Law of Sines and Law of Cosines

Study Sheet

Law of Sines and Law of Cosines

Solving oblique triangles: AAS, ASA, SSA, SAS, SSS, area, and applications

The Law of Sines (AAS and ASA)

Tip

Rounding convention (used throughout). Unless told otherwise, round angles to the nearest tenth of a degree and side lengths / areas to the nearest hundredth. Carry extra digits in intermediate steps and round only at the end.

For triangles without a right angle, the law of sines relates each side to the sine of its opposite angle, and the law of cosines generalizes Pythagoras with a correction term −2abcos⁡C-2ab\cos C. Use sines for ASA/AAS (and SSA, watching for the ambiguous case) and cosines for SSS and SAS. Area is 12absin⁡C\tfrac12ab\sin C.

Every triangle here is oblique (no right angle guaranteed). We name the three angles A,B,CA,B,C and put the side opposite each angle in the matching lowercase letter: side aa is opposite angle AA, and so on.

Concept
Law of Sines

For any triangle ABCABC,

asin⁡A=bsin⁡B=csin⁡C.\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.

When to use it: you know an angle and the side opposite it, plus one more piece.

  • AAS / ASA — two angles and any one side. Find the third angle by A+B+C=180∘A+B+C=180^\circ, then use a ratio.
  • SSA — two sides and an angle opposite one of them (the ambiguous case, next section).
Reminder — Law of Sines:  asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}

Drop the altitude hh from CC to side cc. The two right triangles it makes give h=bsin⁡Ah=b\sin A and h=asin⁡Bh=a\sin B, so bsin⁡A=asin⁡Bb\sin A=a\sin B, which rearranges to asin⁡A=bsin⁡B\dfrac{a}{\sin A}=\dfrac{b}{\sin B}. Dropping the altitude from AA instead brings in cc and CC. The equal ratios say that a larger angle faces a proportionally longer side, which is a quick check on every answer: the longest side must sit across from the largest angle.

Example
AAS: given A=40∘A=40^\circ, B=60∘B=60^\circ, a=10a=10
ABCa = 10b ≈ 13.47c ≈ 15.3240°60°80°
Drawn to scale: the 80∘80^\circ angle faces the longest side.

Third angle: C=180∘−40∘−60∘=80∘C=180^\circ-40^\circ-60^\circ=80^\circ.

b=asin⁡Bsin⁡A=10sin⁡60∘sin⁡40∘≈13.47,c=asin⁡Csin⁡A=10sin⁡80∘sin⁡40∘≈15.32.b=\frac{a\sin B}{\sin A}=\frac{10\sin 60^\circ}{\sin 40^\circ}\approx 13.47,\qquad c=\frac{a\sin C}{\sin A}=\frac{10\sin 80^\circ}{\sin 40^\circ}\approx 15.32.
Side note
Where it goes wrong

Setting up the ratio upside down. In the AAS example, writing b=asin⁡Asin⁡B=10sin⁡40∘sin⁡60∘≈7.42b=\dfrac{a\sin A}{\sin B}=\dfrac{10\sin 40^\circ}{\sin 60^\circ}\approx 7.42 looks tidy but cannot be right: B=60∘B=60^\circ is larger than A=40∘A=40^\circ, so bb must be longer than a=10a=10. Keep each side over the sine of its own opposite angle, bsin⁡B=asin⁡A\dfrac{b}{\sin B}=\dfrac{a}{\sin A}, then multiply by sin⁡B\sin B.

Example
ASA: given A=45∘A=45^\circ, B=105∘B=105^\circ, included side c=20c=20
ABCa ≈ 28.28b ≈ 38.64c = 2045°105°30°
The obtuse angle at BB pushes CC far out; bb is the longest side.

Third angle: C=180∘−45∘−105∘=30∘C=180^\circ-45^\circ-105^\circ=30^\circ. Now cc is opposite the known angle CC, so

a=csin⁡Asin⁡C=20sin⁡45∘sin⁡30∘≈28.28,b=csin⁡Bsin⁡C=20sin⁡105∘sin⁡30∘≈38.64.a=\frac{c\sin A}{\sin C}=\frac{20\sin 45^\circ}{\sin 30^\circ}\approx 28.28,\qquad b=\frac{c\sin B}{\sin C}=\frac{20\sin 105^\circ}{\sin 30^\circ}\approx 38.64.
Tip

Pick-the-law tip. If a problem gives you a complete angle–opposite-side pair (an angle and its opposite side)\big(\text{an angle and its opposite side}\big), reach for the Law of Sines. If it does not, you will need the Law of Cosines.

Reminder — Law of Cosines:  c2=a2+b2−2abcos⁡Cc^{2}=a^{2}+b^{2}-2ab\cos C
Try it
Quick check

A=35∘A=35^\circ, B=65∘B=65^\circ, a=12a=12. Find side bb.

