Rd Sharma Xi 2018 Solutions for Class 11 Humanities Math Chapter 10 Sine And Cosine Formulae And Their Applications are provided here with simple step-by-step explanations. These solutions for Sine And Cosine Formulae And Their Applications are extremely popular among Class 11 Humanities students for Math Sine And Cosine Formulae And Their Applications Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2018 Book of Class 11 Humanities Math Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2018 Solutions. All Rd Sharma Xi 2018 Solutions for class Class 11 Humanities Math are prepared by experts and are 100% accurate.

Page No 10.12:

Question 1:

If in ∆ABC, ∠A = 45°, ∠B = 60° and ∠C = 75°, find the ratio of its sides.

Answer:

Let asinA=bsinB=csinC=k

Then,

asin45°=bsin60°=csin75°=ka12=b32=c1221+3        sin75°=sin30°+45°=sin30°cos45°+sin45°cos30°

On multiplying by 22 , we get:
a : b : c=2 : 6 : 1+3

Hence, the ratio of the sides is 2 : 6 : 1+3.

Page No 10.12:

Question 2:

If in ∆ABC, ∠C = 105°, ∠B = 45° and a = 2, then find b.

Answer:

We know, A+B+C=π  A=π-B+CA=180°-45°+105°=30°Now,According to sine rule, asinA=bsinB .2sin30°=bsin45°       a=2, B=45°212=b124×12=bb=22

Page No 10.12:

Question 3:

In ∆ABC, if a = 18, b = 24 and c = 30 and ∠c = 90°, find sin A, sin B and sin C.

Answer:

Given,∠C = 90°, a = 18, b = 24 and c = 30

According to sine rule, asinA =bsinB=csinC.

 csinC=asinAsinA=asinCc             =18×sin90°30             =1830             =35Also, bsinB=csinCsinB=bsinCc             =24sin90°30            =2430            =45and sinC=sin90°=1

Page No 10.12:

Question 4:

In triangle ABC, prove the following:

a-ba+b=tan A-B2tan A+B2

Answer:

AssumeasinA=bsinB=csinC=k

Consider the LHS of the equation a-ba+b=tan A-B2tan A+B2.
LHS = a-ba+b         =ksinA-sinBksinA+sinB

 sinA-sinB=2sinA-B2cosA+B2, sinA+sinB=2sinA+B2cosA-B2.

 LHS=2sinA-B2cosA+B22sinA+B2cosA-B2             =tanA-B2tanA+B2=RHS   Hence proved.



Page No 10.13:

Question 5:

In triangle ABC, prove the following:

a-b cos C2=c sin A-B2

Answer:

Let asinA=bsinB=csinC=k    ...(1)

Consider the LHS of the equation a-b cos C2=c sin A-B2
LHS=a-bcosC2        =ksinA-sinBcosC2     using 1                 =k×2sinA-B2cosA+B2cosC2       =2ksinA-B2cosA+B2cosπ-A+B2           A+B+C=π       =2ksinA-B2cosA+B2sinA+B2       =ksinA-B2sinA+B            2cosA+B2sinA+B2=sinA+B       =ksinA-B2sinπ-C             A+B+C=π       =ksinCsinA-B2           =CsinA-B2=RHS                                                                   

Hence proved.

Page No 10.13:

Question 6:

In triangle ABC, prove the following:

ca-b=tanA2+tan B2tan A2-tan B2

Answer:

Let asinA=bsinB=csinC=k     ...(1)

We need to prove:
ca-b=tanA2+tan B2tan A2-tan B2

Consider

LHS=ca-b        =ksinCksinA-sinB       using 1        =2sinC2cosC22sinA-B2cosA+B2        =sinπ-A+B2cosC2sinA-B2cosA+B2       A+B+C=π        =cosC2cosA+B2sinA-B2cosA+B2      =cosC2sinA-B2       ...2

RHS=tanA2+tanB2tanA2-tanB2=sinA2cosA2+sinB2cosB2sinA2cosA2-sinB2cosB2=sinA2cosB2+sinB2cosA2sinA2cosB2-sinB2cosA2=sinA+B2sinA-B2=sinπ-C2sinA-B2=cosC2sinA-B2        =LHS    from 2Hence proved.

Page No 10.13:

Question 7:

In triangle ABC, prove the following:

ca+b=1-tan A2 tan B21+tan A2 tan B2

Answer:

Let asinA=bsinB=csinC=k    ...(1)

We need to prove:
ca+b=1-tan A2 tan B21+tan A2 tan B2

Consider

LHS=ca+b        =ksinCksinA+sinB      using1        =2sinC2cosC22sinA+B2cosA-B2       =sinC2cosπ-A+B2sinA+B2cosA-B2        A+B+C = π      =sinC2sinA+B2sinA+B2cosA-B2      =sinC2cosA-B2    ... 2RHS=1-tanA2tanB21+tanA2tanB2       =1-sinA2cosA2sinB2cosB21+sinA2cosA2sinB2cosB2      =cosA2cosB2-sinA2sinB2cosA2cosB2+sinA2sinB2     =cosA+B2cosA-B2    =cosπ-C2cosA-B2         A+B+C = π     =sinC2cosA-B2=LHS       from2Hence proved.

Page No 10.13:

Question 8:

In triangle ABC, prove the following:

a+bc=cos A-B2sin C2

Answer:

Let asinA=bsinB=csinC=k    ...(1)

Then,
Consider the LHS of the equation a+bc=cos A-B2sin C2.
LHS=a+bc        =ksinA+ksinBksinC       using 1        =2sinA+B2cosA-B22sinC2cosC2       =sinA+B2cosA-B2sinC2cosπ-A+B2            A+B+C=π      =sinA+B2cosA-B2sinC2sinA+B2      =cosA-B2sinC2=RHS   Hence proved.     

Page No 10.13:

Question 9:

In any triangle ABC, prove the following:

sin B-C2=b-ca cosA2

Answer:

Let asinA=bsinB=csinC=k

Then,
Consider the RHS of the equation sin B-C2=b-ca cosA2
RHS=b-cacosA2         =ksinB-sinCksinAcosπ-B+C2       A+B+C=π              =2sinB-C2cosB+C2sinA         =sinB-C22cosB+C2sinAsinB+C2        =sinB-C2sinB+CsinA       =sinB-C2sinπ-AsinA       =sinAsinB-C2sinA       =sinB-C2=LHSHence proved.

