Example 1: Using Sine Rule (AAS)

Question: In a triangle ABC, if A=30∘,B=45∘A=30^\circ, B=45^\circ and side a=10a=10, find side b.

Solution: Using the Sine Rule, asin⁡A=bsin⁡B  ⟹  10sin⁡30∘=bsin⁡45∘\frac{a}{\sin A} = \frac{b}{\sin B} \implies \frac{10}{\sin 30^\circ} = \frac{b}{\sin 45^\circ}. b=10⋅sin⁡45∘sin⁡30∘=10⋅(1/2)1/2=102b = \frac{10 \cdot \sin 45^\circ}{\sin 30^\circ} = \frac{10 \cdot (1/\sqrt{2})}{1/2} = 10\sqrt{2}

Example 2: Using Cosine Rule (SAS)

Question: In a triangle ABC, sides a=4,b=5a=4, b=5 and angle C=60∘C=60^\circ. Find the length of side c.

Solution: Using the Cosine Rule, c2=a2+b2−2abcos⁡C=42+52−2(4)(5)cos⁡60∘=16+25−40(1/2)=21c^2 = a^2 + b^2 - 2ab\cos C = 4^2 + 5^2 - 2(4)(5)\cos 60^\circ = 16 + 25 - 40(1/2) = 21. So, c=21c = \sqrt{21}.

Example 3: Using Cosine Rule (SSS)

Question: The sides of a triangle are a=7, b=8, c=5. Find the angle A.

Solution: Using the Cosine Rule for angles: cos⁡A=b2+c2−a22bc=82+52−722(8)(5)=64+25−4980=4080=12\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{8^2+5^2-7^2}{2(8)(5)} = \frac{64+25-49}{80} = \frac{40}{80} = \frac{1}{2}. So, A=60∘A = 60^\circ.

Example 4: Finding the Circumradius (R)

Question: In △ABC\triangle ABC, if a=4a=4 and A=30∘A=30^\circ, find the circumradius R.

Solution: From the Sine Rule, asin⁡A=2R\frac{a}{\sin A} = 2R. So, 2R=4sin⁡30∘=41/2=82R = \frac{4}{\sin 30^\circ} = \frac{4}{1/2} = 8. The circumradius R=4R=4.

Example 5: Proving the Projection Formula

Question: In any triangle ABC, prove that a=bcos⁡C+ccos⁡Ba = b\cos C + c\cos B.

Solution: Using the Cosine Rule: RHS = b(a2+b2−c22ab)+c(a2+c2−b22ac)=a2+b2−c22a+a2+c2−b22a=2a22a=ab(\frac{a^2+b^2-c^2}{2ab}) + c(\frac{a^2+c^2-b^2}{2ac}) = \frac{a^2+b^2-c^2}{2a} + \frac{a^2+c^2-b^2}{2a} = \frac{2a^2}{2a} = a = LHS.

Example 6: Using Half-Angle Sine Formula

Question: In a triangle ABC, if a=13,b=14,c=15a=13, b=14, c=15, find the value of sin⁡(A/2)\sin(A/2).

Solution: First, find the semi-perimeter: s=13+14+152=21s = \frac{13+14+15}{2} = 21. Using the formula, sin⁡(A/2)=(s−b)(s−c)bc=(21−14)(21−15)14×15=7×6210=15=15\sin(A/2) = \sqrt{\frac{(s-b)(s-c)}{bc}} = \sqrt{\frac{(21-14)(21-15)}{14 \times 15}} = \sqrt{\frac{7 \times 6}{210}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}.

Example 7: Using Half-Angle Cosine Formula

Question: For a triangle with sides 13, 14, 15, find cos⁡(A/2)\cos(A/2).

Solution: We have s=21s=21. Using the formula, cos⁡(A/2)=s(s−a)bc=21(21−13)14×15=21×8210=45=25\cos(A/2) = \sqrt{\frac{s(s-a)}{bc}} = \sqrt{\frac{21(21-13)}{14 \times 15}} = \sqrt{\frac{21 \times 8}{210}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}.

Example 8: Using Half-Angle Tangent Formula

Question: For the same triangle (sides 13, 14, 15), find tan⁡(A/2)\tan(A/2).

Solution: Using the formula: tan⁡(A/2)=(s−b)(s−c)s(s−a)=(7)(6)21(8)=14=1/2\tan(A/2) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \sqrt{\frac{(7)(6)}{21(8)}} = \sqrt{\frac{1}{4}} = 1/2.

Example 9: Area with Two Sides and Included Angle

Question: In a △ABC\triangle ABC, if a=4,b=5a=4, b=5 and angle C=30∘C=30^\circ, find the area.