Work it on paper before revealing the answer.

Show the answer

Answer: b=12sin⁡65∘sin⁡35∘≈18.96b=\dfrac{12\sin 65^\circ}{\sin 35^\circ}\approx 18.96.

The Ambiguous Case (SSA)

Concept
SSA: how many triangles?

Given angle AA, its opposite side aa, and an adjacent side bb. Let the “altitude” be h=bsin⁡Ah=b\sin A. Compare aa with hh and with bb:

  • If AA is acute: a<h⇒a<h\Rightarrow 0 triangles; a=h⇒a=h\Rightarrow 1 (right) triangle; h<a<b⇒h<a<b\Rightarrow 2 triangles; a≥b⇒a\ge b\Rightarrow 1 triangle.
  • If AA is obtuse: a≤b⇒a\le b\Rightarrow 0 triangles; a>b⇒a>b\Rightarrow 1 triangle.

To solve: compute sin⁡B=bsin⁡Aa\sin B=\dfrac{b\sin A}{a}. The acute value B1B_1 and its supplement B2=180∘−B1B_2=180^\circ-B_1 are both candidates; keep B2B_2 only if A+B2<180∘A+B_2<180^\circ.

Picture the given pieces physically. Side bb and angle AA are fixed, so vertex CC is fixed; side aa is a stick of known length hinged at CC that must reach the ray along side cc. The shortest distance from CC to that ray is the altitude h=bsin⁡Ah=b\sin A. A stick shorter than hh misses the ray; one exactly hh long touches it once, at a right angle; one between hh and bb crosses the ray twice, both times in front of AA; one at least as long as bb reaches back past AA on one side, so only one crossing counts.

When h<a<bh<a<b the side of length aa can reach the base line at two points B1B_1 and B2B_2, giving two different triangles.

Example
Full SSA with two triangles: a=8a=8, b=10b=10, A=40∘A=40^\circ

Here AA is acute and h=bsin⁡A=10sin⁡40∘≈6.43h=b\sin A=10\sin 40^\circ\approx 6.43. Since h<a<bh<a<b (6.43<8<106.43<8<10), there are two triangles.

sin⁡B=bsin⁡Aa=10sin⁡40∘8≈0.8035  ⇒  B1≈53.5∘  or  B2=180∘−53.5∘=126.5∘.\sin B=\frac{b\sin A}{a}=\frac{10\sin 40^\circ}{8}\approx 0.8035 \;\Rightarrow\; B_1\approx 53.5^\circ \ \text{ or }\ B_2=180^\circ-53.5^\circ=126.5^\circ.

Both keep the angle sum below 180∘180^\circ, so both survive.

Triangle 1: B1≈53.5∘B_1\approx 53.5^\circ,   C1=180∘−40∘−53.5∘≈86.5∘\;C_1=180^\circ-40^\circ-53.5^\circ\approx 86.5^\circ,

c1=asin⁡C1sin⁡A=8sin⁡86.5∘sin⁡40∘≈12.42.c_1=\frac{a\sin C_1}{\sin A}=\frac{8\sin 86.5^\circ}{\sin 40^\circ}\approx 12.42.

Triangle 2: B2≈126.5∘B_2\approx 126.5^\circ,   C2=180∘−40∘−126.5∘≈13.5∘\;C_2=180^\circ-40^\circ-126.5^\circ\approx 13.5^\circ,

c2=asin⁡C2sin⁡A=8sin⁡13.5∘sin⁡40∘≈2.90.c_2=\frac{a\sin C_2}{\sin A}=\frac{8\sin 13.5^\circ}{\sin 40^\circ}\approx 2.90.
Side note
Where it goes wrong

Trusting the calculator's single answer. For a=8a=8, b=10b=10, A=40∘A=40^\circ the key sin⁡−1(0.8035)\sin^{-1}(0.8035) returns only 53.5∘53.5^\circ, because inverse sine never outputs an angle above 90∘90^\circ. The supplement 126.5∘126.5^\circ has the same sine and also fits beside A=40∘A=40^\circ, so stopping at the calculator's value loses the second triangle, the one with c2≈2.90c_2\approx 2.90.

Tip

SSA warning. Always compute h=bsin⁡Ah=b\sin A first. Never accept a single answer from your calculator's sin⁡−1\sin^{-1} without checking whether the supplementary angle also gives a valid triangle.

Try it
Quick check

A=30∘A=30^\circ, b=12b=12, a=7a=7. How many triangles are there? Find every possible value of BB.

