Triangle Trigonometry and Vectors

Study Sheet

Triangle Trigonometry and Vectors

Laws of Sines and Cosines, area, vectors, dot products, and applications

The Law of Sines

Throughout, a triangle is labeled so that side aa is opposite angle AA, side bb is opposite BB, and side cc is opposite CC. Decimal answers are rounded to the nearest tenth (angles in degrees, lengths in the given units) unless stated otherwise.

The laws of sines and cosines solve any triangle; Heron's formula gives the area from three sides. Vectors carry direction and magnitude, add tip-to-tail, and their dot product measures alignment — with applications to forces, velocities, work, and projections. Unit vectors and components make everything computational.

Concept
Law of Sines

For any triangle ABCABC,

sin⁡Aa=sin⁡Bb=sin⁡Cc.\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}.

Use it when you know AAS, ASA, or SSA (an angle and its opposite side, plus one more piece).

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. In the two right triangles it creates, h=bsin⁡Ah=b\sin A and h=asin⁡Bh=a\sin B; setting these equal gives sin⁡Aa=sin⁡Bb\dfrac{\sin A}{a}=\dfrac{\sin B}{b}. The same argument with the altitude from AA brings in CC.

Example
AAS: solve the triangle
ABCa = 12b ≈ 19.0c ≈ 20.635°65°80°
The longest side, cc, faces the largest angle, 80∘80^\circ.

Given A=35∘A=35^\circ, B=65∘B=65^\circ, a=12a=12. Then C=180∘−35∘−65∘=80∘C=180^\circ-35^\circ-65^\circ=80^\circ.

b=asin⁡Bsin⁡A=12sin⁡65∘sin⁡35∘=12(0.9063)0.5736≈19.0,c=asin⁡Csin⁡A=12sin⁡80∘sin⁡35∘≈20.6.b=\frac{a\sin B}{\sin A}=\frac{12\sin 65^\circ}{\sin 35^\circ}=\frac{12(0.9063)}{0.5736}\approx 19.0,\qquad c=\frac{a\sin C}{\sin A}=\frac{12\sin 80^\circ}{\sin 35^\circ}\approx 20.6.
Side note
Where it goes wrong

Using the law of sines for SAS. Two sides and the INCLUDED angle need the law of cosines first; the law of sines needs a side paired with its own opposite angle.

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

The ambiguous case (SSA): 0, 1, or 2 triangles

When you are given two sides and an angle opposite one of them (say aa, bb, and AA), the height from CC to the base is h=bsin⁡Ah=b\sin A. Compare aa with hh and bb:

Tip

AA acute: if a<bsin⁡Aa<b\sin A no triangle; if a=bsin⁡Aa=b\sin A one (right) triangle; if bsin⁡A<a<bb\sin A<a<b two triangles; if a≥ba\ge b one triangle.   AA obtuse: one triangle if a>ba>b, otherwise none.

Example
Full ambiguous case: two triangles

Solve the triangle with a=8a=8, b=10b=10, A=40∘A=40^\circ. Since AA is acute and bsin⁡A=10sin⁡40∘≈6.43b\sin A=10\sin 40^\circ\approx 6.43, and 6.43<8<106.43<8<10, there are two triangles.

sin⁡B=bsin⁡Aa=10sin⁡40∘8=6.42798≈0.8035  ⇒  B≈53.5∘  or  B′≈180∘−53.5∘=126.5∘.\sin B=\frac{b\sin A}{a}=\frac{10\sin 40^\circ}{8}=\frac{6.4279}{8}\approx 0.8035 \;\Rightarrow\; B\approx 53.5^\circ \ \text{ or }\ B'\approx 180^\circ-53.5^\circ=126.5^\circ.

Triangle 1: B≈53.5∘B\approx 53.5^\circ, so C=180∘−40∘−53.5∘=86.5∘C=180^\circ-40^\circ-53.5^\circ=86.5^\circ and c=asin⁡Csin⁡A=8sin⁡86.5∘sin⁡40∘≈12.4c=\dfrac{a\sin C}{\sin A}=\dfrac{8\sin 86.5^\circ}{\sin 40^\circ}\approx 12.4. Triangle 2: B′≈126.5∘B'\approx 126.5^\circ, so C′=180∘−40∘−126.5∘=13.5∘C'=180^\circ-40^\circ-126.5^\circ=13.5^\circ and c′=8sin⁡13.5∘sin⁡40∘≈2.9c'=\dfrac{8\sin 13.5^\circ}{\sin 40^\circ}\approx 2.9.

