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Law of sines and law of cosines

Solving any triangle from three pieces of information.

Law of cosines: c² = a² + b² − 2ab cos C, which is Pythagoras with a correction for the angle. Law of sines: a/sin A = b/sin B = c/sin C. Picture it: two sides and the angle between them pin the third side; the figure is drawn from those numbers. Think it: the law of cosines is the dot product in disguise: |a − b|² = |a|² + |b|² − 2 a·b.

Àwọn ààtò àwọn ìṣàmúlò-ètò · triangle 5 7 with angle 60

Geometry · Triangles

Triangle 5 7 with angle 60

5,\ 7

Àwọn ìṣàmúlò-ètò

a = 5,\ b = 7,\ C = 60^\circ
Two sides and the included angle (SAS): find the third side with the law of cosines.
c^2 = a^2 + b^2 - 2ab\cos C = 39
Law of cosines.
c = \sqrt{39} \approx 6.2450
P = a + b + c = \sqrt{39} + 12
Perimeter.
s = \tfrac{P}{2} = \frac{\sqrt{39}}{2} + 6,\quad A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{35 \sqrt{3}}{4} \approx 15.155
Heron's formula for the area.
\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{3 \sqrt{39}}{26} \Rightarrow A \approx 43.898^\circ
Law of cosines for angle A (opposite side a).
\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{\sqrt{39}}{26} \Rightarrow B \approx 76.102^\circ
Law of cosines for angle B (opposite side b).
\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{1}{2} \Rightarrow C \approx 60.000^\circ
Law of cosines for angle C (opposite side c).
A + B + C = 180.0^\circ
The angles add to 180°: a acute, scalene triangle.

Fi àwọn àgbèwọlé hàn

A = \frac{35 \sqrt{3}}{4} \approx 15.155,\quad P = \sqrt{39} + 12,\quad \angle \approx 43.9^\circ, 76.1^\circ, 60.0^\circ

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Bii o ṣe le: Law of sines and law of cosines

Two sides and the included angle (SAS): find the third side with the law of cosines.
Law of cosines.
Perimeter.
Heron's formula for the area.
Law of cosines for angle A (opposite side a).
Law of cosines for angle B (opposite side b).
Law of cosines for angle C (opposite side c).
The angles add to 180°: a acute, scalene triangle.

Àwọn Àtòjọ-ẹ̀yàn

Why radians instead of degrees?

A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.

Why does sin x = 1/2 have infinitely many solutions?

Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.

How do I remember the exact values?

Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.

Diẹ̀ nínú Trigonometry

The unit circle Trigonometric equations Trigonometric identities Degrees and radians Right-triangle trigonometry (SOH-CAH-TOA)Graphs of sine, cosine and tangent Inverse trigonometric functions