Show the answer

Answer: two triangles, B≈59.0∘B\approx 59.0^\circ or B≈121.0∘B\approx 121.0^\circ. First the altitude: h=bsin⁡A=12⋅12=6h=b\sin A=12\cdot\tfrac12=6. Since 6<7<126<7<12, i.e. h<a<bh<a<b, there are two triangles. Then sin⁡B=bsin⁡Aa=12⋅0.57=67≈0.8571\sin B=\dfrac{b\sin A}{a}=\dfrac{12\cdot 0.5}{7}=\dfrac67\approx 0.8571, so B1≈59.0∘B_1\approx 59.0^\circ and B2=180∘−59.0∘=121.0∘B_2=180^\circ-59.0^\circ=121.0^\circ. Both leave room for CC: 30+121.0<18030+121.0<180.

The Law of Cosines (SAS and SSS)

Concept
Law of Cosines
CABaca cos Cb − a cos Ca sin CC
Put CC at the origin: B=(acos⁡C, asin⁡C)B=(a\cos C,\ a\sin C) and A=(b,0)A=(b,0).

For any triangle ABCABC,

a2=b2+c2−2bccos⁡A,b2=a2+c2−2accos⁡B,c2=a2+b2−2abcos⁡C.a^2=b^2+c^2-2bc\cos A,\quad b^2=a^2+c^2-2ac\cos B,\quad c^2=a^2+b^2-2ab\cos C.

When to use it: there is no complete angle–opposite-side pair.

  • SAS — two sides and the included angle: find the third side directly.
  • SSS — all three sides: solve an angle by rearranging, e.g. cos⁡A=b2+c2−a22bc\cos A=\dfrac{b^2+c^2-a^2}{2bc}.

Tip: in SSS, find the largest angle first (opposite the longest side); a negative cosine tells you it is obtuse.

Place CC at the origin with side bb along the xx-axis, so A=(b,0)A=(b,0) and B=(acos⁡C, asin⁡C)B=(a\cos C,\ a\sin C). The distance formula gives c2=(b−acos⁡C)2+(asin⁡C)2=b2−2abcos⁡C+a2(cos⁡2C+sin⁡2C)=a2+b2−2abcos⁡Cc^2=(b-a\cos C)^2+(a\sin C)^2=b^2-2ab\cos C+a^2(\cos^2C+\sin^2C)=a^2+b^2-2ab\cos C. The term −2abcos⁡C-2ab\cos C is the correction to Pythagoras: it vanishes at C=90∘C=90^\circ, is subtracted when CC is acute (the far side is shorter than a right triangle's would be), and becomes an addition when CC is obtuse, because then cos⁡C<0\cos C<0.

Example
SAS: given a=5a=5, b=7b=7, included angle C=45∘C=45^\circ
ABCa = 5b = 7c ≈ 4.9545°45.6°89.4°
The included angle CC sits between the two known sides.
c2=a2+b2−2abcos⁡C=25+49−2(5)(7)cos⁡45∘≈24.50  ⇒  c≈4.95.c^2=a^2+b^2-2ab\cos C=25+49-2(5)(7)\cos 45^\circ\approx 24.50\;\Rightarrow\; c\approx 4.95.

Then find another angle with the Law of Cosines:

cos⁡A=b2+c2−a22bc=49+24.50−252(7)(4.95)≈0.6999  ⇒  A≈45.6∘,\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{49+24.50-25}{2(7)(4.95)}\approx 0.6999\;\Rightarrow\; A\approx 45.6^\circ,

and B=180∘−45∘−45.6∘≈89.4∘B=180^\circ-45^\circ-45.6^\circ\approx 89.4^\circ.

Example
SSS: given a=8a=8, b=11b=11, c=15c=15
ABCa = 8b = 11c = 15103.1°31.3°45.6°
The longest side, c=15c=15, faces the only obtuse angle.

Longest side is cc, so angle CC is largest. Solve it first:

cos⁡C=a2+b2−c22ab=64+121−2252(8)(11)=−40176≈−0.2273  ⇒  C≈103.1∘ (obtuse).\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{64+121-225}{2(8)(11)}=\frac{-40}{176}\approx -0.2273\;\Rightarrow\; C\approx 103.1^\circ \ (\text{obtuse}).

Next,

cos⁡B=a2+c2−b22ac=64+225−1212(8)(15)=168240=0.7  ⇒  B≈45.6∘,\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{64+225-121}{2(8)(15)}=\frac{168}{240}=0.7\;\Rightarrow\; B\approx 45.6^\circ,

and A=180∘−103.1∘−45.6∘≈31.3∘A=180^\circ-103.1^\circ-45.6^\circ\approx 31.3^\circ. (Check: 31.3+45.6+103.1=180.031.3+45.6+103.1=180.0.)