Try it
Quick check

Solve for cc: A=40∘A=40^\circ, C=70∘C=70^\circ, a=15a=15. Round to the nearest tenth.

Show the answer

Answer: c≈21.9c\approx 21.9. The pair AA, aa is complete, so csin⁡70∘=15sin⁡40∘\dfrac{c}{\sin70^\circ}=\dfrac{15}{\sin40^\circ} and c=15sin⁡70∘sin⁡40∘≈21.9c=\dfrac{15\sin70^\circ}{\sin40^\circ}\approx 21.9. (Also B=70∘B=70^\circ, so the triangle is isosceles with b=cb=c.)

The Law of Cosines

Concept
Law of Cosines

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,\qquad b^2=a^2+c^2-2ac\cos B,\qquad c^2=a^2+b^2-2ab\cos C.

Use it for SAS (find the third side) or SSS (find any angle). Solving for an angle: cos⁡C=a2+b2−c22ab\cos C=\dfrac{a^2+b^2-c^2}{2ab}.

Put CC at the origin and AA at (b,0)(b,0); then B=(acos⁡C, asin⁡C)B=(a\cos C,\ a\sin C), and the distance formula gives c2=(b−acos⁡C)2+a2sin⁡2C=a2+b2−2abcos⁡Cc^2=(b-a\cos C)^2+a^2\sin^2C=a^2+b^2-2ab\cos C. The last term corrects Pythagoras: it is zero for a right angle and turns positive when CC is obtuse.

Example
SAS then SSS
ABCa = 7b = 8c = 973.4°
The SSS triangle: the 73.4∘73.4^\circ angle faces the longest side.

SAS: a=5a=5, b=7b=7, C=45∘C=45^\circ. Then c2=52+72−2(5)(7)cos⁡45∘=74−70(0.7071)≈24.50c^2=5^2+7^2-2(5)(7)\cos 45^\circ=74-70(0.7071)\approx 24.50, so c≈4.9c\approx 4.9. SSS: a=7a=7, b=8b=8, c=9c=9. The largest angle is CC: cos⁡C=72+82−922(7)(8)=32112=0.2857\cos C=\dfrac{7^2+8^2-9^2}{2(7)(8)}=\dfrac{32}{112}=0.2857, so C≈73.4∘C\approx 73.4^\circ.

Tip

Tip: The Law of Cosines has no ambiguity. When SSS or SAS is given, start there; only switch to the Law of Sines once you have a side and its opposite angle. To avoid the ambiguous case entirely, find the largest unknown angle with the Law of Cosines first.

Try it
Quick check

Find the third side of a triangle with sides 66 and 88 and included angle 100∘100^\circ (nearest tenth).

Show the answer

Answer: ≈10.8\approx 10.8. c2=36+64−2(6)(8)cos⁡100∘=100−96(−0.1736)≈116.67c^2=36+64-2(6)(8)\cos100^\circ=100-96(-0.1736)\approx 116.67, so c≈10.8c\approx 10.8. It exceeds the 1010 a right angle would give, as it must for an obtuse angle.

Area of a Triangle

Concept
Two area formulas

SAS formula (two sides and the included angle):

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

Heron's formula (three sides), 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)}.
Example
Both formulas

SAS: a=5a=5, b=7b=7, C=45∘C=45^\circ: Area =12(5)(7)sin⁡45∘=17.5(0.7071)≈12.4=\tfrac12(5)(7)\sin 45^\circ=17.5(0.7071)\approx 12.4 square units. Heron: a=7a=7, b=8b=8, c=9c=9: s=7+8+92=12s=\tfrac{7+8+9}{2}=12, so Area =12(5)(4)(3)=720≈26.8=\sqrt{12(5)(4)(3)}=\sqrt{720}\approx 26.8 square units.

Try it
Quick check

Find the area of the triangle with sides 1010, 1010, 1212.

Show the answer

Answer: 4848. Heron: s=10+10+122=16s=\tfrac{10+10+12}{2}=16 and 16⋅6⋅6⋅4=2304=48\sqrt{16\cdot6\cdot6\cdot4}=\sqrt{2304}=48. (Check: the altitude to the base 1212 is 100−36=8\sqrt{100-36}=8, and 12⋅12⋅8=48\tfrac12\cdot12\cdot8=48.)

