circ, \angle C = 60^\circ \Rightarrow \angle B = 45^\circ

    Also given, ΔBADΔBCD=3=c.x.sinθa.x.sin(45θ)\frac{\Delta BAD}{\Delta BCD} = \sqrt{3} = \frac{c.x.\sin\theta}{a.x.\sin(45^\circ - \theta)} where BD=xBD = x

    csinθasin(45θ)=3\Rightarrow \frac{c\sin\theta}{a\sin(45^\circ - \theta)} = \sqrt{3}

    sinCsinθsinAsin(45θ)=3\Rightarrow \frac{\sin C\sin\theta}{\sin A\sin(45^\circ - \theta)} = \sqrt{3}

    32sinθ3+122sin(45θ)=3\Rightarrow \frac{\frac{\sqrt{3}}{2}\sin\theta}{\frac{\sqrt{3} + 1}{2\sqrt{2}}\sin(45^\circ - \theta)} = \sqrt{3}

    2sinθ=(3+1)sin(45θ)\Rightarrow \sqrt{2}\sin\theta = (\sqrt{3} + 1)\sin(45^\circ - \theta)

    2sinθ=(3+1)(12cosθ12sinθ)\Rightarrow \sqrt{2}\sin\theta = (\sqrt{3} + 1)\left(\frac{1}{\sqrt{2}}\cos\theta - \frac{1}{\sqrt{2}}\sin\theta\right)

    2sinθ=(3+1)(cosθsinθ)\Rightarrow 2\sin\theta = (\sqrt{3} + 1)(\cos\theta - \sin\theta)

    tanθ=13θ=30\tan\theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = 30^\circ

  9. We find the largest angle which is opposite to side 7,7, if any angle can be obtuse angle then this one can.

    cosθ=32+52722.3.5=1530=12\cos \theta = \frac{3^2 + 5^2 - 7^2}{2.3.5} = -\frac{15}{30} = -\frac{1}{2}

    θ=120\Rightarrow \theta = 120^\circ

  10. Given A=45,B=75=60\angle A = 45^\circ, \angle B = 75^\circ \Rightarrow \angle = 60^\circ

    We know that asinA=bsinB=csinC=k\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k

    a=12k,b=3+122k,c=3k2a = \frac{1}{\sqrt{2}}k, b = \frac{\sqrt{3} + 1}{2\sqrt{2}}k, c = \frac{\sqrt{3}k}{2}

    a+c2=12k+32k=1+32ka + c\sqrt{2} = \frac{1}{\sqrt{\sqrt{2}}}k + \frac{\sqrt{3}}{\sqrt{2}}k = \frac{1 + \sqrt{3}}{\sqrt{2}}k

    2b=3+122b = \frac{\sqrt{3} + 1}{\sqrt{2}}

    a+2c=2b\Rightarrow a + \sqrt{2}c = 2b

  11. The diagram is given below:

    Problem 11

    ΔBCD+ΔACD=ΔABC\Delta BCD + \Delta ACD = \Delta ABC

    123CDsin30+124CDsin60=123.4\Rightarrow \frac{1}{2}3CD\sin30^\circ + \frac{1}{2}4CD\sin60^\circ = \frac{1}{2}3.4

    34CD+3CD=6CD=243+43\frac{3}{4}CD + \sqrt{3}CD = 6 \Rightarrow CD = \frac{24}{3 + 4\sqrt{3}}

  12. The smallest angle will be opposite to smallest side i.e. 44 cm. Similarly, greatest angle will be opposite to greatest side i.e. 66 cm.