Side note
Where it goes wrong

Combining terms in the wrong order. In the SAS example, 25+49−2(5)(7)cos⁡45∘25+49-2(5)(7)\cos 45^\circ is not (74−70)cos⁡45∘≈2.83(74-70)\cos 45^\circ\approx 2.83. The cosine multiplies only 2ab2ab: compute 70cos⁡45∘≈49.5070\cos 45^\circ\approx 49.50 first, then subtract it from 7474 to get c2≈24.50c^2\approx 24.50. A quick sense check catches the slip: two sides of 55 and 77 around a 45∘45^\circ angle cannot close with a side of length 2.83≈1.68\sqrt{2.83}\approx 1.68.

Tip

Which law? SAS or SSS ⇒\Rightarrow Law of Cosines. AAS, ASA, or SSA ⇒\Rightarrow Law of Sines. When you must chase down remaining parts, finish with whichever law keeps the arithmetic cleanest.

Try it
Quick check

Two sides of a triangle are b=6b=6 and c=10c=10, and the angle between them is A=120∘A=120^\circ. Find the third side aa.

Show the answer

Answer: a=14a=14. The givens are SAS, so use a2=b2+c2−2bccos⁡Aa^2=b^2+c^2-2bc\cos A. With cos⁡120∘=−12\cos 120^\circ=-\tfrac12 the correction term is added: a2=36+100−2(6)(10)(−12)=136+60=196a^2=36+100-2(6)(10)\left(-\tfrac12\right)=136+60=196, so a=14a=14. As expected for an obtuse angle, aa is longer than the 136≈11.66\sqrt{136}\approx 11.66 a right angle would give.

Area of a Triangle

Concept
Two area formulas

SAS (two sides and the included angle):

Area=12 absin⁡C=12 bcsin⁡A=12 acsin⁡B.\text{Area}=\tfrac12\,ab\sin C=\tfrac12\,bc\sin A=\tfrac12\,ac\sin B.

Heron's formula (three sides, SSS): with semiperimeter s=a+b+c2s=\dfrac{a+b+c}{2},

Area=s(s−a)(s−b)(s−c).\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}.

Use the SAS formula when you have an included angle; use Heron's when you only have the three sides.

The SAS formula is base times height over two with the height written in trig. Take bb as the base; the height from BB is the leg opposite CC in a right triangle whose hypotenuse is aa, so h=asin⁡Ch=a\sin C and the area is 12b (asin⁡C)\tfrac12 b\,(a\sin C). Heron's formula needs no angle at all: it comes from finding cos⁡C\cos C with the Law of Cosines, turning it into sin⁡C\sin C with sin⁡2C=1−cos⁡2C\sin^2C=1-\cos^2C, and factoring the result.

Example
SAS area: a=6a=6, b=9b=9, C=55∘C=55^\circ
ABCa = 6b = 955°h ≈ 4.91
Base b=9b=9, height asin⁡Ca\sin C: area =12⋅9⋅4.91=\tfrac12\cdot 9\cdot 4.91.
Area=12 absin⁡C=12(6)(9)sin⁡55∘≈22.12.\text{Area}=\tfrac12\,ab\sin C=\tfrac12(6)(9)\sin 55^\circ\approx 22.12.
Example
Heron's formula: a=7a=7, b=8b=8, c=9c=9
s=7+8+92=12,Area=12(12−7)(12−8)(12−9)=12⋅5⋅4⋅3=720≈26.83.s=\frac{7+8+9}{2}=12,\qquad \text{Area}=\sqrt{12(12-7)(12-8)(12-9)}=\sqrt{12\cdot 5\cdot 4\cdot 3}=\sqrt{720}\approx 26.83.
Reminder — Heron's formula:  A=s(s−a)(s−b)(s−c),s=a+b+c2A=\sqrt{s(s-a)(s-b)(s-c)},\qquad s=\frac{a+b+c}{2}
Try it
Quick check

(a) Two sides of a triangle are 55 and 88 and the angle between them is 150∘150^\circ. Find the area. (b) Find the area of the triangle with sides 1313, 1414, 1515.

Show the answer

(a) Area =10=10. Included angle, so 12(5)(8)sin⁡150∘=20⋅12=10\tfrac12(5)(8)\sin 150^\circ=20\cdot\tfrac12=10. (An obtuse angle is fine: sin⁡150∘=sin⁡30∘\sin 150^\circ=\sin 30^\circ.) (b) Area =84=84. Only sides are known, so use Heron: s=13+14+152=21s=\tfrac{13+14+15}{2}=21 and 21⋅8⋅7⋅6=7056=84\sqrt{21\cdot 8\cdot 7\cdot 6}=\sqrt{7056}=84.