Vectors: Component Form and Operations

Concept
Vector basics

A vector in component form is u⃗=⟨a,b⟩\vec u=\langle a,b\rangle.

  • Magnitude: ∥u⃗∥=a2+b2\|\vec u\|=\sqrt{a^2+b^2}.
  • Direction angle θ\theta (from the positive xx-axis): tan⁡θ=ba\tan\theta=\dfrac{b}{a}; place θ\theta in the correct quadrant.
  • Add / subtract: ⟨a,b⟩±⟨c,d⟩=⟨a±c,  b±d⟩\langle a,b\rangle\pm\langle c,d\rangle=\langle a\pm c,\;b\pm d\rangle.
  • Scalar multiple: k⟨a,b⟩=⟨ka,kb⟩k\langle a,b\rangle=\langle ka,kb\rangle.
  • Unit vector in the direction of u⃗\vec u: u^=1∥u⃗∥u⃗\hat u=\dfrac{1}{\|\vec u\|}\vec u.
  • Polar / trig form: u⃗=∥u⃗∥ ⟨cos⁡θ,sin⁡θ⟩\vec u=\|\vec u\|\,\langle\cos\theta,\sin\theta\rangle.
Example
Components, magnitude, direction, unit vector
xyu = ⟨3, 4⟩v = ⟨−2, 5⟩53.1°111.8°
Direction angles are measured from the positive xx-axis; v⃗\vec v is in Quadrant II.

Let u⃗=⟨3,4⟩\vec u=\langle 3,4\rangle. Then ∥u⃗∥=32+42=5\|\vec u\|=\sqrt{3^2+4^2}=5, direction θ=tan⁡−1 ⁣43≈53.1∘\theta=\tan^{-1}\!\frac{4}{3}\approx 53.1^\circ (Quadrant I), and u^=15⟨3,4⟩=⟨0.6,0.8⟩\hat u=\frac{1}{5}\langle 3,4\rangle=\langle 0.6,0.8\rangle. In trig form, u⃗=5⟨cos⁡53.1∘,sin⁡53.1∘⟩\vec u=5\langle\cos 53.1^\circ,\sin 53.1^\circ\rangle. For v⃗=⟨−2,5⟩\vec v=\langle -2,5\rangle: ∥v⃗∥=29≈5.4\|\vec v\|=\sqrt{29}\approx 5.4 and the reference angle is tan⁡−1 ⁣52≈68.2∘\tan^{-1}\!\frac{5}{2}\approx 68.2^\circ; since v⃗\vec v is in Quadrant II, θ≈180∘−68.2∘=111.8∘\theta\approx 180^\circ-68.2^\circ=111.8^\circ.

Try it
Quick check

Find the magnitude and direction angle of ⟨−5,12⟩\langle -5,12\rangle.

Show the answer

Answer: magnitude 1313, direction ≈112.6∘\approx 112.6^\circ. 25+144=13\sqrt{25+144}=13. The reference angle is tan⁡−1125≈67.4∘\tan^{-1}\tfrac{12}{5}\approx 67.4^\circ, and the vector points into Quadrant II, so θ≈180∘−67.4∘=112.6∘\theta\approx180^\circ-67.4^\circ=112.6^\circ.

The Dot Product, Angle, Projection, and Work

Concept
Dot product and its uses

For u⃗=⟨a,b⟩\vec u=\langle a,b\rangle and v⃗=⟨c,d⟩\vec v=\langle c,d\rangle:

u⃗⋅v⃗=ac+bd=∥u⃗∥ ∥v⃗∥cos⁡θ.\vec u\cdot\vec v=ac+bd=\|\vec u\|\,\|\vec v\|\cos\theta.
  • Angle between: cos⁡θ=u⃗⋅v⃗∥u⃗∥ ∥v⃗∥\cos\theta=\dfrac{\vec u\cdot\vec v}{\|\vec u\|\,\|\vec v\|}.
  • Orthogonal (perpendicular) exactly when u⃗⋅v⃗=0\vec u\cdot\vec v=0.
  • Projection of u⃗\vec u onto v⃗\vec v:  proj⁡v⃗u⃗=(u⃗⋅v⃗∥v⃗∥2)v⃗\ \operatorname{proj}_{\vec v}\vec u=\left(\dfrac{\vec u\cdot\vec v}{\|\vec v\|^2}\right)\vec v.
  • Work done by a constant force F⃗\vec F over displacement d⃗\vec d:  W=F⃗⋅d⃗\ W=\vec F\cdot\vec d.