Page No 10.13:

Question 10:

In triangle ABC, prove the following:

a2-c2b2=sin A-Csin A+C

Answer:

Let asinA=b sinB=csinC=k
Then,
Consider the LHS of the equation a2-c2b2=sin A-Csin A+C.
LHS=ksinA2-ksinC2ksinB2        =k2sin2A-sin2Ck2sin2B        =sinA+CsinA-Csin2B            sin2A-sin2C=sinA+CsinA-C        =sinA+CsinA-CSin2π-A+C     A+B+C=π        =sinA+CsinA-Csin2A+C         =sinA-CsinA+C=RHS        Hence proved.

Page No 10.13:

Question 11:

In triangle ABC, prove the following:
b sin B-c sin C=a sin B-C

Answer:

Let asinA=b sinB=csinC=k
Then,

Consider the LHS of he equation b sin B-c sin C=a sin B-C.
LHS=ksinBsinB-ksinCsinC 
       =ksin2B-sin2C=ksinB+CsinB-C              sin2B-sin2C=sinB+CsinB-C=ksinπ-AsinB-C              A+B+C=π=ksinAsinB-C                         a=ksinA=asinB-C=RHSHence proved.

Page No 10.13:

Question 12:

In triangle ABC, prove the following:
a2 sin B-C=b2-c2 sin A

Answer:

Let asinA=b sinB=csinC=k
Then,

Consider the RHS of the equation a2 sin B-C=b2-c2 sin A.

    RHS=k2sinAsin2B-sin2C           =k2sinAsinB+CsinB-C       sin2B-sin2C=sinB+CsinB-C        =k2sinAsinπ-AsinB-C      A+B+C=π        =k2sinAsinAsinB-C        =k2sin2AsinB-C        =a2sinB-C =LHS                     a=ksinAHence proved.

Page No 10.13:

Question 13:

In triangle ABC, prove the following:

sin A-sin Bsin A+sin B=a+b-2aba-b

Answer:

Consider the LHS of the equation sin A-sin Bsin A+sin B=a+b-2aba-b.

  LHS=sinA-sinBsinA+sinB        =sinA-sinBsinA+sinB×sinA-sinBsinA-sinB        =sinA+sinB-2×sinAsinBsinA-sinB

Let asinA=b sinB=csinC=k

Then,
LHS=ak+bk-2×akbkak-bk          =1ka+b-2×ab1ka-b       =a+b-2aba-b=RHSHence proved.

Page No 10.13:

Question 14:

In triangle ABC, prove the following:
a sin B-sin C + sin C-sin A+c sin A-sin B=0

Answer:

Let asinA=bsinB=csinC=k

Then,

Consider the LHS of the equation a sin B-sin C + sin C-sin A+c sin A-sin B=0.

 LHS =asinB-sinC+bsinC-sinA+csinA-sinB         =ksinAsinB-sinC+ksinBsinC-sinA+ksinCsinA-sinB             =ksinAsinB-ksinAsinC+ksinBsinC-ksinBsinA+ksinCsinA-ksinCsinB         =0=RHSHence proved.

Page No 10.13:

Question 15:

In triangle ABC, prove the following:

a2 sin B-Csin A+b2 sin C-Asin B+c2 sin A-Bsin C=0

Answer:

Let asinA=bsinB=csinC=k

Then,

Consider the LHS of the equation a2 sin B-Csin A+b2 sin C-Asin B+c2 sin A-Bsin C=0.

LHS=a2sinB-CsinA+b2sinC-AsinB+c2sinA-BsinC       =k2sin2AsinB-CsinA+k2sin2BsinC-AsinB+k2sin2CsinA-BsinC               =k2sinAsinB-C+k2sinBsinC-A+k2sinCsinA-B           =k2sinAsinBcosC-sinCcosB+sinBsinCcosA-sinAcosC+sinCsinAcosB-sinBcosA          =k2sinAsinBcosC-sinAsinCcosB+sinBsinCcosA-sinAsinBcosC+sinAsinCcosB-sinCsinBcosA       =0=RHSHence proved.

Page No 10.13:

Question 16:

In triangle ABC, prove the following:
a2 cos2 B-cos2 C +b2 cos2 C-cos2 A+c2 cos2 A-cos2 B=0

Answer:

Let asinA=bsinB=csinC=k

Then,

Consider the LHS of the equation a2 cos2 B-cos2 C +b2 cos2 C-cos2 A+c2 cos2 A-cos2 B=0.

LHS=a2cos2B-cos2C+b2cos2C-cos2A+c2cos2A-cos2B        =k2sin2A1-sin2B-1+sin2C+k2sin2B1-sin2C-1+sin2A+k2sin2C1-sin2A-1+sin2B            =k2sin2Asin2C-sin2B+k2sin2Bsin2A-sin2C+k2sin2Csin2B-sin2A        =k2sin2Asin2C-sin2Asin2B+sin2Asin2B-sin2Bsin2C+sin2Csin2B-sin2Csin2A        =k2×0=0=RHSHence proved.

Page No 10.13:

Question 17:

In triangle ABC, prove the following:
b cos B + c cos C = a cos B-C

Answer:

Let asinA=bsinB=csinC=k

Then,

Consider the LHS of the equation b cos B + c cos C = a cos B-C.

  LHS=bcosB+ccosC        =ksinBcosB+sinCcosC                       =k22sinBcosB+2sinCcosC        =k2sin2B+sin2C      ...1 RHS= acosB-C         =ksinAcosB-C                                  =k22sinAcosB-C         =k2sinA+B-C+sinA-B+C           2sinAcosB=sinA+B+sinA-B         =k2sinπ-C-C+sinπ-B-B            sinπ-A=sinA, A+B+C=π         =k2sin2C+sin2B         =k2sin2B+sin2C=LHS       from1Hence proved.

Page No 10.13:

Question 18:

In triangle ABC, prove the following:

cos 2Aa2-cos 2Bb2-1a2-1b2

Answer:

Let asinA=bsinB=csinC=k

Then,

Consider the LHS of the equation cos 2Aa2-cos 2Bb2-1a2-1b2.