Solution: Using the formula Δ=12absin⁡C=12(4)(5)sin⁡30∘=10(1/2)=5\Delta = \frac{1}{2}ab\sin C = \frac{1}{2}(4)(5)\sin 30^\circ = 10(1/2) = 5 sq. units.

Example 10: Area with Heron's Formula

Question: The sides of a triangle are 13, 14, and 15. Find its area.

Solution: Semi-perimeter s=21s = 21. Using Heron's formula: Δ=s(s−a)(s−b)(s−c)=21(8)(7)(6)=84\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(8)(7)(6)} = 84 sq. units.

Example 11: Finding Inradius (r)

Question: For a triangle with sides 13, 14, 15, find the inradius r.

Solution: We have s=21s=21 and Δ=84\Delta=84. Using the formula r=Δ/s=84/21=4r = \Delta/s = 84/21 = 4.

Example 12: Finding an Exradius (r1r_1)

Question: For a triangle with sides a=13, b=14, c=15, find the exradius r1r_1.

Solution: We have s=21,Δ=84s=21, \Delta=84. s−a=21−13=8s-a = 21-13=8. Using the formula r1=Δs−a=84/8=10.5r_1 = \frac{\Delta}{s-a} = 84/8 = 10.5.

Example 13: Using Napier's Analogy

Question: In a △ABC\triangle ABC, if b=3,c=1b=\sqrt{3}, c=1 and A=30∘A=30^\circ, find angle B.

Solution: We have B+C=150∘B+C = 150^\circ. Using Napier's Analogy: tan⁡(B−C2)=b−cb+ccot⁡(A/2)=3−13+1cot⁡(15∘)=(2−3)(2+3)=1\tan(\frac{B-C}{2}) = \frac{b-c}{b+c}\cot(A/2) = \frac{\sqrt{3}-1}{\sqrt{3}+1}\cot(15^\circ) = (2-\sqrt{3})(2+\sqrt{3})=1. This means B−C2=45∘\frac{B-C}{2}=45^\circ, so B−C=90∘B-C=90^\circ. Solving B+C=150B+C=150 and B−C=90B-C=90 gives 2B=240  ⟹  B=120∘2B=240 \implies B=120^\circ.

Example 14: Solving a Triangle (AAS)

Question: In a triangle ABC, A=45∘,B=60∘,a=2A=45^\circ, B=60^\circ, a=2. Find the remaining angle and sides.

Solution: Angle C=180−(45+60)=75∘C = 180 - (45+60) = 75^\circ. By Sine Rule, bsin⁡60=2sin⁡45  ⟹  b=6\frac{b}{\sin 60} = \frac{2}{\sin 45} \implies b = \sqrt{6}. And csin⁡75=2sin⁡45  ⟹  c=3+1\frac{c}{\sin 75} = \frac{2}{\sin 45} \implies c = \sqrt{3}+1.

Example 15: Proving an Identity with Sine Rule

Question: In any triangle ABC, prove that a(sin⁡B−sin⁡C)+b(sin⁡C−sin⁡A)+c(sin⁡A−sin⁡B)=0a(\sin B - \sin C) + b(\sin C - \sin A) + c(\sin A - \sin B) = 0.

Solution: Substitute a=2Rsin⁡A,b=2Rsin⁡B,c=2Rsin⁡Ca=2R\sin A, b=2R\sin B, c=2R\sin C. The expression becomes 2R[sin⁡Asin⁡B−sin⁡Asin⁡C+sin⁡Bsin⁡C−sin⁡Bsin⁡A+sin⁡Csin⁡A−sin⁡Csin⁡B]2R[\sin A\sin B - \sin A\sin C + \sin B\sin C - \sin B\sin A + \sin C\sin A - \sin C\sin B], which simplifies to 0.

Example 16: Right-Angled Triangle Radii

Question: For a right-angled triangle with sides 5, 12, 13, find R and r.

Solution: For a right-angled triangle, the circumradius is half the hypotenuse: R=13/2=6.5R = 13/2 = 6.5. The inradius is r=(a+b−c)/2=(5+12−13)/2=2r = (a+b-c)/2 = (5+12-13)/2 = 2.

Example 17: Proving an Identity with Radii

Question: Prove that 1r1+1r2+1r3=1r\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}.

Solution: LHS = s−aΔ+s−bΔ+s−cΔ=3s−(a+b+c)Δ=3s−2sΔ=sΔ=1r\frac{s-a}{\Delta} + \frac{s-b}{\Delta} + \frac{s-c}{\Delta} = \frac{3s-(a+b+c)}{\Delta} = \frac{3s-2s}{\Delta} = \frac{s}{\Delta} = \frac{1}{r} = RHS.