Applications: Surveying, Navigation, and Bearings

Concept
Setting up word problems
  • Bearing is an angle measured clockwise from due north, e.g. N40∘40^\circE means 40∘40^\circ east of north. A three-digit bearing like 120∘120^\circ is measured clockwise from north.
  • Draw the picture, label every known length and angle, then decide: is there a complete angle–opposite-side pair (Law of Sines) or not (Law of Cosines)?
  • In a change-of-course problem, the interior angle at the turning point is 180∘180^\circ minus the change in bearing.
  • Multi-triangle problems: solve one triangle to get a shared side, then use that side in the next triangle.

Word problems hide an ordinary triangle. The work is in the drawing: a bearing is an angle at a north line, so draw a short north arrow at every point where a bearing is measured, and the angles inside the triangle then come from parallel north lines (co-interior angles add to 180∘180^\circ). Once each vertex angle and side of the triangle is labelled, the choice between the two laws is the same as in a bare triangle.

Example
Surveying: distance across a pond
ABT500 mAT ≈ 469.14 m40°65°75°pond
The baseline ABAB is measured; the distance across the pond is computed.

Points AA and BB are 500500 m apart on one shore. A tree TT across the pond makes ∠TAB=40∘\angle TAB=40^\circ and ∠TBA=65∘\angle TBA=65^\circ. Find ATAT.

∠T=180∘−40∘−65∘=75∘,AT=ABsin⁡Bsin⁡T=500sin⁡65∘sin⁡75∘≈469.14 m.\angle T=180^\circ-40^\circ-65^\circ=75^\circ,\qquad AT=\frac{AB\sin B}{\sin T}=\frac{500\sin 65^\circ}{\sin 75^\circ}\approx 469.14\text{ m}.
Example
Navigation: change of course
NN30°70°140°20 mi30 mid ≈ 47.11 miO
Bearings are measured clockwise from north; the turn leaves a 140∘140^\circ angle inside the triangle.

A ship sails 2020 mi on bearing N30∘30^\circE, then 3030 mi on bearing N70∘70^\circE. How far is it from the start?

The change in bearing is 70∘−30∘=40∘70^\circ-30^\circ=40^\circ, so the interior angle at the turn is 180∘−40∘=140∘180^\circ-40^\circ=140^\circ. With the two legs as the known sides,

d2=202+302−2(20)(30)cos⁡140∘≈2219.25  ⇒  d≈47.11 mi.d^2=20^2+30^2-2(20)(30)\cos 140^\circ\approx 2219.25\;\Rightarrow\; d\approx 47.11\text{ mi}.
Tip

Final reminders. Sketch first. Watch SSA for two answers. Use the Law of Cosines to find the largest angle when all three sides are known. State units and round only at the end.

Try it
Quick check

Solve the triangle completely: A=55∘A=55^\circ, B=45∘B=45^\circ, included side c=18c=18. Find CC, aa, and bb.

Work it on paper before revealing the answer.

Show the answer

Answer: C=80∘C=80^\circ; a≈14.97a\approx 14.97; b≈12.92b\approx 12.92.

Going Deeper: Advanced Triangle Laws

Everything below builds on the two laws you already know. The rounding convention from the top of the sheet still applies: angles to the nearest tenth of a degree, lengths and areas to the nearest hundredth.

Concept
The Extended Law of Sines: circumradius and inradius

The common ratio in the Law of Sines is not just a number—it equals the diameter of the circumscribed circle (the circle through all three vertices, radius RR):

asin⁡A=bsin⁡B=csin⁡C=2R.\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R.

Two more links tie area to these radii. Writing KK for the area and s=a+b+c2s=\dfrac{a+b+c}{2} for the semiperimeter,

R=abc4K,r=Ks,R=\frac{abc}{4K},\qquad\qquad r=\frac{K}{s},

where rr is the radius of the inscribed circle (the circle tangent to all three sides). The second identity comes from splitting the triangle into three smaller triangles, one per side, each with height rr: K=12ra+12rb+12rc=rsK=\tfrac12 r a+\tfrac12 r b+\tfrac12 r c=rs.

Example
Circumradius and inradius of the 77–88–99 triangle
ABCRr7
The 77–88–99 triangle with its circumcircle (R≈4.70R\approx 4.70) and incircle (r≈2.24r\approx 2.24).

From the earlier Heron computation, a=7a=7, b=8b=8, c=9c=9 give s=12s=12 and

K=720≈26.83.K=\sqrt{720}\approx 26.83.

Therefore

R=abc4K=7⋅8⋅94720=5044720≈4.70,r=Ks=72012≈2.24.R=\frac{abc}{4K}=\frac{7\cdot 8\cdot 9}{4\sqrt{720}}=\frac{504}{4\sqrt{720}}\approx 4.70, \qquad r=\frac{K}{s}=\frac{\sqrt{720}}{12}\approx 2.24.