LHS=cos2Aa2-cos2Bb2        =1-2sin2Aa2-1-2sin2Bb2           =1-2a2k2a2-1-2b2k2b2               =k2-2a2k2a2-k2-2b2k2b2     =k2b2-2a2b2-k2a2+2a2b2a2b2    =k2b2-a2k2a2b2    =1a2-1b2=RHSHence proved.

Page No 10.13:

Question 19:

In triangle ABC, prove the following:

cos2 B-cos2 Cb+c+cos2 C-cos2 Ac+a+cos2 A-cos2 Ba+b=0

Answer:

Let asinA=bsinB=csinC=k

Then,

Consider the LHS of the equation cos2 B-cos2 Cb+c+cos2 C-cos2 Ac+a+cos2 A-cos2 Ba+b=0.

LHS=cos2B-cos2Cb+c+cos2C-cos2Ac+a+cos2A-cos2Ba+bNow,cos2B-cos2Cb+c=cos2B-cos2CksinB+sinC                             =cosB+cosCcosB-cosCksinB+sinC         cos2B-cos2C = cosB+cosCcosB-cosC                             =2cosB+C2cosB-C2-2sinB+C2sinB-C22ksinB+C2sinB-C2                              =-2cosB+C2sinB-C2k=-sinB-sinCk=sinC-sinBk 
Also,

cos2C-cos2Ac+a=cos2C-cos2AksinC+sinA                              =cosC+cosAcosC-cosAksinC+sinA                              =2cosC+A2cosC-A2-2sinC+A2sinC-A22ksinC+sinAk                              =-2cosC+A2cosC-A2k=-sinC-sinAk=sinA-sinCk
Similarly,
cos2A-cos2Ba+b=sinB-sinAk

Thus,

LHS=sinA-sinCk+sinB-sinAk+sinC-sinBk        =0 = RHS 


Hence, in any triangle ABC, cos2 B-cos2 Cb+c+cos2 C-cos2 Ac+a+cos2 A-cos2 Ba+b=0.

Page No 10.13:

Question 20:

In ∆ABC, prove that:

a sinA2 sin B-C2+b sin B2 sin C-A2+c sin C2 sin A-B2=0 .

Answer:

Consider
asinA2sinB-C2+bsinB2sinC-A2+csinC2sinA-B2
=ksinAsinA2sinB-C2+sinBsinB2sinC-A2+sinCsinC2sinA-B2=ksinπ-B+CsinA2sinB-C2+sinπ-C+A sinB2sinC-A2+sinπ-A+BsinC2sinA-B2       A+B+C=π=ksinB+CsinA2sinB-C2+sinA+CsinB2sinC-A2+sinA+BsinC2sinA-B2=k2sinB+C2cosB-C2sinA2sinB-C2+2sinA+C2cosC-A2sinB2sinC-A2+2sinA+B2cosA-B2sinC2sinA-B2=2ksinB+C2sinA2sinA2sinB-C2+sinA+C2sinB2sinB2sinC-A2+sinA+B2sinC2sinC2sinA-B2=2ksinB+C2sinB-C2sin2A2+sinA+C2sinC-A2sin2B2+sinA+B2sinA-B2sin2C2=2ksin2A2sin2B2-sin2C2+2ksin2B2sin2C2-sin2A2+2ksin2C2sin2A2-sin2B2=2ksin2A2sin2B2-sin2A2sin2C2+sin2B2sin2C2-sin2A2sin2B2+sin2A2sin2C2-sin2C2sin2B2=k0=0

Hence proved.

Page No 10.13:

Question 21:

In ∆ABC, prove that:

b sec B+c sec Ctan B+tan C=c sec C+a sec Atan C+tan A=a sec A+b sec Btan A+tan B

Answer:

Let ABC be any triangle.

Suppose asinA=bsinB=csinC=k

Now,
bsecB+csecCtanB+tanC=bcosB+ccosCsinBcosB+sinCcosC                              =bcosC+ccosBsinBcosC+sinCcosB                             =ksinBcosC+ksinCcosBsinBcosC+sinCcosB                                = ksinBcosC+sinCcosBsinBcosC+sinCcosB=k    ...1
Also,
csecC+asecAtanC+tanA=ccosC+acosAsinCcosC+sinAcosA                             =ccosA+acosCsinCcosA+sinAcosC                             =ksinCcosA+ksinAcosCsinCcosA+sinAcosC                                   =ksinCcosA+sinAcosCsinCcosA+sinAcosC =k     ...2andasecA+bsecBtanA+tanB=acosA+bcosBsinAcosA+sinBcosB                             =acosB+bcosAsinAcosB+sinBcosA                             =ksinAcosB+sinBcosAsinAcosB+sinBcosA =k          ...3               

From (1), (2) and (3), we get:

b sec B+c sec Ctan B+tan C=c sec C+a sec Atan C+tan A=a sec A+b sec Btan A+tan B

Hence proved.

Page No 10.13:

Question 22:

In triangle ABC, prove the following:
a cos A+bcos B+c cos C=2b sin A sin C=2 c sin A sin B 

Answer:

LetasinA=bsinB=csinC=k    ...1Consider the LHS of the equation a cos A+bcos B+c cos C.acosA+bcosB+ccosC=ksinAcosA+sinBcosB+sinCcosC                                                   =k22sinAcosA+2sinAcosA+2sinCcosC                                           =k2sin2A+sin2B+sin2C                                           =k22sinA+BcosA-B+2sinCcosC                                           =k22sinπ-CcosA-B+2sinCcosC                                            =k22sinCcosA-B+2sinCcosC                                            =2ksinC2cosA-B+cosC      
                                   =ksinCcosA-B+cosπ-A+B=ksinCcosA-B-cosA+B=ksinC2sinAsinB=2ksinAsinBsinC    ...(1)



Now,on putting ksinC=C in equation (1), we get:2csinAsinBand on putting ksinB=b in equation (1), we get:2bsinAsinC

So, from (1), we have
a cos A+bcos B+c cos C=2b sin A sin C=2 c sin A sin B .

Hence proved.