Example 18: Equilateral Triangle Radii

Question: For an equilateral triangle of side 'a', find the ratio R:r.

Solution: For an equilateral triangle, R=a/3R=a/\sqrt{3} and r=a/(23)r=a/(2\sqrt{3}). The ratio R:r=a3:a23=2:1R:r = \frac{a}{\sqrt{3}} : \frac{a}{2\sqrt{3}} = 2:1.

Example 19: Finding Sides from Angles and Perimeter

Question: The angles of a triangle are 30∘,60∘,90∘30^\circ, 60^\circ, 90^\circ. If the perimeter is 3+33+\sqrt{3}, find the sides.

Solution: The ratio of sides a:b:c=sin⁡A:sin⁡B:sin⁡C=1/2:3/2:1=1:3:2a:b:c = \sin A : \sin B : \sin C = 1/2 : \sqrt{3}/2 : 1 = 1:\sqrt{3}:2. Let the sides be x,x3,2xx, x\sqrt{3}, 2x. Perimeter = x(3+3)x(3+\sqrt{3}). Equating this to the given perimeter gives x=1x=1. The sides are 1,3,21, \sqrt{3}, 2.

Example 20: Finding an Angle from Radii

Question: In a triangle ABC, if r1=r+r2+r3r_1 = r+r_2+r_3, prove that it is a right-angled triangle.

Solution: Using the formulas for radii, Δs−a=Δs+Δs−b+Δs−c\frac{\Delta}{s-a} = \frac{\Delta}{s} + \frac{\Delta}{s-b} + \frac{\Delta}{s-c}. This simplifies to 1s−a−1s=1s−b+1s−c\frac{1}{s-a} - \frac{1}{s} = \frac{1}{s-b} + \frac{1}{s-c}, which after further algebra leads to b2+c2=a2b^2+c^2=a^2, proving the triangle is right-angled at A.

Example 21: Area from One Side and Angles

Question: In a triangle ABC, A=60∘,B=45∘A=60^\circ, B=45^\circ and side b=23b=2\sqrt{3}. Find the area.

Solution: Angle C=75∘C = 75^\circ. Using Δ=b2sin⁡Asin⁡C2sin⁡B=(23)2sin⁡60∘sin⁡75∘2sin⁡45∘=12(3/2)((3+1)/22)2(1/2)=9+332\Delta = \frac{b^2 \sin A \sin C}{2\sin B} = \frac{(2\sqrt{3})^2 \sin 60^\circ \sin 75^\circ}{2\sin 45^\circ} = \frac{12 (\sqrt{3}/2) ((\sqrt{3}+1)/2\sqrt{2})}{2(1/\sqrt{2})} = \frac{9+3\sqrt{3}}{2}.

Example 22: Area using Half-Angle Formulas

Question: In a triangle with sides a=3, b=5, c=6, find the area using half-angle formulas.

Solution: s=7s=7. sin⁡(A/2)=(2)(1)30=1/15\sin(A/2) = \sqrt{\frac{(2)(1)}{30}} = 1/\sqrt{15}. cos⁡(A/2)=7(4)30=14/15\cos(A/2) = \sqrt{\frac{7(4)}{30}} = \sqrt{14/15}. sin⁡A=2sin⁡(A/2)cos⁡(A/2)=2(115)(1415)=21415\sin A = 2\sin(A/2)\cos(A/2) = 2(\frac{1}{\sqrt{15}})(\frac{\sqrt{14}}{\sqrt{15}}) = \frac{2\sqrt{14}}{15}. Area Δ=12bcsin⁡A=12(5)(6)(21415)=214\Delta = \frac{1}{2}bc\sin A = \frac{1}{2}(5)(6)(\frac{2\sqrt{14}}{15}) = 2\sqrt{14}.

Example 23: Proving a Sine Rule Identity

Question: In △ABC\triangle ABC, prove that a2−c2b2=sin⁡(A−C)sin⁡(A+C)\frac{a^2-c^2}{b^2} = \frac{\sin(A-C)}{\sin(A+C)}.

Solution: RHS = sin⁡Acos⁡C−cos⁡Asin⁡Csin⁡B\frac{\sin A \cos C - \cos A \sin C}{\sin B}. Using Sine rule and Cosine rule to substitute for all terms and simplifying leads to the LHS.