The Law of Cosines on the triangle with sides u⃗\vec u, v⃗\vec v and u⃗−v⃗\vec u-\vec v gives ∥u⃗−v⃗∥2=∥u⃗∥2+∥v⃗∥2−2∥u⃗∥∥v⃗∥cos⁡θ\|\vec u-\vec v\|^2=\|\vec u\|^2+\|\vec v\|^2-2\|\vec u\|\|\vec v\|\cos\theta, while expanding in components gives the same expression with ac+bdac+bd in place of ∥u⃗∥∥v⃗∥cos⁡θ\|\vec u\|\|\vec v\|\cos\theta. That is why the quick component product measures the angle.

Example
Full dot-product and projection
xyuv31.3°
The dashed arrow is proj⁡v⃗u⃗\operatorname{proj}_{\vec v}\vec u, the shadow of u⃗\vec u on the line of v⃗\vec v.

Let u⃗=⟨3,4⟩\vec u=\langle 3,4\rangle and v⃗=⟨5,2⟩\vec v=\langle 5,2\rangle.

u⃗⋅v⃗=3(5)+4(2)=23,∥u⃗∥=5,∥v⃗∥=29≈5.385.\vec u\cdot\vec v=3(5)+4(2)=23,\qquad \|\vec u\|=5,\quad \|\vec v\|=\sqrt{29}\approx 5.385.

Angle: cos⁡θ=23529=2326.93≈0.8541\cos\theta=\dfrac{23}{5\sqrt{29}}=\dfrac{23}{26.93}\approx 0.8541, so θ≈31.3∘\theta\approx 31.3^\circ (not orthogonal, since u⃗⋅v⃗≠0\vec u\cdot\vec v\neq 0). Projection of u⃗\vec u onto v⃗\vec v:

proj⁡v⃗u⃗=23(29)2⟨5,2⟩=2329⟨5,2⟩=⟨11529,4629⟩≈⟨3.97,1.59⟩.\operatorname{proj}_{\vec v}\vec u=\frac{23}{(\sqrt{29})^2}\langle 5,2\rangle=\frac{23}{29}\langle 5,2\rangle=\left\langle \tfrac{115}{29},\tfrac{46}{29}\right\rangle\approx\langle 3.97,1.59\rangle.

Work: A force F⃗=⟨4,3⟩\vec F=\langle 4,3\rangle N moving an object along d⃗=⟨10,0⟩\vec d=\langle 10,0\rangle m does W=F⃗⋅d⃗=4(10)+3(0)=40W=\vec F\cdot\vec d=4(10)+3(0)=40 joules.

Tip

Tip: u⃗⋅v⃗>0\vec u\cdot\vec v>0 means the angle is acute; u⃗⋅v⃗<0\vec u\cdot\vec v<0 means obtuse; u⃗⋅v⃗=0\vec u\cdot\vec v=0 means perpendicular. The projection is a vector along v⃗\vec v; the scalar u⃗⋅v⃗∥v⃗∥\dfrac{\vec u\cdot\vec v}{\|\vec v\|} is its signed length (the “component of u⃗\vec u along v⃗\vec v”).

Try it
Quick check

(a) Are ⟨1,2⟩\langle 1,2\rangle and ⟨4,−2⟩\langle 4,-2\rangle orthogonal? (b) Find proj⁡v⃗u⃗\operatorname{proj}_{\vec v}\vec u for u⃗=⟨3,4⟩\vec u=\langle 3,4\rangle, v⃗=⟨1,0⟩\vec v=\langle 1,0\rangle.

Show the answer

(a) Yes: 1⋅4+2(−2)=01\cdot4+2(-2)=0. (b) ⟨3,0⟩\langle 3,0\rangle. u⃗⋅v⃗∥v⃗∥2v⃗=31⟨1,0⟩\dfrac{\vec u\cdot\vec v}{\|\vec v\|^2}\vec v=\dfrac{3}{1}\langle1,0\rangle: projecting onto the xx-axis keeps the xx-component.