Page No 10.13:

Question 23:

a cos B cos C+cos A=b cos C cos A+cos B=c cos A cos B+cos C

Answer:

Suppose asinA=bsinB=csinC=k

Consider:
acosBcosC+cosA=ksinAcosBcosC+cosA        =ksinAcosBcosC+cosAsinA=k12cosCsinA+B+sinA-B+sinAcosA=k12sinA+BcosC+sinA-BcosC+sinAcosA=k1212sinA+B+C+sinA+B-C+sinA-B+C+sinA-B-C+sinAcosA=k14sinπ+sinπ-2C+sinπ-2B-sinπ-2A+sin2A2           A+B+C=π=k4sin2C+sin2B+sin2A  ....1and bcosAcosC+cosB=ksinBcosAcosC+sinBcosB=k12cosAsinB+C+sinB-C+sin2B2=k12sinB+CcosA+sinB-CcosA+sin2B2=k14sinB+C+A+sinB+C-A+sinB-C+A+sinB-C-A+sin2B2=k4sinπ+sinπ-2A+sinπ-2C-sinπ-2B+sin2B2               A+B+C=π=k4sin2A+sin2C+sin2B   ...2Similarly,ccosAcosB+cosC=k4sin2A+sin2B+sin2C     ...3


From (1), (2) and (3), we get:

a cos B cos C+cos A=b cos C cos A+cos B=c cos A cos B+cos C

Hence proved.

Page No 10.13:

Question 24:

In ∆ABC, prove that a cos C-cos B=2 b-c cos2A2.

Answer:

ConsideracosC-cosB=ksinAcosC-sinAcosB         asinA=bsinB=csinC=k=k22sinAcosC-2sinAcosB=k2sinA+C+sinA-C-sinA+B-sinA-B=k2sinπ-B+sinA-C-sinπ-C-sinA-B    A+B+C=π=k2sinB-sinC+sinA-C-sinA-B=k22sinB-C2cosB+C2+2sinA-C-A+B2cosA-C+A-B2=ksinB-C2cosπ2-A2+cos2A-π-A2=ksinB-C2sinA2+sin3A2=ksinB-C22sinA2+3A22cos3A2-A22=2ksinB-C2sinAcosA2=4ksinB-C2sinA2cos2A2   ...1Now,Consider2b-ccos2A2=2ksinB-sinCcos2A2=2k2sinB-C2cosB+C2cos2A2=4ksinB-C2cosπ2-A2cos2A2=4ksinB-C2sinA2cos2A2    ...2   From 1 & 2, we geta cos C-cos B=2 b-c cos2A2Hence proved.

Page No 10.13:

Question 25:

In ∆ABC, prove that if θ be any angle, then b cosθ = c cos (A − θ) + a cos (C + θ).

Answer:

Suppose  asinA=bsinB=csinC=k.     ...(1)

Consider the RHS of the equation b cosθ = c cos (A − θ) + a cos (C + θ).

RHS=ccosA-θ+acosC+θ         =ksinCcosA-θ+ksinAcosC+θ       from 1           =k22sinCcosA-θ+2sinAcosC+θ         =k2sinA+C-θ+sinC+θ-A+sinA+C+θ+sinA-C-θ             =k2sinπ-B-θ+sinC+θ-A+sinπ-B+θ-sinC+θ-A                       A+B+C=π         =k2sinB+θ+sinB-θ         =k2sinBcosθ+sinθcosB+sinBcosθ-sinθcosB         =k22sinBcosθ         =ksinBcosθ         =bcosθ=LHSHence proved.

Page No 10.13:

Question 26:

In ∆ABC, if sin2 A + sin2 B = sin2 C. show that the triangle is right-angled.

Answer:

In ∆ ABC,
Given, sin2A+sin2B=sin2C   ......1

Suppose asinA=bsinB=csinC=k .

sinA=ak, sinB=bk, sinC=ck

On putting these values in equation (1), we get:

a2k2+b2k2=c2k2a2+b2=c2 

Thus, ∆ ABC is right-angled.



Page No 10.14:

Question 27:

In ∆ABC, if a2, b2 and c2 are in A.P., prove that cot A, cot B and cot C are also in A.P.

Answer:

Let sinAa=sinBb=sinCc=k

Then,
sinA=ka, sinB=kb, sinC=kc
 a2, b2 and c2 are in A.P.
2b2=a2+c2 2a2+c2-b2=22b2-b2=2b2=b2+b2+c2-a2-c2+a22a2+c2-b2=b2+c2-a2+a2+b2-c22a2+c2-b22abc=b2+c2-a22abc+a2+b2-c22abc2cosBkb=cosAka+cosCkc2cosBsinB=cosAsinA+cosCsinC2cotB=cotA+cotCcotA,cotB and cotC are in AP.

Page No 10.14:

Question 28:

The upper part of a tree broken by the wind makes an angle of 30° with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. Using sine rule, find the height of the tree.

Answer:

Suppose BD be the tree and the upper part of the tree is broken over by the wind at point A.


The total height of the tree is x+y.In ABC,C=30° and B=90°.A=60°.So, on using sine rule, we get:ABsin30°=BCsin60°=ACsin90°xsin30°=15sin60°=ysin90°   So, xsin30°=15sin60°x12=1532x=153=53Also,15sin60°=ysin90°1532=yy=303=103So, the height of the tree is x+y=53+103 m                                                        =153m

Page No 10.14:

Question 29:

At the foot of a mountain, the elevation of it summit is 45°; after ascending 1000 m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain.

Answer:


Suppose, AB is a mountain of height t + x.

In DFC,sin30°=x1000 x=1000×12=500 mand tan30°=xyy=xtan30°=5003In ABC,tan45°=t+xy+zt+x=y+z   ...1In ADE,tan60°=tzt=3z      ...2From 1 and 2, we have 3z+x=y+zz3-1=5003-1z=500 m t=3z=5003

Hence, height of the mountain = t+x=5003+500=5003+1 m .

Page No 10.14:

Question 30:

A person observes the angle of elevation of the peak of a hill from a station to be α. He walks c metres along a slope inclined at an angle β and finds the angle of elevation of the peak of the hill to be ϒ. Show that the height of the peak above the ground is c sin α sin γ-βsin γ-α.

Answer:

Suppose, AB is a peak whose height above the ground is t+x.


