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1.	In any `DeltaABC`, prove that `(b^(2)-c^(2))cotA+(c^(2)+a^(2))cotB+(a^(2)-b^(2))cotC=0`.
Answer» By the sine rule, we have `a/("sin A")=b/("sin B")=c/("sin C")="k (sky)"` `rArr" "("sin A")/a=("sin B")/b=("sin C")/c="k (sky)"` `rArr" "sinA=ka, sin B=kband sin C=kc.` `:." "(b^(2)-c^(2))cot A=((b^(2)-c^(2))cosA)/("sin A")=((b^(2)-c^(2)))/("ka")((b^(2)+c^(2)-a^(2)))/("2ac")` `=1/("2kabc")(b^(2)-c^(2)-a^(2))=((b^(4)-c^(4)-a^(2)b^(2)+a^(2)c^(2)))/("2kabc")` Similarly, `(c^(2)-a^(2))cosB=((c^(4)-a^(4)-b^(2)c^(2)+a^(2)b^(2)))/("2kabc")` `"And, "(a^(2)-b^(2))cotC=((a^(4)-b^(4)-a^(2)c^(2)+c^(2)c^(2)))/("2kabc")` `:." LHS"=(b^(2)-c^(2))cotA+c^(2)-a^(2)cotB+(a^(2)-b^(2))cotC` `=1/("2kabc")[(b^(4)-c^(4)-a^(2)b^(2)+a^(2)c^(2))+(c^(4)+a^(4)-b^(2)c^(2)+a^(2)b^(2))+(a^(4)-b^(4)-a^(2)c^(2)+b^(2)c^(2))]` `=1/("2kabc")xx0=0="RHS".` Hence, `(b^(2)-c^(2))cotA+(c^(2)-a^(2))cotB(a^(2)-b^(2))cotC=0`.

2.	In any triangle `A B C ,`prove that: ` a^3sin(B-C)+b^3sin(C-A)+c^3sin(A-B)=0`
Answer» By the sine rule, we have `a/("sin A")=b/("sin B")=c/("sin C")="k (sky)"` `rArr" "a=ksinA, b-ksinBandc=ksinC.` `:." "a^(3)sin(B-C)=k^(3)sin^(3)Asin(B-C)` `=k^(3)sin^(2)AsinAsin(B-C)` `=k^(3)sin^(2)A[sin{pi-(B+C)}sin(B-C)]" "[becauseA+(B+C)=pi]` `=k^(3)sin^(3)A[sin(B+C)sin(B-C)]` `=k^(3)sin^(2)A(sin^(2)B-sin^(2)C).` Similarly, `b^(3)sin(C-A)=k^(3)sin^(2)B(sin^(2)C-sin^(2)A)`. And, `c^(3)sin(A-B)=k^(3)sin^(2)C(sin^(2)A-sin^(2)B).` `:." "a^(3)sin(B-C)+b^(3)sin(C-A)+c^(3)sin(A-B)` `=k^(3)sin^(2)A(sin^(2)B-sin^(2)C)+k^(3)sin^(2)B(sin^(2)C-sin^(2)A)+k^(3)sin^(2)C(sin^(2)A-sin^(2)B)` `=0`

3.	In any triangle `A B C ,`prove that: `(b-c)/(b+c)=(tan((b-C)/2))/(tan((B+C)/2))`
Answer» By the sine rule, we have `a/("sin A")=b/("sin B")=c/("sin C")="k (sky)"` `rArr" a=ksin A, b=k sin B and c = ksinC"`. `:." LHS"=((b-c))/((b+c))=("k sin B-k sin C")/("k sin B+k sin C")=("k(sin B - sin C)")/("k(sin B + sin C)")` `=("(sin B - sin C)")/("(sin B + sin C)")=("2cos"((B+C))/2"sin"((B-C))/2)/("2sin"((B+C))/2"cos"((B-C))/2)` `=("tan"1/2(B-C))/("tan"1/2(B+C))="RHS"`. `"Hence,"((b-c))/((b+c))=("tan"1/2(B-C))/("tan"1/2(B+C))`