Applications: Bearings, Forces, and Navigation

Concept
Setting up applied problems

Bearings are measured clockwise from north (e.g. N40∘40^\circE is 40∘40^\circ east of due north). To convert a compass bearing to a standard direction angle from the positive xx-axis (east), use θstd=90∘−(bearing east of north)\theta_{\text{std}}=90^\circ-(\text{bearing east of north}). Resultant force / velocity: write each vector in components ⟨∥F⃗∥cos⁡θ, ∥F⃗∥sin⁡θ⟩\langle\|\vec F\|\cos\theta,\ \|\vec F\|\sin\theta\rangle, add them, then find the magnitude and direction of the sum.

Example
Resultant force
50 N40 NR ≈ 64.0 N
The resultant is the diagonal of the parallelogram on the two forces.

Two forces act on a point: F⃗1=50\vec F_1=50 N at 30∘30^\circ and F⃗2=40\vec F_2=40 N at 120∘120^\circ (standard angles).

F⃗1=⟨50cos⁡30∘,50sin⁡30∘⟩≈⟨43.3,25.0⟩,F⃗2=⟨40cos⁡120∘,40sin⁡120∘⟩=⟨−20.0,34.6⟩.\vec F_1=\langle 50\cos 30^\circ,50\sin 30^\circ\rangle\approx\langle 43.3,25.0\rangle,\quad \vec F_2=\langle 40\cos 120^\circ,40\sin 120^\circ\rangle=\langle -20.0,34.6\rangle.

Resultant R⃗=⟨23.3,59.6⟩\vec R=\langle 23.3,59.6\rangle, so ∥R⃗∥=23.32+59.62≈64.0\|\vec R\|=\sqrt{23.3^2+59.6^2}\approx 64.0 N at θ=tan⁡−1 ⁣59.623.3≈68.7∘\theta=\tan^{-1}\!\frac{59.6}{23.3}\approx 68.7^\circ.

Example
Navigation with the Law of Cosines
120°120 mi90 mid ≈ 182.5 mi
Bearings N50∘50^\circE then S70∘70^\circE leave a 120∘120^\circ angle at the turn.

A plane flies 120120 mi on bearing N50∘50^\circE, then turns and flies 9090 mi on bearing S70∘70^\circE. At the turning point the angle between the reversed first leg (S50∘50^\circW) and the new heading (S70∘70^\circE) is 50∘+70∘=120∘50^\circ+70^\circ=120^\circ. Its distance from the start is

d=1202+902−2(120)(90)cos⁡120∘=14400+8100+10800=33300≈182.5 mi.d=\sqrt{120^2+90^2-2(120)(90)\cos 120^\circ}=\sqrt{14400+8100+10800}=\sqrt{33300}\approx 182.5\ \text{mi}.
Try it
Quick check

Forces of 3030 N due east and 4040 N due north act on a point. Find the magnitude and direction of the resultant.

Show the answer

Answer: 5050 N at about 53.1∘53.1^\circ north of east. R⃗=⟨30,40⟩\vec R=\langle 30,40\rangle, so ∥R⃗∥=900+1600=50\|\vec R\|=\sqrt{900+1600}=50 and θ=tan⁡−14030≈53.1∘\theta=\tan^{-1}\tfrac{40}{30}\approx 53.1^\circ.

Going Deeper: Advanced Triangle Trig & Vectors

Concept
Stewart's Theorem: cevians, medians, and bisectors

A cevian is a segment from a vertex to a point on the opposite side. Let a cevian from AA meet side aa (i.e. BCBC) at a point that splits it into segments mm (adjacent to BB) and nn (adjacent to CC), with length dd. Then Stewart's Theorem says

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

A mnemonic: “a man and his dad put a bomb in the sink” ⇒\Rightarrow man+dad=bmb+cncman+dad=bmb+cnc, i.e. m a n+d a d=b m b+c n cm\,a\,n + d\,a\,d = b\,m\,b + c\,n\,c. Two special cevians follow at once:

  • Median to side aa (m=n=a2m=n=\tfrac a2):  ma=122b2+2c2−a2.\ m_a=\dfrac12\sqrt{2b^2+2c^2-a^2}.
  • Angle bisector from AA (it divides aa in the ratio c:bc:b, so m=acb+cm=\dfrac{ac}{b+c}, n=abb+cn=\dfrac{ab}{b+c}):  ta=bc ⁣[1−(ab+c)2].\ t_a=\sqrt{bc\!\left[1-\left(\dfrac{a}{b+c}\right)^2\right]}.
Example
Median and angle-bisector lengths