In DFC,sinβ=xc x=csinβand tanβ=xyy=xtanβ=csinβsinβ×cosβ=ccosβ    ...1In ADE,tanγ=tzz=t cotγ            ...2In ABC,tanα=t+xy+zt+x=ccosβtanα+tcotγtanα      from 1 and 2t-tcotγtanα=ccosβtanα-csinβ    x=csinβt1-sinαcosγcosαsinγ=ccosβsinα-cosαsinβcosαtsinγcosα-sinαcosγcosαsinγ=csinα-βcosαtsinγ-βcosαsinγ=csinα-βcosαt=csinγsinα-βsinγ-β     ...3Now, AB=t+x=csinγsinα-βsinγ-β+csinβ    using 3                 =csinγsinα-βsinγ-β+sinβ                 =csinγsinα-β+sinβsinγ-βsinγ-β                 =csinγsinαcosβ-sinβsinγcosα+sinβsinγcosα-sinβcosγsinαsinγ-β                 =csinγsinαcosβ-sinβcosγsinαsinγ-β                 =csinαsinγ-βsinγ-β                 =csinαsinγ-βsinγ-βHence proved.

Page No 10.14:

Question 31:

If the sides a, b and c of ∆ABC are in H.P., prove that sin2A2, sin2B2 and sin2C2 are in H.P.

Answer:

 sin2A2, sin2B2 and sin2C2 is a H.P.1sin2A2, 1sin2B2 and 1sin2C2 is an A.P.1sin2B2-1sin2A2=1sin2C2-1sin2B2sin2A2-sin2B2sin2A2sin2B2=sin2B2-sin2C2sin2B2sin2C2sinA+B2sinA-B2sin2A2=sinB+C2sinB-C2sin2C2cosC2sinA-B2sin2A2=cosA2sinB-C2sin2C2         As, A+B+C=πsin2C2cosC2sinA-B2=sin2A2cosA2sinB-C22sinC2sinC2cosC2sinA-B2=2sinA2sinA2cosA2sinB-C2sinC2sinC sinA-B2=sinA2sinAsinB-C2                  sin2θ=2sinθcosθsinC cosA+B2sinA-B2=sinA cosB+C2 sinB-C2           As, A+B+C=πsinCsinA-sinB2=sinAsinB-sinC2                   sinC-sinD=2cosC+D2sinC-D2sinCsinA-sinB=sinAsinB-sinCckak-bk=akbk-ck         sinAa=sinBb=sinCc=k sayca-cb=ab-ac2ac=ab+bc2b=1c+1a         multiplying both the sides by abca, b, c are in H.P.



Page No 10.25:

Question 1:

In ABC, if a=5, b=6 and C=60°, show that its area is 1532 sq. units.

Answer:

Given: a=5, b=6, c=60°Area of a triangle=12absinC                              =12×5×6×sin60°=15×32sq. unitsHence proved.

Page No 10.25:

Question 2:

In ABC, if a=2, b=3 and c=5, show that its area is 126 sq. units.

Answer:

Given: a=2, b=3, c=5 cosC=a2+b2-c22abcosC=2+3-52×6=0cosC=0cosC=cos90°C=90°Thus, sinC=sin90°=1Hence,  Area of ABC=12absinC=126×1=62sq.units .

Page No 10.25:

Question 3:

The sides of a triangle are a = 4, b = 6 and c = 8. Show that 8 cos A+16 cos B+4 cos C=17.

Answer:

Given: a=4, b=6 and c=8.
Then,
cosB=a2+c2-b22ac=16+64-362×4×8=1116cosA=b2+c2-a22bc=36+64-162×6×8=78cosC=b2+a2-c22ab=16+36-642×4×6=-14Now, 8cosA+16cosB+4cosC=8×78+16×1116-4×148cosA+16cosB+4cosC=7+11-1=17

Hence proved.

Page No 10.25:

Question 4:

In ∆ ABC, if a = 18, b = 24 and c = 30, find cos A, cos B and cos C.

Answer:

GIven: a=18, b=24 and c=30 .cosA=b2+c2-a22bc=576+900-3242×24×30=11521140=45cosB=a2+c2-b22ac=324+900-5762×18×30=6481080=35cosC=a2+b2-c22ab=576+324-9002×24×18=0

Hence, cosA=45, cosB=35, cosC=0

Page No 10.25:

Question 5:

In ∆ABC, prove the following:
b c cos A-a cos C=c2-a2

Answer:

Let ABC be any triangle.

Consider bccosA-acosC=bccosA-abcosC                                 =bcb2+c2-a22bc-aba2+b2-c22ab                                     =b2+c2-a2-a2-b2+c22                                 =2c2-a22                                 =c2-a2  

Hence proved.

Page No 10.25:

Question 6:

In ∆ABC, prove the following:
c a cos B-b cos A=a2-b2

Answer:

Consider

cacosB-bcosA=cacosB-cbcosA                                =caa2+c2-b22ac-cbb2+c2-a22bc                                =a2+c2-b2-b2-c2+a22                                =2a2-b22                                =a2-b2

Hence proved.

Page No 10.25:

Question 7:

In ∆ABC, prove the following:
2 bc cos A+ca cos B+ab cos C=a2+b2+c2

Answer:

LHS = 2 bc cos A+ca cos B+ab cos C

On using the cosine law, we get:
LHS=2bcb2+c2-a22bc+caa2+c2-b22ac+aba2+b2-c22ab
     =b2+c2-a2+a2+c2-b2+a2+b2-c2=a2+b2+c2=RHS

Hence proved.

Page No 10.25:

Question 8:

In ∆ABC, prove the following:
c2-a2+b2 tan A=a2-b2+c2 tan B=b2-c2+a2 tan C

Answer:

We know that cosA=b2+c2-a22bc,sinAa=sinBb=sinCc=k

So,c2-a2+b2tanA=c2-a2+b2sinAcosA                               =c2-a2+b2sinA2bcb2+c2-a2                               =2bcsinA                               =2kabc    ...1a2-b2+c2tanB=a2-b2+c2sinBcosB                               =a2-b2+c2sinB2aca2+c2-b2                               =2acsinB                               =2kabc        ...2b2-a2+c2tanC=b2-a2+c2sinCcosC                               =b2-c2+a2sinC2aba2+b2-c2                               =2absinC                               =2kabc                   ....3

From (1), (2) and (3), we get:
c2-a2+b2 tan A=a2-b2+c2 tan B=b2-c2+a2 tan C

Page No 10.25:

Question 9:

In ∆ABC, prove the following:
c-b cos Ab-c cos A=cos Bcos C

Answer:

Let ABC be any triangle.