    Let a=4a = 4 cm, b=5b = 5 cm and c=6c = 6 cm. Also, let opposite angles are A,BA, B and C.C.

    cosA=b2+c2a22bc=4560=34\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{45}{60} = \frac{3}{4}

    cosC=a2+b2c22ab=540=18\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{5}{40} = \frac{1}{8}

    cos2A=2cos2A1=2.9161=18=cosC\cos 2A = 2\cos^2A - 1 = 2.\frac{9}{16} - 1 = \frac{1}{8} = \cos C

  13. The diagram is given below:

    Problem 13

    A+B+C=180=10B+B+BB=15\angle A + \angle B + \angle C = 180^\circ = 10\angle B + \angle B + \angle B \Rightarrow \angle B = 15^\circ

    A=150\Rightarrow \angle A = 150^\circ

    Let AB=AC=xAB = AC = x

    cos150=x2+x2a22.x.x\therefore \cos 150^\circ = \frac{x^2 + x^2 - a^2}{2.x.x}

    3x2=2x2a2x=12+3a\Rightarrow -\sqrt{3}x^2 = 2x^2 - a^2 \Rightarrow x = \sqrt{\frac{1}{2 + \sqrt{3}}}a

  14. Let angles are A=2k,B=3k,C=7k2k+3k+7k=180k=15A = 2k, B=3k, C=7k \therefore 2k + 3k + 7k = 180^\circ \Rightarrow k = 15^\circ

    asinA=bsinB=csinC=l\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = l

    sinA=12,sinB=12,sinC=2+32\sin A = \frac{1}{2}, \sin B= \frac{1}{\sqrt{2}}, \sin C = \frac{\sqrt{2 + \sqrt{3}}}{2}

    Now the ratios of sides can be proven.

  15. Clearly, the sides are in the ratio of 7:6:57:6:5

    cosA=62+52722.6.5=15\therefore \cos A = \frac{6^2 + 5^2 - 7^2}{2.6.5} = \frac{1}{5}

    cosB=72+52622.7.5=1935\cos B = \frac{7^2 + 5^2 - 6^2}{2.7.5} = \frac{19}{35}

    cosC=72+62522.7.6=57\cos C = \frac{7^2 + 6^2 - 5^2}{2.7.6} = \frac{5}{7}

    cosA:cosB:cosC=7:19;25\therefore \cos A:\cos B:\cos C = 7:19;25

  16. tanC2=tanπ(A+B)2=cotA+B2=1tanA2tanB2tanA2+tanB2\tan \frac{C}{2} = \tan\frac{\pi - (A + B)}{2} = \cot\frac{A + B}{2} = \frac{1 - \tan\frac{A}{2}\tan\frac{B}{2}}{\tan\frac{A}{2} + \tan\frac{B}{2}}

    =156.203756+2037= \frac{1 - \frac{5}{6}.\frac{20}{37}}{\frac{5}{6} + \frac{20}{37}}

    =122222305222=122305=25= \frac{\frac{122}{222}}{\frac{305}{222}} = \frac{122}{305} = \frac{2}{5}

    sinA=2tanA21+tan2A2=2.561+2536\sin A = \frac{2\tan\frac{A}{2}}{1 + \tan^2\frac{A}{2}} = \frac{2.\frac{5}{6}}{1 + \frac{25}{36}}

    =6061= \frac{60}{61}

    sinB=2tanB21+tan2B2=2.20371+4001369\sin B = \frac{2\tan\frac{B}{2}}{1 + \tan^2\frac{B}{2}} = \frac{2.\frac{20}{37}}{1 + \frac{400}{1369}}

    =14801769= \frac{1480}{1769}

    sinB=2tanC21+tan2C2=2251+425\sin B = \frac{2\tan\frac{C}{2}}{1 + \tan^2\frac{C}{2}} = \frac{2\frac{2}{5}}{1 + \frac{4}{25}}

    =2029= \frac{20}{29}

    a+c=k(sinA+sinC)[asinA=bsinB=csinC=k]a + c = k(\sin A + \sin C)[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k]

    =k(6061+2029)=1740+12201769=29601769=2b.= k\left(\frac{60}{61} + \frac{20}{29}\right) = \frac{1740 + 1220}{1769} = \frac{2960}{1769} = 2b.