4.	In any triangle `A B C ,`prove that:`(b-c)cot A/2+(c-a)cot B/2+(a-b)cot C/2`=0
Answer» By the sine rule, we have `a/("sin A")=b/("sin B")=c/("sin C")="k (sky)"` `rArr" "a=ksinA, b=ksin B and C=sinC.` `:." "(b-c)cot""A/2=k(sinB-sinC)cot""A/2` `=2kcos""((B+C))/2sin""((B-C))/2(cos(A/2))/(sin(A/2))` `=2kcos(pi/2-A/2)sin""((B-C))/2(cos(A/2))/(sin(A/2))` `=2ksin""A/2sin""((B-C))/2(cos(A/2))/(sin(A/2))` `=2ksin""((B-C))/2cos""A/2` `=2ksin((B-C))/2cos{pi/2-((B+C))/2}` `=2ksin""((B-C))/2sin""((B+C))/2` `=k(cosC-cosB)`. Similarly, we have `(c-a)cot""B/2=k(cosA-cosC)and (a-b)cos""C/2=k(cosB-cosA).` `:." "(b-c)cos""A/2+(c+a)cos""B/2+(a-b)cot""C/2` `=k[(cosC-cosB)+(cosA-cosC)+(cosB-cosA)]=0`.

5.	Solve the triangle in which a = 2 cm, b = 1 cm and `c=sqrt3`cm.
Answer» Correct Answer - `/_=90^(@),/_B=30" and "/_C=60^(@)`

6.	If in a ` A B C , /_A=45^0, /_B=60^0, a n d /_C=75^0`, find the ratio of its sides.
Answer» By the sine formula, we have `a:b:c=sinA:sinB:sinC` `=sin45^(@):sin60^(@):sin75^(@)` `=1/sqrt2:sqrt3/2:((sqrt3+1))/(2sqrt2)=2:sqrt6:(sqrt3+1).`

7.	In a `DeltaABC," if "/_A=30^(@)"and "b : c =2: sqrt3, " find "/_B`.
Answer» Let b = 2k and c `= sqrt3k`. Then, `cosA=(b^(2)+c^(2)-a^(2))/(2bc)` `rArr" "cos30^(@)=(4k^(2)+3k^(2)-a^(2))/(4sqrt3k^(2))` `rArr" "sqrt3/2xx4sqrt3k^(2)=7k^(2)-a^(2)` `rArr" "k^(2)=a^(2)rArrk=a.` Thus, b = 2a and `c=sqrt3a`. `"Now,"a/("sin A")=b/("sin B")=c/("sin C")=(2a)/("sin B")` `rArr" "sinB=(2axx1/2xx1/a)=1` `rArr" "B=90^(@)`.

8.	In any `DeltaABC`, prove that `((a-b))/c"cos"C/2="sin"((A-B))/2`
Answer» By the sine rule, we have `a/("sin A")=b/("sin B")=c/("sin C")="k(sya)"` `rArr" a=k sin A, b = k sin B and c = k sin C".` `:." LHS"=((a-b))/c"cos"C/2` `=(("k sin A-k sin B"))/(" sin C")"cos"C/2=("k(sin A-sinB)")/("k sin C")"cos"C/2` `=("sin A - sin B")/("sin C")"cos"C/2=("2cos"((A+B))/2"sin"((A-B))/2)/("2sin"C/2"cos"C/2)"cos"C/2" "[because("sin C- sin D")="2cos"((C+D))/2"sin"((C-D))/2]` `=("cos"(pi/2-C/2)"sin"((A-B))/2)/("sin"C/2)" "[because(A+B)/2=(pi/2-C/2)]` `=("sin"C/2"sin"((A-B))/2)/("sin"C/2)="sin"((A-C))/2="RHS."` `"Hence, "((a-b))/c"cos"C/2="sin"((A-B))/2.`