LHS=c-bcosAb-ccosA        =acosB+bcosA-bcosAacosC+ccosA-ccosA           Using projection formula:c=acosB+bcosA, b=acosC+ccosA        =acosBacosC        =cosBcosC=RHS

Hence proved.

Page No 10.25:

Question 10:

In ∆ABC, prove that:
 a cos B+cos C-1+b cos C+cos A-1+ccos A+cos B-1=0

Answer:

Consider the LHS of the given equation.LHS=acosB+cosC-1+bcosC+cosA-1+ccosA+cosB-1        =acosB+bcosC+acosC+bcosA+ccosA+ccosB-a+b+c        =acosB+bcosA+bcosC+ccosB+acosC+ccosA-a+b+c            =c+a+b-a+b+c        ..Using projection formula : a=bcosC+ccosB, b=acosC+ccosA, c=acosB+bcosA       =0=RHS    Hence proved.

Page No 10.25:

Question 11:

a cos A + b cos B + c cos C = 2b sin A sin C

Answer:

By sine rule, we know thatasin A=bsin B=csin C=k saya=k sin A, b=k sin B, c=k sin CNow,LHS=a cos A+b cos B+c cos C        =k sin A cos A+k sin B cos B+k sin C cos C        =k2 2 sin A cos A+2 sin B cos B+2 sin C cos C        =k2 sin 2A+sin 2B+2 sin C cos C        =k2 2 sin 2A+2B2cos2A-2B2+2 sin C cos C        =k2 2 sin A+B cos A-B+2 sin C cos C        =k2 2 sin π-C cos A-B+2 sin C cos C                A+B+C=π        =k2 2 sin C cos A-B+2 sin C cos C        =k2× 2 sin Ccos A-B+cos C        =k sin C2 cos A-B+C2cos A-B-C2        =k sin C2 cos π-B-B2cos B+C-A2                A+B+C=π        =k sin C2 cos π-2B2cos π-2A2                A+B+C=π        =k sin C2 cos π2-Bcos π2-A        =2k sin Csin B sin A        =2 k sin B sin A sin C        =2b sin A sin C        =RHS LHS=RHS

Hence, a cos A + b cos B + c cos C = 2b sin A sin C.

Page No 10.25:

Question 12:

In ∆ABC, prove the following:
a2=b+c2-4 bc cos2A2

Answer:

RHS=b+c2-4bccos2A2         =b2+c2+2bc-4bc1+cosA2         =b2+c2+2bc-2bc1+cosA         =b2+c2+2bc1-1-cosA         =b2+c2-2bccosA         =b2+c2-2bcb2+c2-a22bc     cosA=b2+c2-a22bc        =b2+c2-b2-c2+a2        =a2=LHS

Hence proved.

Page No 10.25:

Question 13:

In ∆ABC, prove the following:
4bc cos2A2+ca cos2B2+ab cos2 C2=a+b+c2

Answer:

LHS=4bc cos2A2+ca cos2B2+ab cos2 C2=4bc1+cosA2+ca1+cosB2+ab1+cosC2=2bc+2bccosA+2ca+2cacosB+2ab+2abcosC=2ab+bc+ca+2bcb2+c2-a22bc+2cac2+a2-b22ca+2aba2+b2-c22ab

=2ab+bc+ac+b2+c2-a2+c2+a2-b2+a2+b2-c2=a2+b2+c2+2ab+2bc+2ac=a+b+c2=RHS

Hence proved.

Page No 10.25:

Question 14:

In ∆ABC, prove the following:
sin3 A cos B-C+sin3 B cos C-A+sin3 C cos A-B=3 sin A sin B sin C

Answer:

Let asinA=bsinB=csinC=k    ...1LHS=sin3AcosB-C+sin3BcosC-A+sin3CcosA-B        =sin2AsinAcosB-C+sin2BsinBcosC-A+sin2AsinAcosA-B        =a2k2sinAcosB-C+b2k2sinBcosC-A+c2k2sinAcosA-B       from 1        =12k2a22sinAcosB-C+b22sinBcosC-A+c22sinAcosA-B        =12k2a22sinπ-B+CcosB-C+b22sinπ-A+CcosC-A+c22sinπ-B+CcosA-B        =12k2a22sinB+CcosB-C+b22sinC+AcosC-A+c22sinA+BcosA-B        =12k2a2sin2B+sin2C+b2sin2C+sin2A+c2sin2A+sin2B        =12k22a2sinBcosB+sinCcosC+2b2sinCcosC+sinAcosA+2c2sinAcosA+sinBcosB        =12k32a2ksinBcosB+ksinCcosC+2b2ksinCcosC+ksinAcosA+2c2ksinAcosA+ksinBcosB        =1k3a2bcosB+ccosC+2b2ccosC+acosA+2c2acosA+acosB        =1k3abacosB+bcosA+bcbcosC+ccosB+acacosC+ccosA        =1k3abc+bca+acb        =3abc×1k3        =3sinAsinBsinC×1k3×k3        =3sinAsinBsinC        =RHSHence proved.

Page No 10.25:

Question 15:

In  ABC, b+c12=c+a13=a+b15. Prove that cos A2=cos B7=cos C11.

Answer:

Let b+c12=c+a13=a+b15=kb+c=12k, c+a=13k, a+b=15kb+c+c+a+a+b=12k+13k+15k2a+b+c=40ka+c+b=20ka+12k=20k    b+c=12ka=8kAlso, c+a=13kc=13k-a=13k-8k=5kand a+b=15kb=15k-a=15k-8k=7kNow,cosA=b2+c2-a22bc=k249+25-64k22×35=17cosB=a2+c2-b22ac=k264+25-492×40k2=12cosC=a2+b2-c22ab=k264+49-252×56k2=1114  cosA:cosB:cosC=17:12:1114=2:7:11 cosA2=cosB7=cosC11

Hence proved.

