Solution of triangles

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Solution of triangles

  1. 1. Solution of TrianglesTRIGONOMETRY: FORMULAS, IDENTITIES, SOLUTIONOF TRIANGLESTrigonometric functions of an acute angle. The basic definitions ofthe various trigonometric functions are given in terms of the acute angles ofa right triangle. See Fig. 1. Shown is a right triangle in which C is the rightangle, the side opposite being the hypotenuse c. In terms of this righttriangle of Fig. 1 the definitions are as follows:Trigonometric functions of complementary angles. The acute anglesA and B of the right triangle ABC are complementary, that is A + B = 90o.From Fig. 1 we haveIn a triangle ABC, the angles are denoted by capital letters, A, Band C and the lengths of the sides opposite these angles aredenoted by a, b, c respectively. Semi-perimeter of the triangle iswritten as s =( a + b + c )/ 2 , and its area by S or Δ. Let R be theradius of the circumcircle of the ΔABC.
  2. 2. Basic FormulaeIf the three angles A, B, C are given, we can only find the ratios ofthe sides a, b, c by using the sine rule (since there are infinitesimilar triangles possible). In trigonometry, the law of sines (also known as the sine law, sine formula, or sine rule) is an equation relating the lengths of the sides of an arbitrary triangle to the sines of its angles. According to the law, where a, b, and c are the lengths of the sides of a triangle, and A, B, and C are the opposite angles (see the figure to the right). Sometimes the law is stated using the reciprocal of this equation: The law of sines can be used to compute the remaining sides of a triangle when two angles and a side are known—a technique known as triangulation. It can also be used when two sides and one of the non-enclosed angles are known. In some such cases, the formula gives two possible values for the enclosed angle, leading to an ambiguous case. The law of sines is one of two trigonometric equations commonly applied to find lengths and angles in a general triangle, the other being the law of cosines(i) Sine rule A c b B C D a
  3. 3. Case i When △ABC is an acute angled triangle:AD is per. on BC , in △ABD sinB = AD/AB ⇨ csinB = AD....(i)In △ ACD , AD = bsinC......(ii), from (i) & (ii) b/sinB = c/sinCSimilarly a/sinA = c/sinCCase (ii) When △ABC is an obtuse angled triangle A C b D B C aAD is per. on BC , in △ABD sin<ABD = AD/AB=sin(1800-B)=sinB ⇨c sinB = AD....(i)In △ ACD , AD = bsinC......(ii), from (i) & (ii) b/sinB = c/sinCSimilarly a/sinA = c/sinCCase(iii)When △ABC is rightangled triangleSinC = 1, SinA = BC/AB = a/c & sinB = AC/AB = b/c∴ sin A/a = sin B/b = sinC/c = 1/2R = 2Δ/abc. [R=abc/4△].radius of circumcircle or R=a/2sinA=b/2sinB=c/2sinC,△=(1/2)bcsinA and radius of incircle is r=△/s or r =(s-a)tan(A/2)r =(s-b)tan(B/2), r=(s-c)tan(C/2) andr=4Rsin(A/2)sinB/2sinC/2Example 1. In △ABC, if a=2, b=3 and sinA =2/3 , find <B.[hint: by using law of sine , we get = , <B=п/2]In △ABC, if the angles are in the ratio of 1:2:3. Prove that the correspondingSides are in the ratio of 1:√3:2.