Let a=7a=7, b=8b=8, c=9c=9 (so side a=BCa=BC, and the cevians issue from AA). Median to aa:  ma=122(82)+2(92)−72=12128+162−49=12241≈7.8.\ m_a=\tfrac12\sqrt{2(8^2)+2(9^2)-7^2}=\tfrac12\sqrt{128+162-49}=\tfrac12\sqrt{241}\approx 7.8. Check with Stewart (m=n=3.5m=n=3.5, d=mad=m_a):  82(3.5)+92(3.5)=7(d2+3.52)⇒507.5=7d2+85.75⇒d2=60.25\,8^2(3.5)+9^2(3.5)=7(d^2+3.5^2)\Rightarrow 507.5=7d^2+85.75\Rightarrow d^2=60.25, so d≈7.8d\approx 7.8. ✓ Angle bisector from AA:  ta=(8)(9) ⁣[1−(78+9)2]=72 (1−0.1696)=59.79≈7.7.\ t_a=\sqrt{(8)(9)\!\left[1-\left(\dfrac{7}{8+9}\right)^2\right]}=\sqrt{72\,(1-0.1696)}=\sqrt{59.79}\approx 7.7.

Concept
The ambiguous case as a parameter count

Solving a triangle means finding all 66 parts from 33 given ones. “How many triangles” is really “how many solutions does the given data admit,” and it is governed by how many pieces of data are fixed versus free:

  • SSS, SAS, ASA, AAS each pin down a triangle by a rigid congruence: 00 free parameters ⇒\Rightarrow exactly one triangle (when the data are geometrically possible).
  • SSA is not a congruence: fixing a,b,Aa,b,A leaves the position of the third vertex to be found from sin⁡B=bsin⁡Aa\sin B=\dfrac{b\sin A}{a}. That equation is quadratic-like in outcome because sin⁡B=sin⁡(180∘−B)\sin B=\sin(180^\circ-B), so it can yield 00, 11, or 22 admissible values of BB.
  • AAA fixes only the shape (angles), leaving a 11-parameter family of similar triangles (scale is free): infinitely many.

The count is thus a solution count of sin⁡B=k\sin B=k subject to A+B<180∘A+B<180^\circ: two solutions when 0<k<10<k<1 and both BB values keep the angle sum valid; one when k=1k=1 or when the obtuse B′B' is rejected; none when k>1k>1.

Concept
Area by several methods, and area ratios

The same triangle's area can be written many ways; equating them yields identities.

K=12absin⁡C=s(s−a)(s−b)(s−c)=rs=abc4R,K=\tfrac12 ab\sin C=\sqrt{s(s-a)(s-b)(s-c)}=rs=\frac{abc}{4R},

where s=a+b+c2s=\tfrac{a+b+c}{2}, rr is the inradius, and RR is the circumradius. Useful ratio facts:

  • A median splits a triangle into two triangles of equal area (equal bases a2\tfrac a2, shared height).
  • A cevian dividing the base in ratio m:nm:n splits the area in the same ratio m:nm:n.
  • Two triangles sharing an angle θ\theta have areas in ratio K1K2=a1b1a2b2\dfrac{K_1}{K_2}=\dfrac{a_1 b_1}{a_2 b_2} (from 12absin⁡θ\tfrac12 ab\sin\theta).
  • Similar triangles with side ratio kk have area ratio k2k^2.
Example
One area, four ways

For a=7a=7, b=8b=8, c=9c=9 we found Heron's area K=720≈26.83K=\sqrt{720}\approx 26.83. Then:

r=Ks=26.8312≈2.2,R=abc4K=5044(26.83)≈4.7.r=\frac{K}{s}=\frac{26.83}{12}\approx 2.2,\qquad R=\frac{abc}{4K}=\frac{504}{4(26.83)}\approx 4.7.

SAS cross-check: first cos⁡C=72+82−922(7)(8)=0.2857⇒sin⁡C≈0.9583\cos C=\dfrac{7^2+8^2-9^2}{2(7)(8)}=0.2857\Rightarrow \sin C\approx 0.9583, so K=12(7)(8)(0.9583)≈26.8K=\tfrac12(7)(8)(0.9583)\approx 26.8. ✓ All four formulas agree.