  17. It is much easier to prove this problem in reverse.

    Given, 1a+c+1b+c=3a+b+c\frac{1}{a + c} + \frac{1}{b + c} = \frac{3}{a + b + c}

    Upon solving a2+b2c2=aba^2 + b^2 - c^2 = ab

    a2+b2c22ab=12=cos60=cosC\Rightarrow \frac{a^2 + b^2 - c^2}{2ab} = \frac{1}{2} = \cos60^\circ = \cos C

  18. Let the sides be a,b,c.a, b, c. We know that Δ=12a.α\Delta = \frac{1}{2}a.\alpha because area = 12×\frac{1}{2}\times base ×\times altitude

    Δ=12αα=2Δa1α=a2Δ\Delta = \frac{1}{2}\alpha \Rightarrow \alpha = \frac{2\Delta}{a} \Rightarrow \frac{1}{\alpha} = \frac{a}{2\Delta}

    1α2+1β2+1γ2=a2+b2+c24Δ2\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{a^2 + b^2 + c^2}{4\Delta^2}

    cotA+cotB+cotCΔ=cosAΔsinA+cosBΔsinB+cosCΔsinC\frac{\cot A + \cot B + \cot C}{\Delta} = \frac{\cos A}{\Delta\sin A} + \frac{\cos B}{\Delta\sin B} + \frac{\cos C}{\Delta\sin C}

    Δ=12bcsinAsinA=2Δbc\Delta = \frac{1}{2}bc\sin A \Rightarrow \sin A = \frac{2\Delta}{bc}

    cosAΔsinA=bccosA2Δ2=b2+c2a24Δ2\therefore \frac{\cos A}{\Delta\sin A} = \frac{bc\cos A}{2\Delta^2} = \frac{b^2 + c^2 - a^2}{4\Delta^2}

    cosAΔsinA+cosBΔsinB+cosCΔsinC=a2+b2+c24Δ2\therefore \frac{\cos A}{\Delta\sin A} + \frac{\cos B}{\Delta\sin B} + \frac{\cos C}{\Delta\sin C} = \frac{a^2 + b^2 + c^2}{4\Delta^2}

    Hence proven.

  19. Given, ab=2+3=tan75=sin75cos75\frac{a}{b} = 2 + \sqrt{3} = \tan75^\circ = \frac{\sin75^\circ}{\cos75^\circ}

    sinAsinB=sin(90+15)sin(7515)\frac{\sin A}{\sin B} = \frac{\sin(90^\circ + 15^\circ)}{\sin(75^\circ - 15^\circ)}

    A=105,B=15\Rightarrow A = 105^\circ, B= 15^\circ which satisfied A+B+C=180A + B + C = 180^\circ

  20. cosC=12=a2+b2c22ab(1+32)+4c2=2(1+3)\cos C = \frac{1}{2} = \frac{a^2 + b^2 - c^2}{2ab} \Rightarrow (1 + \sqrt{3}^2) + 4 - c^2 = 2(1 + \sqrt{3})

    c2=6c=6\Rightarrow c^2 = 6 \Rightarrow c = \sqrt{6}

    cosA=b2+c2a22bc=4+6(1+3)246\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{4 + 6 - (1 + \sqrt{3})^2}{4\sqrt{6}}

    =6+236=6+2A=75= \frac{6 + 2\sqrt{3}}{\sqrt{6}} = \sqrt{6} + \sqrt{2} \Rightarrow A = 75^\circ

    Thus, B=45\angle B = 45^\circ

  21. Greatest angle will be opposite to greatest side i.e. x2+xy+y2\sqrt{x^2 + xy + y^2}

    cosθ=x2+y2x2xyy22.x.y=12θ=120\cos\theta = \frac{x^2 + y^2 - x^2 - xy - y^2}{2.x.y} = -\frac{1}{2} \Rightarrow \theta = 120^\circ

  22. Given sides are 2x+3,x2+3x+32x + 3, x^2 + 3x + 3 and x2+2x.x^2 + 2x. Since lengths of sides is a positive quantity, therefore x2+2x>0x>0x^2 + 2x > 0 \Rightarrow x > 0

    This leads to the fact that x2+3x+3x^2 + 3x + 3 will be greatest side.