9.	In any triangle `A B C ,`prove that: `("sin"(B-C))/("sin"(B+C))=(b^2-c^2)/(a^2)`
Answer» By the sine rule, we have `a/("sin A")=b/("sin B")=c/("sin C")="k(say)"` `rArr" a=ksin A, b=ksin B and c=ksin C"`. `:." RHS"=((b^(2)-c^(2)))/a^(2)=(k^(2)sin^(2)B-k^(2)sin^(2)C)/(k^(2)sin^(2)A)` `=((sin^(2)B-sin^(2)C))/(sin^(2)A)=(sin(B+C)" sin "(B-C))/(sin^(2)(B+C))` `[{:(because,(A+B+C=pirArrA=pi-(B+C))),(therefore,sin A =sin{pi-(B+C)}=sin(B+C)):}]` `=("sin (B-C)")/("sin (B+C)")="LHS"`. `:." RHS=LHS"`. `"Hence, "("sin (B-C)")/("sin (B+C)")=((b^(2)-c^(2)))/a^(2)`

10.	In any `DeltaABC`, prove that `"a sin (B - C) + bsin(C - A) + csin(A - B)=0`.
Answer» By the sine rule, we have `a/("sin A")=b/("sin B")=c/("sin C")` `rArr" "("sin A")/a=("sin ")/b=("sin C")/c="k (say)"` `rArr" sin A = ka, sin B = kb and sin C = kc"`. Substituting the values of sin B and C and using cosine formulae, we have `"a sin (B - C) = a(sin B cos C - cos B sin C)"` `=a["kb"((a^(2)+b^(2)-c^(2)))/(2ab)"-kc"((c^(2)+a^(2)-b^(2)))/(2ac)]` `=k/2[(a^(2)+b^(2)-c^(2))-(c^(2)+a^(2)-b^(2))=k(b^(2)-c^(2))]`. Similarly, b `"sin(C - A)=b(sin C cos A - cos C sin A)"=k(c^(2)-a^(2))`. `:." a sin(B - C)+b sin(C - A) + csin(A - B)"` `=k(b^(2)-c^(2))+k(c^(2)+a^(2))+k(a^(2)-b^(2))` `=k(b^(2)-c^(2)-a^(2)+a^(2)-b^(2))=kxx0=0`.

11.	At the foot of a mountain, the angle of elevation of its summit is `45^(@)`. After ascending 1 km towards the mountain up an incline of `30^(@)`, the elevatioon changes to `60^(@)` (as shown in the given figure). Find the height of the mountain. [Given: `sqrt3=1.73.`]
Answer» Correct Answer - `1.365 km` `(SD)/(FD)=tan45^(@)=1rArrSD=FD=hkm` `rArr" "/_FSD=/_SFD=45^(@)` `/_ASC=180^(@)-(60^(@)+90^(@))=30^(@)` `rArr/_FSA=(45^(@)-30^(@))=15^(@)and/_SFA=15^(@).` By the sine formula on `DeltaSAF`, we have `(AF)/(sin/_FSA)=(FS)/(sin/_FAS)rArr1/(sin15^(@))=(sqrt2h)/(sin15^(@))=(sqrt2h)/(sin150^(@))rArr(2sqrt2)/((sqrt3-1))=2sqrt2h[becauseFS=sqrt(h^(2)+h^(2))=sqrt2h"and sin"15^(@)=((sqrt3-1))/(2sqrt2)]` `rArrh=1/((sqrt3-1))xx((sqrt3+1))/((sqrt3+1))=((sqrt3+1))/2=((1.73+1))/2=(2.73)/2=1.365km.`

12.	In a triangle `A B C ,`if `acosA=bcosB ,`show that the triangle is either isosceles or right angled.
Answer» By the sine rule, we have `a/("sin A")=b/("sin B")=c/("sin C")="k (sky)"` `rArr" "a=ksinAandb=ksinB` `:." "acosA=bcosB` `rArr" "ksinAcosA=ksinBcosB` `rArr" "1/2(sin2A)=1/2(sin2B)` `rArr" "sin2A=sin2B` `rArr" "sin2A-sin2B=0` `rArr" "2cos(A+B)sin(A-B)=0` `rArr" "cos(A+B)=0orsin(A-B)=0` rArr" "(A+B)=pi/2or(A-B)=0` `rArr" "/_C=pi/2orA=B.` Hence, `DeltaABC` is right-angled or isosceles.