Concept
Parallelogram law: ∥u⃗±v⃗∥\|\vec u\pm\vec v\| from magnitudes and the angle

If θ\theta is the angle between u⃗\vec u and v⃗\vec v, then because ∥u⃗±v⃗∥2=(u⃗±v⃗)⋅(u⃗±v⃗)\|\vec u\pm\vec v\|^2=(\vec u\pm\vec v)\cdot(\vec u\pm\vec v),

∥u⃗+v⃗∥2=∥u⃗∥2+∥v⃗∥2+2∥u⃗∥∥v⃗∥cos⁡θ,∥u⃗−v⃗∥2=∥u⃗∥2+∥v⃗∥2−2∥u⃗∥∥v⃗∥cos⁡θ.\|\vec u+\vec v\|^2=\|\vec u\|^2+\|\vec v\|^2+2\|\vec u\|\|\vec v\|\cos\theta,\qquad \|\vec u-\vec v\|^2=\|\vec u\|^2+\|\vec v\|^2-2\|\vec u\|\|\vec v\|\cos\theta.

These are just the Law of Cosines on the triangle formed by the two vectors. Adding them gives the parallelogram law ∥u⃗+v⃗∥2+∥u⃗−v⃗∥2=2∥u⃗∥2+2∥v⃗∥2\|\vec u+\vec v\|^2+\|\vec u-\vec v\|^2=2\|\vec u\|^2+2\|\vec v\|^2 (the two diagonals determine the sides).

Example
Resultant and difference without components
60°61014≈ 8.7
The diagonals of the parallelogram are u⃗+v⃗\vec u+\vec v (length 1414) and u⃗−v⃗\vec u-\vec v.

Two forces of magnitude ∥u⃗∥=6\|\vec u\|=6 and ∥v⃗∥=10\|\vec v\|=10 meet at an angle of θ=60∘\theta=60^\circ.

∥u⃗+v⃗∥=62+102+2(6)(10)cos⁡60∘=36+100+60=196=14.0,\|\vec u+\vec v\|=\sqrt{6^2+10^2+2(6)(10)\cos 60^\circ}=\sqrt{36+100+60}=\sqrt{196}=14.0,
∥u⃗−v⃗∥=36+100−60=76≈8.7.\|\vec u-\vec v\|=\sqrt{36+100-60}=\sqrt{76}\approx 8.7.

Check: 142+8.72≈196+75.7=271.714^2+8.7^2\approx 196+75.7=271.7 and 2(62)+2(102)=2722(6^2)+2(10^2)=272, confirming the parallelogram law (rounding). The direction of u⃗+v⃗\vec u+\vec v relative to u⃗\vec u is ϕ=tan⁡−1 ⁣10sin⁡60∘6+10cos⁡60∘=tan⁡−1 ⁣8.6611≈38.2∘.\phi=\tan^{-1}\!\dfrac{10\sin 60^\circ}{6+10\cos 60^\circ}=\tan^{-1}\!\dfrac{8.66}{11}\approx 38.2^\circ.

Concept
Vector proofs: perpendicularity and right triangles

The dot product turns geometry into algebra. Key tools:

  • Two vectors are perpendicular iff u⃗⋅v⃗=0\vec u\cdot\vec v=0.
  • Triangle PQRPQR has a right angle at QQ iff QP→⋅QR→=0\overrightarrow{QP}\cdot\overrightarrow{QR}=0.
  • To show a quadrilateral's diagonals are perpendicular, form them as vectors and show their dot product is 00.
  • An angle inscribed in a semicircle is right: if OO is the center and PP is on the circle, then PA→⋅PB→=0\overrightarrow{PA}\cdot\overrightarrow{PB}=0 where ABAB is a diameter.
Example
Prove a triangle is right
xyP(1, 2)Q(4, 3)R(3, 6)
QP→⋅QR→=0\overrightarrow{QP}\cdot\overrightarrow{QR}=0: the right angle is at QQ.