    cosθ=(2x+3)2+(x2+2x)2(x2+3x+3)22(2x+3)(x2+2x)\cos\theta = \frac{(2x + 3)^2 + (x^2 + 2x)^2 - (x^2 + 3x + 3)^2}{2(2x + 3)(x^2 + 2x)}

    =4x2+12x+9+x4+4x3+4x2x49x296x36x218x4x3+14x2+12x= \frac{4x^2 + 12x + 9 + x^4 + 4x^3 + 4x^2 - x^4 - 9x^2 - 9 - 6x^3 - 6x^2 - 18x}{4x^3 + 14x^2 + 12x}

    =2x37x26x2(2x3+7x2+6x)=12=cos120= \frac{-2x^3 -7x^2 - 6x}{2(2x^3 + 7x^2 + 6x)} = -\frac{1}{2} = \cos120^\circ

  23. Given, 3a=b+c.3a = b + c. We know that s=a+b+c2s=2as = \frac{a + b + c}{2} \Rightarrow s = 2a

    cotB2cotC2=s(sb)Δ.s(sc)Δ\cot\frac{B}{2}\cot\frac{C}{2} = \frac{s(s - b)}{\Delta}.\frac{s(s - c)}{\Delta}

    =s2(sb)(sc)s(sa)(sb)(sc)=ssa=2=\frac{s^2(s - b)(s - c)}{s(s - a)(s - b)(s - c)} = \frac{s}{s - a} = 2

  24. We have to prove that asin(A2+B)=(b+c)sinA2a\sin\left(\frac{A}{2} + B\right) = (b + c)\sin\frac{A}{2}

    b+ca=k(sinB+sinC)sinA\frac{b + c}{a} = \frac{k(\sin B + \sin C)}{\sin A}

    =2sinB+c2cosBC22sinA3cosA2= \frac{2\sin\frac{B + c}{2}\cos\frac{B - C}{2}}{2\sin\frac{A}{3}\cos\frac{A}{2}}

    =cos(B2C2)sinA2= \frac{\cos\left(\frac{B}{2} - \frac{C}{2}\right)}{\sin\frac{A}{2}}

    =cos[B2{π2(A+B2)}]sinA2= \frac{\cos\left[\frac{B}{2} - \left\{\frac{\pi}{2} - \left(\frac{A + B}{2}\right)\right\}\right]}{}\sin\frac{A}{2}

    =sin(A2+B)sinA2= \frac{\sin\left(\frac{A}{2} + B\right)}{\sin\frac{A}{2}}

  25. Numerator of L.H.S. =s(sa)Δ+s(sb)Δ+s(sc)Δ= \frac{s(s - a)}{\Delta} + \frac{s(s - b)}{\Delta} + \frac{s(s - c)}{\Delta}

    =s(sa+sb+sc)Δ=s2Δ=(a+b+c)24Δ= \frac{s(s - a + s - b + s - c)}{\Delta} = \frac{s^2}{\Delta} = \frac{(a + b + c)^2}{4\Delta}

    Denominator of R.H.S. =cosAsinA+cosBsinB+cosCsinC= \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} + \frac{\cos C}{\sin C}

    =b2+c2a22bcsinA+c2+a2b22casinB+a2+b2c22absinC= \frac{b^2 + c^2 - a^2}{2bc\sin A} + \frac{c^2 + a^2 - b^2}{2ca\sin B} + \frac{a^2 + b^2 - c^2}{2ab\sin C}

    [ Δ=12bcsinA=12casinB=12absinC\Delta = \frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B = \frac{1}{2}ab\sin C ]

    =a2+b2+c24Δ= \frac{a^2 + b^2 + c^2}{4\Delta}

    L.H.S. =(a+b+c)2a2+b2+c2= \frac{(a + b + c)^2}{a^2 + b^2 + c^2}

  26. First term of L.H.S. =b2c2a2sin2A=b2c2a22sinAcosA= \frac{b^2 - c^2}{a^2}\sin2A = \frac{b^2 - c^2}{a^2}2\sin A\cos A

    =b2c2a2.2aK.b2+c2a22bc= \frac{b^2 - c^2}{a^2}.2\frac{a}{K}.\frac{b^2 + c^2 - a^2}{2bc}