  4. 4. [hint: angles are Ѳ+2Ѳ+3Ѳ=180⇨ 300, 600 ,900 ⇨ = = ](ii) Cosine ruleCASE(i) cosB = BD/c ⇨ BD = c cosB & CD = bcosCBy Pythagoras thm. AC2 = AD2+CD2 = AD2+(BC – BD)2AC2 =BC2 +(AD2+BD2) – 2. BC.BD= BC2 + AB2– 2. BC.BD⇨ b2 = c2+a2 – 2cacosBCase(ii) cos(1800-B) = BD/c ⇨ BD = -c cosB & CD = bcosC AC2 =AD2+CD2 = AD2+(BC +BD)2AC2 =BC2 +(AD2+BD2) + 2. BC.BD= BC2 + AB2+ 2. BC.BD⇨ b2 =c2+a2 – 2cacosBCASE(iii) ⇨ b2 = c2+a2 ⇨ b2 = c2+a2 – 2cacosB∴cos A =( b2 + c2 - a2 )/2abc cosB =( a2 + c2 - b2 ) /2ac ,cosC = (a2 + b2 - c2)/2ab .(iii) Trigonometric ratios of half-angles:sin A/2 = √[(s-b) (s-c)]/bc, cos B/2 = √[s(s - b)]/ca,tan C/2 = √ [(s - a) (s - b)]/[s(s - c)] .(iv) Projection rule:In case (i) cosB=BD/AB ⇨ BD=cCosB & cosC=CD/AC ⇨ CD=bcosC ∴ a = BC=BD+CD = cCosB+ bcosCIn case (ii) cos(1800-B)= BD/AB ⇨ BD= - cCosB∴ a = BC= CD - BD = cCosB+ bcosCa = b cosC + c cosB, b = c cosA + a cosC, c = a cosB + b cosA.(v) Area of a triangleIn case (i) sinB= AD/AB ⇨ AD=csinB ∴ △ = ½. BC.AD=1/2.acsinBIn case(ii) sin(1800-B)= AD/AB ⇨ AD= - csinBΔ = 1/2 bc sin A = 1/2 ca sinB = 1/2 ab sin C = √[s(s - a) (s - b)(s - c)] = abc/4R = rs.
  5. 5. (vi) Napier’s analogytan (B – C)/2 =( b – c)/(b +c) cot A/2 , tan (C – A)/2 = (c –a)/(c + a) cot B/2 , tan (A – B)/2 = (a – b) /(a + b) cot c/2.**(vii) m-n theoremIf D be the point on the side BC of a triangle ABC which divide theside BC in the ratio m : n, then with respect to mentioned figure,we have:(i) (m + n) cot θ = m cot α – n cot ß.(ii)(m + n) cot θ = n cot B – m cot C.**(viii) Apollonius theoremIn a triangle ABC, AD is median through A, then AB2 + AC2 =2(AD2+BD2).Process of Solution of TrianglesThe three sides a, b, c and the three angles A, B, C are called theelements of the triangle ABC. When any three of these sixelements (except all the three angles) of a triangle are given, thetriangle is known completely; that is the other three elements canbe expressed in terms of the given elements and can beevaluated. This process is called the solution of triangles.(i) If the three sides a, b, c are given, angle A is obtained fromtan A/2 = √[(s - b) (s - c)] / [s(s - a)] or cos A = (b2 + c2 - a2 )/2bc . B and C can be obtained in the similar way.(ii) If two sides b and c and the included angle A are given, thentan (B – c)/2 = (b – c)/ (b + c) cot A/2 gives (B – C)/2. Also( B+ C)/2 = 90o - A/2, so that B and C can be evaluated. The thirdside is given by a = b sin A/sin B or a2 = b2 + c2 – 2bc cosA.