Page No 10.25:

Question 16:

In ABC, if B=60°, prove that a+b+c a-b+c=3ca.

Answer:

Given, B=60°We know that, cosB=a2+c2-b22accos60°=a2+c2-b22ac12  =a2+c2-b22ac     cos60°=12ac=a2+c2-b23ac-2ac=a2+c2-b23ac=a2+c2-b2+2ac3ac=a2+c2+2ac-b23ac=a+c2-b23ac=a+c+ba+c-b3ac=a+b+ca-b+c

Hence proved.

Page No 10.25:

Question 17:

If in ABC, cos2 A+cos2 B+cos2 C=1, prove that the triangle is right-angled.

Answer:

Let ABC be any triangle.
In ABC,
cos2 A+cos2 B+cos2 C=1cos2 A+cos2 B+cos2 π-B+A=1      A+B+C=πcos2 A+cos2 B+cos2 B+A=1cos2 A+cos2 B=1-cos2 B+Acos2 A+cos2 B=sin2 B+Acos2 A+cos2 B=sinAcosB+cosAsinB2cos2 A+cos2 B=sin2Acos2B+cos2Asin2B+2sinAsinBcosAcosBcos2 A1-sin2B+cos2 B1-sin2A=2sinAsinBcosAcosB2cos2 Acos2B=2sinAsinBcosAcosBcos AcosB=sinAsinBcos AcosB-sinAsinB=0cos A+B=0cos A+B=cos90° A+B=90°       C=90°     A+B+C=180°

Hence, ABC is right angled.



Page No 10.26:

Question 18:

In ABC if cos C=sin A2 sin B, prove that the triangle is isosceles.

Answer:

Let ABC be any triangle.

Suppose sinAa=sinBb=sinCc=k
If  cosC=sinA2sinB, then

b2+a2-c22ab=ka2kb    cosC=b2+a2-c22ab

b2+a2-c2=a2b2-c2=0b-cb+c=0b-c=0b=c    b, c > 0

Thus, the lengths of two sides of the ABC are equal.

Hence, ABC is an isosceles triangle.

Page No 10.26:

Question 19:

Two ships leave a port at the same time. One goes 24 km/hr in the direction N 38° E and other travels 32 km/hr in the direction S 52° E. Find the distance between the ships at the end of 3 hrs.

Answer:


After three hours, let the ships be at P and Q respectively.Then,OP=24×3=72 km and OQ=32×3=96 kmFrom figure, we havePOQ=180°-NOP-SOQ             =180°-38°-52°             =90°Now,Since OPQ is a right angled trianglePQ2=OP2+OQ2PQ2=722+962PQ2=5184+9216PQ2=14400PQ=14400=120 kmHence, the distance between the ships after 3 hours is 120 km.

Page No 10.26:

Question 1:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

Find the area of the triangle ∆ABC in which a = 1, b = 2 and C=60°.

Answer:

In ∆ABC, a = 1, b = 2 and C=60°.

∴ Area of the ∆ABC

=12absinC=12×1×2×sin60°=12×2×32=32 square units

Page No 10.26:

Question 2:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In a ∆ABC, if b = 3, c = 1 and A=30°, find a.

Answer:

In ∆ABC, b = 3, c = 1 and A=30°.

Using cosine formula, we have

cosA=b2+c2-a22bccos30°=32+12-a22×3×132=4-a2233=4-a2a2=4-3=1a=1

Page No 10.26:

Question 3:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In a ∆ABC, if cosA=sinB2sinC, then show that c = a.

Answer:

Given: cosA=sinB2sinC

b2+c2-a22bc=b2c      (Using sine rule and cosine rule)

b2+c2-a2=b2

c2=a2

c=a

Page No 10.26:

Question 4:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In a ∆ABC, if b = 20, c = 21 and sinA=35, find a.

Answer:

In ∆ABC, b = 20, c = 21 and sinA=35.

Using cosine rule, we have

cosA=b2+c2-a22bc1-352=202+212-a22×20×21                    cos2A+sin2A=11625=400+441-a284045=841-a2840672=841-a2

a2=841-672=169a=13

Page No 10.26:

Question 5:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In a ∆ABC, if sinA and sinB are the roots of the equation c2x2-ca+bx+ab=0, then find C.

Answer:

It is given that sinA and sinB are the roots of the equation c2x2-ca+bx+ab=0.

sinA+sinB=--ca+bc2                      Sum of roots=-basinA+sinB=a+bcsinA+sinB=ksinA+ksinBksinC                  Sine rule
sinA+sinB=sinA+sinBsinCsinC=1=sin90°C=90°

Page No 10.26:

Question 6:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In ∆ABC, if a = 8, b = 10, c = 12 and C = λA, find the value of λ.

Answer:

Using cosine rule, we have

cosA=b2+c2-a22bccosA=102+122-822×10×12cosA=100+144-64240cosA=180240=34       .....1

Now, using sine rule, we have

asinA=csinC8sinA=12sinλAsinλA=32sinAsinλA=2×34sinA
sinλA=2sinAcosA              Using1sinλA=sin2Aλ=2

Page No 10.26:

Question 7:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

If the sides of a triangle are proportional to 2, 6 and 3-1, find the measure of its greatest angle.

Answer:

Let ∆ABC be the triangle such that a = 2, b = 6 and c = 3-1.

Clearly, b > a > c. Then,

B is the greatest angle of ∆ABC.                  (Greatest side has greatest angle opposite to it)

Using cosine formula, we have

cosB=c2+a2-b22cacosB=3-12+22-622×3-1×2cosB=3+1-23+4-643-1
cosB=2-2343-1=-23-143-1cosB=-12=cos120°B=120°

Hence, the measure of its greatest angle is 120º.

Page No 10.26:

Question 8:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

If in a ∆ABC, cosAa=cosBb=cosCc, then find the measures of angles A, B, C.

Answer:

In ∆ABC,

cosAa=cosBb=cosCccosAksinA=cosBksinB=cosCksinC                      Using sine rulecotA=cotB=cotCA=B=C

⇒ ∆ABC is an equilateral triangle.