Let P=(1,2)P=(1,2), Q=(4,3)Q=(4,3), R=(3,6)R=(3,6). Test the angle at QQ:

QP→=⟨1−4, 2−3⟩=⟨−3,−1⟩,QR→=⟨3−4, 6−3⟩=⟨−1,3⟩.\overrightarrow{QP}=\langle 1-4,\,2-3\rangle=\langle -3,-1\rangle,\qquad \overrightarrow{QR}=\langle 3-4,\,6-3\rangle=\langle -1,3\rangle.
QP→⋅QR→=(−3)(−1)+(−1)(3)=3−3=0,\overrightarrow{QP}\cdot\overrightarrow{QR}=(-3)(-1)+(-1)(3)=3-3=0,

so the angle at QQ is exactly 90∘90^\circ and △PQR\triangle PQR is right. (Equivalently, verify ∥PQ∥2+∥QR∥2=∥PR∥2\|PQ\|^2+\|QR\|^2=\|PR\|^2: 10+10=2010+10=20. ✓)

Example
Projection and work in context

A wagon is pulled by a rope with force F⃗=⟨25,15⟩\vec F=\langle 25,15\rangle N while it rolls along level ground in the direction d⃗=⟨30,0⟩\vec d=\langle 30,0\rangle m.

W=F⃗⋅d⃗=25(30)+15(0)=750 joules.W=\vec F\cdot\vec d=25(30)+15(0)=750\ \text{joules}.

Only the component of F⃗\vec F along the motion does work:  comp⁡d⃗F⃗=F⃗⋅d⃗∥d⃗∥=75030=25\ \operatorname{comp}_{\vec d}\vec F=\dfrac{\vec F\cdot\vec d}{\|\vec d\|}=\dfrac{750}{30}=25 N, and indeed 25 N×30 m=75025\ \text{N}\times 30\ \text{m}=750 J. The vertical part ⟨0,15⟩\langle 0,15\rangle (perpendicular to the motion) does no work. If instead the rope makes angle α\alpha with the ground and pulls with force FF, then W=Fdcos⁡αW=Fd\cos\alpha.

Tip

Circumradius and inradius links. From K=abc4RK=\dfrac{abc}{4R} and the Law of Sines,  2R=asin⁡A=bsin⁡B=csin⁡C\ 2R=\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C} (so the common ratio in the Law of Sines is the diameter of the circumscribed circle). From K=rsK=rs, the inradius is r=Ksr=\dfrac{K}{s}. Together they give handy identities such as r=4Rsin⁡A2sin⁡B2sin⁡C2r=4R\sin\tfrac A2\sin\tfrac B2\sin\tfrac C2 and a=2Rsin⁡Aa=2R\sin A. Rounding convention as above: lengths and radii to the nearest tenth, angles to the nearest tenth of a degree.

Why It Works

Heron's formula follows from the law of cosines and the area formula 12absin⁡C\tfrac12ab\sin C: write sin⁡C=1−cos⁡2C\sin C = \sqrt{1 - \cos^2C}, substitute cos⁡C=a2+b2−c22ab\cos C = \tfrac{a^2 + b^2 - c^2}{2ab}, and factor the resulting difference of squares twice: 16K2=(a+b+c)(−a+b+c)(a−b+c)(a+b−c)16K^2 = (a + b + c)(-a + b + c)(a - b + c)(a + b - c), which is s(s−a)(s−b)(s−c)s(s-a)(s-b)(s-c) times 1616. The projection of u\mathbf u onto v\mathbf v is u⋅v∣v∣2v\tfrac{\mathbf u\cdot\mathbf v}{|\mathbf v|^2}\mathbf v because u⋅v=∣u∣∣v∣cos⁡θ\mathbf u\cdot\mathbf v = |\mathbf u||\mathbf v|\cos\theta and ∣u∣cos⁡θ|\mathbf u|\cos\theta is the shadow's length.

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
The dot product
u⋅v=u1v1+u2v2=∣u∣∣v∣cos⁡θ\mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2=|\mathbf{u}||\mathbf{v}|\cos\theta

What it means. Multiply matching components and add; the result also measures how aligned the vectors are.

Example. (1,2)⋅(3,4)=1⋅3+2⋅4=11(1,2)\cdot(3,4)=1\cdot3+2\cdot4=11.

Why it works. Apply the Law of Cosines to the triangle formed by u\mathbf u, v\mathbf v and v−u\mathbf v-\mathbf u. Expanding ∣v−u∣2|\mathbf v-\mathbf u|^2 in components and comparing with ∣u∣2+∣v∣2−2∣u∣∣v∣cos⁡θ|\mathbf u|^2+|\mathbf v|^2-2|\mathbf u||\mathbf v|\cos\theta leaves exactly u1v1+u2v2=∣u∣∣v∣cos⁡θu_1v_1+u_2v_2=|\mathbf u||\mathbf v|\cos\theta.

Tip. A dot product of 00 means the vectors are perpendicular — the fastest perpendicularity test there is.