Check with the Extended Law of Sines. The largest angle is opposite c=9c=9:

cos⁡C=a2+b2−c22ab=49+64−812(7)(8)=32112≈0.2857  ⇒  C≈73.4∘,\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{49+64-81}{2(7)(8)}=\frac{32}{112}\approx 0.2857\;\Rightarrow\;C\approx 73.4^\circ,

and csin⁡C=9sin⁡73.4∘≈9.39=2R\dfrac{c}{\sin C}=\dfrac{9}{\sin 73.4^\circ}\approx 9.39=2R, so R≈4.70R\approx 4.70 as before.

Concept
Cevians: Stewart's Theorem, medians, and angle bisectors

A cevian is any segment from a vertex to the opposite side. Draw the cevian ADAD from AA to point DD on side a=BCa=BC, and let

BD=m,DC=n,AD=d,m+n=a.BD=m,\qquad DC=n,\qquad AD=d,\qquad m+n=a.

Stewart's Theorem relates the cevian length to the three sides (mnemonic man + dad = bmb + cnc):

b2m+c2n=a (d2+mn).b^2 m+c^2 n=a\,(d^2+mn).

Two special cevians have ready-made length formulas:

  • Median to side aa (here m=n=a2m=n=\tfrac{a}{2}):   ma2=2b2+2c2−a24.\displaystyle m_a^2=\frac{2b^2+2c^2-a^2}{4}.
  • Angle bisector from AA (it splits aa in the ratio c:bc:b, so m=acb+cm=\dfrac{ac}{b+c}, n=abb+cn=\dfrac{ab}{b+c}): @@BLOCK2@@
Example
Median and angle bisector in the 77–88–99 triangle

Take a=7a=7, b=8b=8, c=9c=9 and cevians drawn from vertex AA to side a=BCa=BC.

Median.

ma2=2(82)+2(92)−724=128+162−494=2414=60.25  ⇒  ma≈7.76.m_a^2=\frac{2(8^2)+2(9^2)-7^2}{4}=\frac{128+162-49}{4}=\frac{241}{4}=60.25\;\Rightarrow\;m_a\approx 7.76.

Angle bisector.

ta2=bc[(b+c)2−a2](b+c)2=8⋅9 (172−72)172=72(289−49)289=72⋅240289≈59.79  ⇒  ta≈7.73.t_a^2=\frac{bc\big[(b+c)^2-a^2\big]}{(b+c)^2}=\frac{8\cdot 9\,(17^2-7^2)}{17^2}=\frac{72(289-49)}{289}=\frac{72\cdot 240}{289}\approx 59.79 \;\Rightarrow\;t_a\approx 7.73.

Verify the median by Stewart with m=n=3.5m=n=3.5:

b2m+c2n=64(3.5)+81(3.5)=507.5,a(d2+mn)=7(60.25+12.25)=7(72.5)=507.5. ✓b^2m+c^2n=64(3.5)+81(3.5)=507.5,\qquad a(d^2+mn)=7\big(60.25+12.25\big)=7(72.5)=507.5.\ \checkmark
Concept
Area, revisited: five methods and area ratios

Depending on what you are given, any of these computes the area KK:

  • SAS: K=12absin⁡CK=\tfrac12 ab\sin C (two sides and the included angle).
  • Heron: K=s(s−a)(s−b)(s−c)K=\sqrt{s(s-a)(s-b)(s-c)} (three sides).
  • Inradius: K=rsK=rs (area from the inscribed circle).
  • Circumradius: K=abc4RK=\dfrac{abc}{4R} (area from the circumscribed circle).
  • Two-angles-and-a-side (AAS/ASA): K=a2sin⁡Bsin⁡C2sin⁡AK=\dfrac{a^2\sin B\sin C}{2\sin A}.

Area ratios. A cevian from AA to DD on BCBC makes two triangles ABDABD and ADCADC that share the same height from AA, so

[ABD][ADC]=BDDC=mn.\frac{[ABD]}{[ADC]}=\frac{BD}{DC}=\frac{m}{n}.

For a quadrilateral, draw a diagonal to split it into two triangles and add their areas—the diagonal itself is found with the Law of Cosines.

Example
Surveyed quadrilateral field from several bearings
ABCD608010070110°≈ 115.25
The diagonal ACAC splits the field into an SAS triangle and an SSS triangle.

A four-sided plot ABCDABCD is surveyed. The walk from AA to BB is 6060 m, and from BB to CC is 8080 m; the recorded bearings make the interior angle at BB equal to 110∘110^\circ. The remaining sides are CD=100CD=100 m and DA=70DA=70 m. Find the plot's area.

Step 1 — diagonal ACAC (Law of Cosines in △ABC\triangle ABC).

AC2=602+802−2(60)(80)cos⁡110∘≈13283.39  ⇒  AC≈115.25 m.AC^2=60^2+80^2-2(60)(80)\cos 110^\circ\approx 13283.39\;\Rightarrow\;AC\approx 115.25\text{ m}.