  6. 6. *(iii) If two sides b and c and the angle B (opposite to side b) aregiven, then sin C = c/b sinB, A = 180o – (B + C) and b = b sinA/sinB give the remaining elements. If b < c sin B, there is notriangle possible (Fig. 1) If b = c sinB and B is an acute angle,then only one triangle is possible (Fig. 2) If c sinB < b < c and B isan acute angle, then there are two values of angle C (Fig. 3). If c< b and B is an acute angle, then there is only one triangle (Fig.4).This is, sometimes, called an ambiguous case. In trigonometry, the law of cosines (also known as the cosine formula or cosine rule) relates the lengths of the sides of a plane triangle to the cosine of one of its angles. Using notation as in Fig. 1, the law of cosines says where γ denotes the angle contained between sides of lengths a and b and opposite the side of length c. The law of cosines generalizes the Pythagorean theorem, which holds only for right triangles: if the angle γ is a right angle (of measure 90° or π/2 radians), then cos(γ) = 0, and thus the law of cosines reduces to
  7. 7. The law of cosines is useful for computing the third side of a triangle when two sides and their enclosed angle are known, and in computing the angles of a triangle if all three sides are known. By changing which sides of the triangle play the roles of a, b, and c in the original formula, one discovers that the following two formulas also state the law of cosines:In trigonometry, the law of tangents is a statement about the relationshipbetween the tangents of two angles of a triangle and the lengths of theopposite sides.In Figure , a, b, and c are the lengths of the three sides of the triangle, andα, β, and γ are the angles opposite those three respective sides. The law oftangents states that The law of tangents, although not as commonly known as the law of sines or the law of cosines, is equivalent to the law of sines, and can be used in any case where two sides and a non-included angle, or two angles and a side, are known.ProofTo prove the law of tangents we can start with the law of sines: Let so that It follows that
  8. 8. Using the trigonometric identity, the factor formula for sines specifically we get As an alternative to using the identity for the sum or difference of two sines, one may cite the trigonometric identitySOME QUESTIONS BANKQ.1 In △ABC , prove that (i) =(ii) a sin(B-C) + b sin(C-A) + c sin(A-B) = 0[hint : put a = ksinA, b= ksin B & c= sinC by law of sineOn the R.H.S of (i) k2(sin2B – sin2C) / k2 sin2A ⇨= = =L.H.S of (ii) k[ sinA.sin(B-C)+ sinB.sin(C-A)+ sinC.sin(A-B)]K [sin(B+C). sin(B-C)+ sin(C+A).sin(C-A)+ sin(A+B).sin(A-B)]K[sin2B-sin2C + sin2C-sin2A+ sin2A-sin2B] = 0Q.2 In △ABC, prove that:
  9. 9. (i) = cos(A/2) (ii) a = (b+c) sin(A/2)(iii) (b-c) cot(A/2)+ (c-a) cot(B/2) + (a-b) cot(C/2) =0[HINT: put the values of a,b,c by law of sines and use A+B+C= on theR.H.S. of (i) & (ii) , but in (iii) take L.H.S. , we get K[2cot(A/2) + 2 cot(B/2)+ 2 cot(C/2)] =k[2cos(A/2) + 2cos(B/2) + 2cos(C/2) ]Q.3 In a △ABC, if a= 2, b= 3 and sinA = 2/3 , find angle B.[hint: use a/sinA = b/ sinB = c/sinC ⇨ 2/(2/3) = 3/sinB ⇨ sinB = 1 ⇨ B=900]Q.4 In △ ABC, if acosA = bcosB , show that the triangle is either isoscelesor right angled.[hint: ksinAcosA = k sinBcosB ⇨ sin2A = sin2B ⇨ 2A=2B or 2A = п-2BA=B or A+B=п/2]Q.5 In any △ ABC , P.T.(i) a(bcosC – ccosB) = b2-c2 (ii) (a-b)2 cos2(C/2) + (a+b)2 sin2(C/2) =c2[hint: L.H.S ab{ (a2 + b2 - c2)/2ab} –ac {( a2 + c2 - b2 ) /2ac}= b2-c2(ii) L.H.S a2 [cos2(C/2) + sin2(C/2)]+ b2 [cos2(C/2) + sin2(C/2)]-2ab[cos2(C/2) - sin2(C/2)]= a2+b2 – 2abcosC = c2 ]Q.6 In any △ ABC , P.T.(i) = (ii) 2(asin2(C/2)+csin2(A/2)) = a-b+c(iii) 2(acos2(C/2)+ccos2(B/2)) = a+b+c(iv) (b+c)cosA+(c+a)cosB+(a+b)cosC=a+b+c[hint: (i) R.H.S = = b/c(ii) [a(1-cosC)+ c(1-cosA)]= a-b+c(iii) same as (ii), (iv) + +
  10. 10. =a+b+c by projection formulas ]Q.7 In △ ABC, angle C= 600 , then p.t. + =[hint: cosC = ½ ⇨ (a2 + b2 - c2)/2ab =1/2 ⇨ a2 + b2 – ab= c2Now + = if = if a2 + b2 – ab= c2 ]Q.8 Find the area of a △ABC in which angle A = 600 ,b=4 cm andc= .[△ = ½ . bc sinA = ½ 4 . sin60 = 3 sq. Cm.]Q. 9 In △ ABC, if a=13 , b=14 and c=15, find the following:(i) △ (ii) sin(A/2) (iii) cos(A/2) (iii) tan(A/2) (iv) sinA (v) cosA (vi) tanA[HINT: s= (a+b+c)/2 = 21 , △ = =84Sin(A/2) = =√(1/5), cos(A/2) = =2/√5Tan(A/2) = =1/2 , △ = ½.bc sinA ⇨ sinA = 4/5,cosA=3/5 , tan A = 4/3]Q.10 In △ ABC, bcosB+ ccosC = acos(B-C).[HINT: L.H.S k(sinBcosB + sinC cosC) =k/2 (sin2B+sin2C) =k/2(2sin(B+C)cos(B-C)) = ksin(1800-A)cos(B-C).]Q. 11 a( sinB-sinC) + b (sinC-sinA) + c( sinA-sinB) = 0. [ use law of sine]Q.12 In △ ABC, p.t. =[HINT: L.H.S = , use )= cos(A+B).cos(A-B)]Q. 13 In △ ABC, P.T. (i) = (ii) = [use lawof sin on L.H.S for (i) & (ii) ]
  11. 11. (iii) a(cosC-cosB) = 2(b-c)cos2(A/2) [put 2cos2(A/2)= 1+cosA)onR.H.S & use projection formulas] (iv) asinA - bsinB = csin(A-B)[put a=ksinA & b=ksinB on L.H.S &use formula of sin2A-sin2B=sin(A+B).sin(A-B)] (v) a(bcosC – ccosB) = b2-c2 [ use cosC =(a2+b2-c2)/2ab]Q.14 In △ ABC, (i) if (b+c)/12=(c+a)/13=(a+b)/15 then p.t.cosA/2=cosB/7=cosC/11 [ let (b+c)/12=(c+a)/13=(a+b)/15=k ⇨ a+b+c=20k 2 2⇨ a= 8k, b=7k & c=5k , we get cosA = 1/7, cosB= ½ & cosC = (a +b -c2)/2ab = 11/14 ,cosA: cosB: cosC= 1/7:1/2:11/14 = 2:7:11](ii) p.t. = [use cos A = on L.H.S](iii) a=18,b=24 & c=30,find cosA , cosB , cosC [use cos A =,ans. are 4/5, 3/5 0 resp.](iv) 2(bccosA+cacosB+abcosC) = a2+b2+c2 [use cos A = ](v) p.t. sin2A+ sin2B+ sin2C=0[HINT: 2sinAcosA+ 2sinBcosB+ 2sinCcosCThen put sinA , sinB ,sinC as ak,bk,ck resp.& cos A = ](vi) p.t + + = [use cos A = on L.H.S](vi) a=4,b=6 &c=8,show that 6cosC=4+3cosB(vii) if <B=600 , P.T (a+b+c)(a-b+c)=3ac(viii)p.t. (b2-c2)cotA+(c2-a2)cotB+(a2-b2)cotC=0[hint: (b2-c2)cosA/sinA+(c2-a2)cosB/sinB+(a2-b2)cosC/sinC , usesinA=ak & cos A = ](ix) if cosC= sinA/(2sinB), p.t. triangle is an isosceles.