Example 24: Finding a Side using Medians

Question: In a △ABC\triangle ABC, if a=6,b=10a=6, b=10 and the median from C has length 8, find side c.

Solution: Using Apollonius' theorem for the median to side c: a2+b2=2(mc2+(c/2)2)a^2+b^2 = 2(m_c^2 + (c/2)^2). 62+102=2(82+c2/4)  ⟹  136=2(64+c2/4)  ⟹  68=64+c2/4  ⟹  4=c2/4  ⟹  c2=16  ⟹  c=46^2+10^2 = 2(8^2+c^2/4) \implies 136 = 2(64+c^2/4) \implies 68=64+c^2/4 \implies 4=c^2/4 \implies c^2=16 \implies c=4.

Example 25: Euler's Theorem

Question: Find the distance between the circumcenter and the incenter of a triangle with R=5R=5 and r=2r=2.

Solution: The distance is given by Euler's theorem: d2=R(R−2r)=5(5−2(2))=5(1)=5d^2 = R(R-2r) = 5(5-2(2)) = 5(1)=5. The distance is d=5d=\sqrt{5}.

Example 26: Sides from Radii

Question: In a △ABC\triangle ABC, if r=1,r1=2,r2=3r=1, r_1=2, r_2=3, find the sides a and b.

Solution: From 1/r=1/r1+1/r2+1/r31/r = 1/r_1+1/r_2+1/r_3, we find r3=6r_3=6. Area Δ=rr1r2r3=6\Delta = \sqrt{rr_1r_2r_3} = 6. Now find the semi-perimeter: r=Δ/s  ⟹  1=6/s  ⟹  s=6r=\Delta/s \implies 1=6/s \implies s=6. Then a=s−Δ/r1=6−6/2=3a = s-\Delta/r_1 = 6-6/2=3. b=s−Δ/r2=6−6/3=4b = s-\Delta/r_2 = 6-6/3=4.

Example 27: Ratio of Radii

Question: In a right triangle with sides a, b, and hypotenuse c, prove that r=(s−c)r = (s-c).

Solution: We know r=(a+b−c)/2r=(a+b-c)/2. Also s=(a+b+c)/2s=(a+b+c)/2. Then s−c=(a+b+c)/2−c=(a+b−c)/2=rs-c = (a+b+c)/2 - c = (a+b-c)/2 = r.

Example 28: Connecting Area and Angles

Question: In a triangle, if the area Δ=a2−(b−c)2\Delta = a^2 - (b-c)^2, find tan⁡A\tan A.

Solution: Δ=a2−b2−c2+2bc=−(b2+c2−a2)+2bc=−2bccos⁡A+2bc=2bc(1−cos⁡A)\Delta = a^2-b^2-c^2+2bc = -(b^2+c^2-a^2)+2bc = -2bc\cos A + 2bc = 2bc(1-\cos A). We also know Δ=12bcsin⁡A\Delta = \frac{1}{2}bc\sin A. Equating the two gives 12bcsin⁡A=2bc(1−cos⁡A)  ⟹  cot⁡(A/2)=4\frac{1}{2}bc\sin A = 2bc(1-\cos A) \implies \cot(A/2)=4. From this, we find tan⁡A=2tan⁡(A/2)1−tan⁡2(A/2)=2(1/4)1−1/16=8/15\tan A = \frac{2\tan(A/2)}{1-\tan^2(A/2)} = \frac{2(1/4)}{1-1/16} = 8/15.

Example 29: Ambiguous Case (SSA)

Question: In a triangle, a=4,b=5,A=30∘a=4, b=5, A=30^\circ. How many triangles are possible?

Solution: Using Sine Rule: sin⁡B=bsin⁡Aa=5sin⁡30∘4=5/8\sin B = \frac{b\sin A}{a} = \frac{5\sin 30^\circ}{4} = 5/8. Since 5/8<15/8 < 1, there are two possible angles for B. Since b>ab>a, both angles are valid. There are two possible triangles.

Example 30: Proving a Relation

Question: In a △ABC\triangle ABC, prove that bcos⁡2(C/2)+ccos⁡2(B/2)=sb\cos^2(C/2)+c\cos^2(B/2)=s.

Solution: LHS = bs(s−c)ab+cs(s−b)ac=s(s−c)a+s(s−b)a=sa(2s−b−c)b\frac{s(s-c)}{ab} + c\frac{s(s-b)}{ac} = \frac{s(s-c)}{a} + \frac{s(s-b)}{a} = \frac{s}{a}(2s-b-c). Since 2s=a+b+c2s=a+b+c, this becomes sa(a)=s\frac{s}{a}(a)=s = RHS.