Step 2 — area of △ABC\triangle ABC (SAS).

[ABC]=12(60)(80)sin⁡110∘≈2255.26 m2.[ABC]=\tfrac12(60)(80)\sin 110^\circ\approx 2255.26\text{ m}^2.

Step 3 — area of △ACD\triangle ACD (Heron, three sides now known). With sides 115.25115.25, 100100, 7070 and s≈142.63s\approx 142.63,

[ACD]=s(s−115.25)(s−100)(s−70)≈3476.59 m2.[ACD]=\sqrt{s(s-115.25)(s-100)(s-70)}\approx 3476.59\text{ m}^2.

Total area ≈2255.26+3476.59≈5731.85 m2\approx 2255.26+3476.59\approx 5731.85\text{ m}^2.

Tip
05100 triangles2 triangles1 triangle(1, right)a
Number of triangles as aa grows, for A=30∘A=30^\circ, b=10b=10.

Parameter-driven ambiguous-case counting. Fix an acute angle AA and side bb, then ask how the number of SSA triangles changes as the opposite side aa grows. The two thresholds are a=bsin⁡Aa=b\sin A (grazing / right triangle) and a=ba=b:

a<bsin⁡A⏟0∣a=bsin⁡A⏟1 (right)∣bsin⁡A<a<b⏟2∣a≥b⏟1.\underbrace{a<b\sin A}_{\textbf{0}}\quad\big|\quad\underbrace{a=b\sin A}_{\textbf{1 (right)}}\quad\big|\quad\underbrace{b\sin A<a<b}_{\textbf{2}}\quad\big|\quad\underbrace{a\ge b}_{\textbf{1}}.

For example, with A=30∘A=30^\circ and b=10b=10 the thresholds sit at bsin⁡A=5b\sin A=5 and b=10b=10 (see the number line above).

If instead AA is obtuse, only a>ba>b yields a triangle (exactly one); a≤ba\le b gives none.

Concept
Proving a relationship: a2=b(b+c) ⇒ A=2Ba^2=b(b+c)\ \Rightarrow\ A=2B

Advanced problems often ask you to prove a triangle identity rather than solve for a number. Here is a model proof using the Law of Sines together with a product-to-difference identity.

Claim. If the sides of a triangle satisfy a2=b(b+c)a^2=b(b+c), then A=2BA=2B.

Proof. By the Law of Sines write a=2Rsin⁡Aa=2R\sin A, b=2Rsin⁡Bb=2R\sin B, c=2Rsin⁡Cc=2R\sin C. Substituting into a2=b2+bca^2=b^2+bc and cancelling the common 4R24R^2,

sin⁡2A=sin⁡2B+sin⁡Bsin⁡C.\sin^2 A=\sin^2 B+\sin B\sin C.

Because C=180∘−(A+B)C=180^\circ-(A+B), we have sin⁡C=sin⁡(A+B)\sin C=\sin(A+B), so

sin⁡2A−sin⁡2B=sin⁡B sin⁡(A+B).\sin^2 A-\sin^2 B=\sin B\,\sin(A+B).

Apply the identity sin⁡2A−sin⁡2B=sin⁡(A+B)sin⁡(A−B)\sin^2 A-\sin^2 B=\sin(A+B)\sin(A-B) to the left side:

sin⁡(A+B)sin⁡(A−B)=sin⁡B sin⁡(A+B).\sin(A+B)\sin(A-B)=\sin B\,\sin(A+B).

Since 0∘<A+B<180∘0^\circ<A+B<180^\circ, sin⁡(A+B)≠0\sin(A+B)\neq 0; dividing gives sin⁡(A−B)=sin⁡B\sin(A-B)=\sin B. The two ways this can hold are A−B=BA-B=B or A−B=180∘−BA-B=180^\circ-B. The latter forces A=180∘A=180^\circ, impossible in a triangle, so A−B=BA-B=B, i.e. A=2BA=2B. ■\qquad\blacksquare

Tip

Advanced toolkit, at a glance. Need a radius? 2R=asin⁡A2R=\dfrac{a}{\sin A} and r=Ksr=\dfrac{K}{s}. Need a cevian? Stewart's b2m+c2n=a(d2+mn)b^2m+c^2n=a(d^2+mn), with ready formulas for medians and bisectors. Need an area? Pick from SAS, Heron, rsrs, abc4R\dfrac{abc}{4R}, or split a polygon along a diagonal. Proving an identity? Convert sides to sines with the Law of Sines, then lean on sum/product identities.

Try it
Quick check

a=9a=9, b=6b=6, A=50∘A=50^\circ. Find sin⁡B\sin B and the acute angle BB. How many triangles are there?

Work it on paper before revealing the answer.

