maths.free › Trigonometry › Right-triangle trigonometry (SOH-CAH-TOA)

Right-triangle trigonometry (SOH-CAH-TOA)

Sine, cosine and tangent as side ratios, and solving right triangles.

In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent. Knowing one angle and one side fixes everything. Picture it: the 5-12-13 triangle below is drawn to scale; the ratios are its sides. Think it: because all right triangles with a given angle are similar, the ratio depends only on the angle. That is why sin θ is a function.

Isibonelo esisebenza · hypotenuse of 5 and 12

Hypotenuse of 5 and 12

5,\ 12

Isigaba

  1. c^2 = a^2 + b^2 = 5^2 + 12^2 = 169

    Pythagoras: in a right triangle the square on the hypotenuse equals the sum of the squares on the legs.

  2. c = \sqrt{169} = 13

    Take the square root.

Bonisa impendulo
c = 13

Manje Khetha izinkinga, noma ubhale noma udwebe zakho. Isigaba ngasinye, isithombe, impendulo ifihlakele kuze kube ubuza.

Qhubeka usebenza

I-akhawunti emahhala ifaka iziphawulo kulesifundo ngasinye, irekhodi lokuthi yini oqedile, izinkinga zakho ezixazululiwe endaweni eyodwa, nodokotela owaziyo ongabuza mayelana nalekhasi. I-matematics ivulekile kubo bonke, noma ngabe babhalise noma cha.

Bhala Ngena

Amaphawu asetshenziswa lapha

Chofoza noma yiluphi uphawu lwesibonakaliso ukuze uthole ukuchaza okuphelele, isithombe, kanye nokuthi iyiphi inhlamvu ekhona ekhuluma ngaso.

Indlela: Right-triangle trigonometry (SOH-CAH-TOA)

  1. Label the sides relative to the angle you know: opposite, adjacent, hypotenuse.
  2. Pick the ratio that links the side you know to the side you want.
  3. Solve the resulting equation; use the inverse function when the unknown is the angle.

Imibuzo abantu bebuza

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.

Okuningi Trigonometry