A = B = C = 60º

Page No 10.26:

Question 9:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In any triangle ABC, find the value of asinB-C+bsinC-A+csinA-B.

Answer:

Using sine rule, we have

asinB-C+bsinC-A+csinA-B=ksinAsinB-C+ksinBsinC-A+ksinCsinA-B=ksinπ-B+CsinB-C+ksinπ-C+AsinC-A+ksinπ-A+BsinA-B          
=ksinB+CsinB-C+sinC+AsinC-A+sinA+BsinA-B=ksin2B-sin2C+sin2C-sin2A+sin2A-sin2B=k×0=0

Hence, the required value is 0.

Page No 10.26:

Question 10:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In any ∆ABC, find the value of asinB-sinC.

Answer:

Using sine rule, we have

asinA=bsinB=csinC=ka=ksinA,b=ksinB,c=ksinC

asinB-sinC=ksinAsinB-sinC

=ksinAsinB-sinC

=ksinAsinB-sinC+sinBsinC-sinA+sinCsinA-sinB

=ksinAsinB-sinAsinC+sinBsinC-sinBsinA+sinCsinA-sinCsinB

=k×0 =0

Page No 10.26:

Question 1:

Mark the correct alternative in each of the following:

In any ∆ABC, a2sinB-sinC=

(a) a2+b2+c2                            (b) a2                                           (c) b2                                          (d) 0                                                                                

Answer:

Using sine rule, we have

a2sinB-sinC

=a2bk-ck+b2ck-ak+c2ak-bk=1ka2b-a2c+b2c-b2a+c2a-c2b

This expression cannot be simplified to match with any of the given options.


However, if the quesion is "In any ∆ABC, a2sin2B-sin2C=", then the solution is as follows.

Using sine rule, we have

a2sin2B-sin2C

=a2b2k2-c2k2+b2c2k2-a2k2+c2a2k2-b2k2=1k2a2b2-a2c2+b2c2-b2a2+c2a2-c2b2=1k2×0=0

Hence, the correct answer is option (d).


Disclaimer: The question given in the book in incorrect or there is some printing mistake in the question.

Page No 10.26:

Question 2:

Mark the correct alternative in each of the following:

In a ∆ABC, if a = 2, B=60° and C=75°, then b =

(a) 3                            (b) 6                           (c) 9                           (d) 1+2                                                                              

Answer:

It is given that a = 2, B=60° and C=75°.

In ∆ABC,

A+B+C=180°            Angle sum propertyA+60°+75°=180°A=180°-135°=45°

Using sine rule, we get

2sin45°=bsin60°                       asinA=bsinB=csinCb=2×3212=6

Hence, the correct answer is option (b).

Page No 10.26:

Question 3:

Mark the correct alternative in each of the following:

If the sides of a triangle are in the ratio 1:3:2, then the measure of its greatest angle is

(a) π6                            (b) π3                            (c) π2                            (d) 2π3                                                                                

Answer:

Let ∆ABC be the given triangle such that its sides are in the ratio 1:3:2.

a=k,b=3k,c=2k

Now, a2+b2=k2+3k2=4k2=c2

So, ∆ABC is a right triangle right angled at C.

C=90°

Using sine rule, we have

asinA=bsinB=csinCksinA=3ksinB=2ksin90°sinA=12 and sinB=32A=30° and B=60°

Thus, the measure of its greatest angle is π2.

Hence, the correct answer is option (c).

Page No 10.26:

Question 4:

Mark the correct alternative in each of the following:

In any ∆ABC, 2(bc cosA + ca cosB + ab cosC) =

(a) abc                            (b) a+b+c                           (c) a2+b2+c2                           (d) 1a2+1b2+1c2                                                                                

Answer:

Using cosine rule, we have

2bccosA+cacosB+abcosC=2bcb2+c2-a22bc+2cac2+a2-b22ca+2aba2+b2-c22ab=b2+c2-a2+c2+a2-b2+a2+b2-c2=a2+b2+c2

Hence, the correct answer is option (c).



Page No 10.27:

Question 5:

Mark the correct alternative in each of the following:

In a triangle ABC, a = 4, b = 3, A=60° then c is a root of the equation

(a) c2-3c-7=0                 (b) c2+3c+7=0                (c) c2-3c+7=0                (d) c2+3c-7=0                                                                                                                                                      

Answer:

It is given that a = 4, b = 3 and A=60°.

Using cosine rule, we have

cosA=b2+c2-a22bccos60°=9+c2-162×3×c12=c2-76cc2-7=3cc2-3c-7=0

Thus, c is the root of c2-3c-7=0.

Hence, the correct answer is option (a).

Page No 10.27:

Question 6:

Mark the correct alternative in each of the following:

In a ∆ABC, if c+a+ba+b-c=ab, then the measure of angle C is

(a) π3                            (b) π6                            (c) 2π3                            (d) π2                                                                                

Answer:

Given: c+a+ba+b-c=ab

a+b2-c2=aba2+b2+2ab-c2=aba2+b2-c2=-aba2+b2-c22ab=-12
cosC=-12=cos2π3                Using cosine ruleC=2π3
Thus, the measure of angle C is 2π3.

Hence, the correct answer is option (c).

Page No 10.27:

Question 7:

Mark the correct alternative in each of the following:

In any ∆ABC, the value of  2acsinA-B+C2 is

(a) a2+b2-c2                            (b) c2+a2-b2                            (c) b2-c2-a2                            (d) c2-a2-b2                                                                               

Answer:

In ∆ABC,

A+B+C=π                  Angle sum propertyA+C=π-B

2acsinA-B+C2=2acsinπ-2B2=2acsinπ2-B=2accosB
=2acc2+a2-b22ca               Using cosine rule=c2+a2-b2

Hence, the correct answer is option (b).

Page No 10.27:

Question 8:

Mark the correct alternative in each of the following:

In any ∆ABC, abcosC-ccosB=

(a) a2                            (b) b2-c2                              (c) 0                               (d) b2+c2                                                                               

Answer:

Using cosine rule, we have

abcosC-ccosB=aba2+b2-c22ab-cac2+a2-b22ca=a2+b2-c2-c2-a2+b22=2b2-2c22=b2-c2

Hence, the correct answer is option (b).



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