    =1Kabc[(b2c2)(b2+c2a2)]=1Kabc[b4c4a2(b2c2)]= \frac{1}{Kabc}[(b^2 - c^2)(b^2 + c^2 - a^2)] = \frac{1}{Kabc}[b^4 - c^4 -a^2(b^2 - c^2)]

    Similarly, second term =1Kabc[c4a4b2(c2a2)]= \frac{1}{Kabc}[c^4 - a^4 - b^2(c^2 - a^2)]

  27. First term of L.H.S =a3cos(BC)=a2[acos(BC)]= a^3\cos(B - C) = a^2[a\cos(B - C)]

    =Ra2[2sinAcos(BC)]=Ra2[2sin(B+C)cos(BC)]=Ra2[sin2B+sin2C]= Ra^2[2\sin A\cos(B - C)] = Ra^2[2\sin(B + C)\cos(B - C)] = Ra^2[\sin2B + \sin2C]

    =Ra2[2sinBcosB+2sinCcosC]=a2[bcosB+ccosC]=Ra^2[2\sin B\cos B + 2\sin C\cos C] = a^2[b\cos B + c\cos C]

    Similarly, second term =b2[acosA+ccosC]= b^2[a\cos A + c\cos C]

    and third term =c2[acosA+bcosB]= c^2[a\cos A + b\cos B]

    Adding, ab[acosB+bcosA]+ca[ccosA+acosC]+bc[bcosC+ccosB]ab[a\cos B + b\cos A] + ca[c\cos A + a\cos C] + bc[b\cos C + c\cos B]

    =3abc== 3abc = R.H.S.

  28. L.H.S. =cos2BC2K2[sinB+sinC]2+sin2BC2K2[sinBsinC]2= \frac{\cos^2\frac{B - C}{2}}{K^2[\sin B + \sin C]^2} + \frac{\sin^2\frac{B - C}{2}}{K^2[\sin B - \sin C]^2}

    =1K2(cos2BC24sin2B+C2cos2BC2+sinBC24cos2B+C2sin2BC2)= \frac{1}{K^2}\left(\frac{\cos^2\frac{B - C}{2}}{4\sin^2\frac{B + C}{2}\cos^2\frac{B - C}{2}} + \frac{\sin^\frac{B - C}{2}}{4\cos^2\frac{B + C}{2}\sin^2\frac{B - C}{2}}\right)

    =14k2(1sin2B+C2+1cos2B+C2)= \frac{1}{4k^2}\left(\frac{1}{\sin^2\frac{B + C}{2}} + \frac{1}{\cos^2\frac{B + C}{2}}\right)

    =14K2(1cos2A2+1sin2A2)= \frac{1}{4K^2}\left(\frac{1}{\cos^2\frac{A}{2}} + \frac{1}{\sin^2\frac{A}{2}}\right)

    =1k2.14sin2A2cos2A2=1a2= \frac{1}{k^2}.\frac{1}{4\sin^2\frac{A}{2}\cos^2\frac{A}{2}} = \frac{1}{a^2}

  29. First term of L.H.S. =acosBcosC=2RsinAcosBcosC= \frac{a}{\cos B\cos C} = \frac{2R\sin A}{\cos B\cos C}

    =2Rsin(B+C)cosBcosC=2R(tanB+tanC)= \frac{2R\sin(B + C)}{\cos B\cos C} = 2R(\tan B + \tan C)

    Similarly, second term =2R(tanC+tanA)= 2R(\tan C + \tan A) and third term =2R(tanA+tanB)= 2R(\tan A + \tan B)

    L.H.S. =4R[tanA+tanB+tanC]= 4R[\tan A + \tan B + \tan C]

    =4R.tanAtanBtanC[A+B+C=πtanA+tanB+tanC=tanAtanBtanC]= 4R.\tan A\tan B\tan C[\because A + B + C = \pi \therefore \tan A + \tan B + \tan C = \tan A\tan B\tan C]

    =2.atanBtanCsecA== 2.a\tan B\tan C\sec A = R.H.S.