  12. 12. [put sinA=ak ,sinB=bk⇨ 2cosAsinC=sinB ⇨ 2 )kc=kb ](x) p.t (a-b)2cos2(C/2)+(a+b)2sin2(C/2)=c2[hint: a2(cos2(C/2)+sin2(C/2)) + b2 (cos2(C/2)+sin2(C/2))-2ab(cos2(C/2)-sin2(C/2)) =a2+b2-2abcosC=c2 ](xi) p.t a2=(b+c)2-4bccos2(A/2)(xii) p.t. (c2+b2-a2)tanA=(a2+c2-b2)tanB=(a2+b2-c2)tanC[hint: (c2+b2-a2)sinA/cosA=(a2+c2-b2)sinB/cosB= (a2+b2-c2)sinC/cosC law of sin & cosine formulas]We have studied that a triangle has six parts or six elements vizthree sides and three angles. From geometry, we know that whenany three elements are given of which necessarily a side is given,the triangle is completely determined i.e, remaining three elementscan be determined. The process of determining the unknownelements knowing the known elements is known as the solution ofa triangle. In practice, there are four different cases for which thesolution is discussed as under.Case IWhen all three sides are given.To solve a triangle given the three sides a, b, c.To determine angles A, B and C.The angles A, B and C are determined by using the followingrelations:
  13. 13. Example:The sides of a triangle are 20, 30, and 21. Find the greatest angle.Suggested answer:The greatest angle is opposite to the side whose length is 30.Let a = 20, b = 30, c = 21.We have to find angle B.Case 2:When two sides and the angle included between these are given.Example:If b = 251, C = 147, A = 47o, find the remaining angles.(Use Napiers rule)Suggested answer:
  14. 14. = 0.268 x 2.2998Case 3:To solve a triangle having given two angles and a side.Let the given parts be denoted by B, C, a, then the third angle A canbe found from the relation.Similarly, c can be found from the equation.log c = log a + log sin C - log sin AExample:
  15. 15. Solve the triangle ABC, given a =18, A = 25o, B = 108o.Suggested answer:C = 180o - (25 + 108o) = 47oTaking log on both sides, we getlog b = log a + log (sin B) - log (sin A)= log 18 + log (sin 180o) - log (25o)= log 18 + log (sin 72o) - log (25o)= log 18 + log (0.9511) - log (0.4226)(from Trigonometric table)= 1.2553 - 1 + 0.9782 + 1 - 0.6295log b = 1.6076Taking antilog, we get b = 40.52.Similarly usinglog c = log a + log (sin C) - log (sin A),we can evaluate the value of C as 31.23o.Case 4:When two sides b, c and an angle B opposite to one of the givensides are given.
  16. 16. log sin C = log c + log B - log b which determines C.The equation (i) in general leads to two solutions say C1 and C2 (saysuch that C1 + C2 = 180), suppose C1 is acute.Corresponding to C1 and C2, we have two values a1, a2 (say of theremaining side a). They are given byFrom equation (i), if an acute angle C1 is a solution of (i) then180o - C1 is also a solution and there is an ambiguity. However ifFrom geometry, C must be acute. For if C is 90o or more than Bthen C must be greater than 90o. This is impossible, since a trianglecan have only one right angle or more in size.If b < c, there may be two triangles having the given elements b, cand B. One having an acute angle C1 and second having an angleFig (a) Fig (b)
  17. 17. Fig (c)Figures (a) and (b) show its two triangles.Figure (c) shows two triangles superimposed and suggests ageometric method of drawing the two triangles.From figure (c), it is evident that if b < c sinB, there is no triangle.The following rules may be used to solve a triangle with given b, cand B.Step 1:Use (i) to find log (sin C).Step 2:If log (sin C) comes out as positive, sin C would have to be greaterthan 1 and there is no solution.Step 3:If log (sin C) comes out as negative or zero, find the correspondingacute angle C.Then, find A = 180o - (B + C) and finally get a by using the law ofsines (this completes the solution of ).Step 4:If bo - C instead of C.Example:If in a DABC, a = 97, b = 119, A = 50o, find B and c given that:
  18. 18. log (sin (70o, 1)) 9.9730318 .Suggested answer: B = 70o057" or (180o - 70o057") i.e., 109o5903"Since a < b, both the values of B is admissible and then case isambiguous.

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