Show the answer

Answer: sin⁡B≈0.5107\sin B\approx 0.5107, B≈30.7∘B\approx 30.7^\circ; one triangle (since a>ba>b).

Why It Works

Law of sines: drop the altitude hh from CC; in the two right triangles, h=bsin⁡A=asin⁡Bh = b\sin A = a\sin B, so asin⁡A=bsin⁡B\tfrac{a}{\sin A} = \tfrac{b}{\sin B}. Law of cosines: place CC at the origin and AA on the xx-axis; then B=(acos⁡C,asin⁡C)B = (a\cos C, a\sin C), and the distance formula gives c2=(b−acos⁡C)2+(asin⁡C)2=a2+b2−2abcos⁡Cc^2 = (b - a\cos C)^2 + (a\sin C)^2 = a^2 + b^2 - 2ab\cos C after using sin⁡2+cos⁡2=1\sin^2 + \cos^2 = 1. When C=90∘C = 90^\circ the correction term vanishes and Pythagoras returns.

Reminder — Distance and midpoint:  d=(x2−x1)2+(y2−y1)2,M=(x1+x22, y1+y22)d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, \qquad M=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)
Reminder — The Pythagorean theorem:  a2+b2=c2a^2+b^2=c^2
Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Law of Sines
asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}

What it means. In any triangle each side is proportional to the sine of its opposite angle.

Example. asin⁡30∘=10sin⁡90∘⇒a=10⋅12=5\dfrac{a}{\sin 30^\circ}=\dfrac{10}{\sin 90^\circ}\Rightarrow a=10\cdot\tfrac12=5.

Why it works. Drop the height hh to side cc. Then h=bsin⁡Ah=b\sin A from one right triangle and h=asin⁡Bh=a\sin B from the other, so bsin⁡A=asin⁡Bb\sin A=a\sin B, which rearranges to asin⁡A=bsin⁡B\tfrac{a}{\sin A}=\tfrac{b}{\sin B}.

Tip. Use it when you have an angle paired with its opposite side (AAS, ASA, SSA). The SSA case can give two triangles — check whether a second angle also fits.

Tip
Law of Cosines
c2=a2+b2−2abcos⁡Cc^{2}=a^{2}+b^{2}-2ab\cos C

What it means. Pythagoras with a correction term for the angle not being right.

Example. Sides 3,43,4 with included angle 60∘60^\circ: c2=9+16−2⋅3⋅4⋅12=13c^2=9+16-2\cdot3\cdot4\cdot\tfrac12=13.

Why it works. Place CC at the origin with aa along the xx-axis. The other vertex sits at (bcos⁡C, bsin⁡C)(b\cos C,\ b\sin C), and the distance formula to (a,0)(a,0) gives c2=(bcos⁡C−a)2+(bsin⁡C)2c^2=(b\cos C-a)^2+(b\sin C)^2. Expanding and using sin⁡2+cos⁡2=1\sin^2+\cos^2=1 leaves a2+b2−2abcos⁡Ca^2+b^2-2ab\cos C.

Tip. When C=90∘C=90^\circ, cos⁡C=0\cos C=0 and it collapses to a2+b2=c2a^2+b^2=c^2. Use it for SSS and SAS, where the Law of Sines cannot start.

Tip
Area from two sides and the included angle
Area=12absin⁡C\text{Area}=\tfrac12 ab\sin C

What it means. Half the product of two sides times the sine of the angle between them.

Example. Sides 6,86,8 with included angle 30∘30^\circ: Area =12⋅6⋅8⋅12=12=\tfrac12\cdot6\cdot8\cdot\tfrac12=12.

Why it works. Area is 12(base)(height)\tfrac12(\text{base})(\text{height}). Taking aa as the base, the height from the far vertex is bsin⁡Cb\sin C, giving 12absin⁡C\tfrac12 ab\sin C.

Tip. The angle must be between the two sides. With all three sides instead, use Heron's formula.

Tip
Heron's formula
A=s(s−a)(s−b)(s−c),s=a+b+c2A=\sqrt{s(s-a)(s-b)(s-c)},\qquad s=\frac{a+b+c}{2}

What it means. The area of a triangle from its three sides alone — no angle or height needed.

Example. Sides 3,4,53,4,5: s=6s=6, so A=6⋅3⋅2⋅1=6A=\sqrt{6\cdot 3\cdot 2\cdot 1}=6.

Why it works. Start from A=12absin⁡CA=\tfrac12 ab\sin C, replace sin⁡C\sin C with 1−cos⁡2C\sqrt{1-\cos^2C}, and substitute cos⁡C\cos C from the Law of Cosines. The algebra factors into the four bracketed terms.

Tip. ss is the SEMI-perimeter — half the perimeter. Forgetting the halving is the usual slip.