  30. We have to prove that (bc)cosA2=asinBC2(b - c)\cos\frac{A}{2} = a\sin\frac{B - C}{2}

    bca=sinBsinCsinA=2cosB+C2sinBC22sinA2cosA2\frac{b - c}{a} = \frac{\sin B - \sin C}{\sin A} = \frac{2\cos\frac{B + C}{2}\sin \frac{B - C}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}}

    =2sinA2sinBC22sinA2cosA2=sinBC2cosA3= \frac{2\sin\frac{A}{2}\sin\frac{B - C}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}} = \frac{\sin\frac{B - C}{2}}{\cos\frac{A}{3}}

  31. We have to prove that tan(A2+B)=c+bcbtanA2\tan\left(\frac{A}{2} + B\right) = \frac{c + b}{c - b}\tan \frac{A}{2}

    c+bcb=sinC+sinBsinCsinB\frac{c + b}{c - b} = \frac{\sin C + \sin B}{\sin C - \sin B}

    =2sinB+C2cosCB22cosB+C2sinCB2= \frac{2\sin\frac{B + C}{2}\cos\frac{C - B}{2}}{2\cos\frac{B + C}{2}\sin\frac{C - B}{2}}

    =tanB+C2tanCB2=cotA2tanπBAB2= \frac{\tan\frac{B + C}{2}}{\tan\frac{C - B}{2}} = \frac{\cot\frac{A}{2}}{\tan\frac{\pi - B - A - B}{2}}

    =tan(A2+B)tanA3= \frac{\tan\left(\frac{A}{2} + B\right)}{\tan\frac{A}{3}}

  32. We have to prove that tanAB2=aba+bcotC2\tan\frac{A - B}{2} = \frac{a - b}{a + b}\cot\frac{C}{2}

    aba+b=sinAsinBsinA+sinB=2cosA+B2sinAB22sinA+B2cosAB2\frac{a - b}{a + b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{2\cos\frac{A + B}{2}\sin\frac{A - B}{2}}{2\sin\frac{A + B}{2}\cos\frac{A - B}{2}}

    =tanAB2tanA+B2=tanAB2cotC2= \frac{\tan\frac{A - B}{2}}{\tan \frac{A + B}{2}} = \frac{\tan\frac{A - B}{2}}{\cot\frac{C}{2}}

  33. L.H.S. =(b+c)cosA+(c+a)cosB+(a+b)cosC= (b + c)\cos A + (c + a)\cos B + (a + b)\cos C

    =(acosB+bcosA)+(bcosC+ccosB)+(acosC+ccosA)= (a\cos B + b\cos A) + (b\cos C + c\cos B) + (a\cos C + c\cos A)

    =c+a+b== c + a + b = R.H.S.

  34. First term of L.H.S. =cos2Bcos2Cb+c=1k[(cosB+cosC)(cosBcosC)sinB+sinC]= \frac{\cos^2B - \cos^2C}{b + c} = \frac{1}{k}\left[\frac{(\cos B + \cos C)(\cos B - \cos C)}{\sin B + \sin C}\right]

    =1k[2cosB+C2cosBC2.2sinB+C2sinCB22sinB+C2cosBC3]= \frac{1}{k}\left[\frac{2\cos\frac{B+C}{2}\cos\frac{B - C}{2}.2\sin\frac{B + C}{2}\sin\frac{C - B}{2}}{2\sin\frac{B + C}{2}\cos\frac{B - C}{3}}\right]

    =1k[2cosB+C2sinCB2]=1k[sinCsinB]= \frac{1}{k}\left[2\cos\frac{B + C}{2}\sin\frac{C - B}{2}\right] = \frac{1}{k}[\sin C - \sin B]

    Similarly, second term =1k[sinAsinC]= \frac{1}{k}[\sin A - \sin C] and third term =1k[sinBsinA]= \frac{1}{k}[\sin B - \sin A]

    Thus, L.H.S. = R.H.S. = 0

  35. First term of L.H.S. =a3sin(BC)=Ra2.2sinAsin(BC)=Ra2.2sin(B+C)sin(BC)= a^3\sin(B - C) = Ra^2.2\sin A\sin(B - C) = Ra^2.2\sin(B + C)\sin(B - C)

    =Ra2[cos2Ccos2B]=Ra2(1sin2C1+sin2B)=R[(2RsinB)2(2RsinC)2]= Ra^2[\cos 2C - \cos 2B] = Ra^2(1 - \sin^2C - 1 + \sin^2B) = R[(2R\sin B)^2 - (2R\sin C)^2]

    =R[b2c2]= R[b^2 - c^2]

    Similarly, second term =R[c2a2]= R[c^2 - a^2] and third term =R[a2b2]= R[a^2 - b^2]

    Thus, L.H.S. =0== 0 = R.H.S.

  36. Consider first term i.e. (b+ca)tanA2(b + c - a)\tan \frac{A}{2}

    b+ca=2s2a=2(sa)b + c - a = 2s - 2a = 2(s - a)

    tanA2=(sb)(sc)s(sa)\tan\frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}}

    (b+ca)tanA2=2(sa)(sb)(sa)s\therefore (b + c - a)\tan \frac{A}{2} = 2\sqrt{\frac{(s - a)(s - b)(s - a)}{s}}

    Similarly, (c+ab)tanB2=2(sa)(sb)(sa)s=(a+bc)tanC2(c + a - b)\tan \frac{B}{2} = 2\sqrt{\frac{(s - a)(s - b)(s - a)}{s}} = (a + b - c)\tan\frac{C}{2}

  37. 1tanA2tanB2=1=(sb)(sc)s(sa).(sa)(sc)s(sb)1 - \tan\frac{A}{2}\tan\frac{B}{2} = 1 =- \sqrt{\frac{(s - b)(s - c)}{s(s - a)}.\frac{(s - a)(s - c)}{s(s - b)}}

    =1scs=cs=2ca+b+c== 1 - \frac{s - c}{s} = \frac{c}{s} = \frac{2c}{a + b + c} = R.H.S.

  38. L.H.S. =cos2Aa2cos2Bb2= \frac{\cos2A}{a^2} - \frac{\cos2B}{b^2}

    =12sin2Aa212sin2Bb2= \frac{1 - 2\sin^2A}{a^2} - \frac{1 - 2\sin^2B}{b^2}

    =12.a24r2a212.b24r2b2= \frac{1 - 2.\frac{a^2}{4r^2}}{a^2} - \frac{1 - 2.\frac{b^2}{4r^2}}{b^2}

    =1a21b2== \frac{1}{a^2} - \frac{1}{b^2} = R.H.S.

  39. We have to prove that a2(cos2Bcos2C)+b2(cos2Ccos2A)+c2(cos2Acos2B)=0a^2(\cos^2B - \cos^2C) + b^2(\cos^2C - \cos^2A) + c^2(\cos^2A - \cos^2B) = 0

    L.H.S. =a2(sin2Csin2B)+b2(sin2Asin2C)+c2(sin2Bsin2A)= a^2(\sin^2C - \sin^2B) + b^2(\sin^2A - \sin^2C) + c^2(\sin^2B - \sin^2A)

    =4R2sinA(sin2Csin2B)+4R2sin2B(sin2Asin2C)+4R2sin2C(sin2Bsin2A)= 4R^2\sin^A(\sin^2C - \sin^2B) + 4R^2\sin^2B(\sin^2A - \sin^2C) + 4R^2\sin^2C(\sin^2B - \sin^2A)

    =0== 0 = R.H.S.

  40. First term of L.H.S. =a2sin(BC)sinB+sinC= \frac{a^2\sin(B - C)}{\sin B + \sin C}