Question on triangles?

Question

If I have the length of all sides of a triangle how do I work out one of its angles?

Answer 1

you use the cosine formula
a^2=b^2+c^2-2bc cos A

Answer 2

Your question is not specific.
The solution will not be difficult because the lengths are already given. If some of the lengths are missing, use the Pythagorean theorem. You can always draw lines to create a right triangle. Then with a right triangle AOB, AB (squared) = OA (squared) + OB (squared).
Case A (For RIGHT TRIANGLES only):
Given: Right triangle AOB.
Angle O = 90 degrees
Angle B = Tangent (OA/OB)
Angle A = 90 - Angle B
Case B (For EQUILATERAL TRIANGLES only ):
Given: Equilateral trial ABC.
Draw a perpendicular line from point A intersecting line BC
Angle C = Cos(0.5BC/AC)
Since an equilateral triangle is also an equiangular, Angle C = Angle A = Angle B.
Case C (For ALL OTHER IRREGULAR TRIANGLES only):
Use other laws as explained by our colleagues here.
However, you can always derive the dimensions by drawing perpendicular lines. You can easily achieve this by drawing the polygon using a scale or simply a ruler for approximation. Use millimeters for accurate reading and lesser errors in your final answer.
I hope that you like this practical solution or non-bookish approach.

Answer 3

Using Cosine Rule.

Cosine rule:
a^2 = b^2 + c^2 - 2bcCosA

where "a" is the length opposite to the angle A.
where "b" and "c" are the two sides making the angle A.

Answer 4

Assuming it's a right-angle triangle (to check : the square of one side will equal the sum of the squares of the other two sides), then use SIN, COSIN and TANGENT :

For any angle :

SIN (of the angle) = Opposite / Hypotenuse
COSIN (of the angle) = Adjacent / Hypotenuse
TANGENT (of the angle) = Opposite / Adjacent

Where:
opposite = the length of the side opposite the angle in question

adjacent = the length of the side touching the angle in question but NOT the longest side of the triangle

hypotenuse = the length of longest side of the triangle

Answer 5

Use the Cosine Rule:-

a^2 = b^2 + c^2 - 2bcCosA
CosA = (a^2 - b^2 - c^2) / (-2bc)
CosB = (b^2 - a^2 - c^2) / (-2ac)
CosC = )c^2 - a^2 - b^2) / (-2ab)
Where A,B and C are the angles to be found #
and 'a, b & c' are the respective opposite sides.

Answer 6

by using the cosine law
assume that the triangle is named as ABC
cosA=b^2+c^2-a^2/2bc
cosB=a^2+c^2-b^2/2ac
cosC=a^2+b^2-c^2/2ba
WHERE "a" is the side opposite to angle "A"
"b" is the side opposite to angle "B"
"c" is the side opposite to angle "C"

Answer 7

cosine of an angle is equal to the leg adjacent to that angle divided by the hypotenuse.

Cos A = x/y. So, if you know x and y take the inverse cosine of (x/y) to get your angle measure A.

Answer 8

in a triangle with sides a,b and
c and angles A,B and C
let s=(a+b+c)/2

sinA/2=sqrt{(s-b)(s-c)/bc}
cosA/2=sqrt{s(s-a)/bc}
tanA/2=sqrt{(s-b)(s-c)/s(s-a)}

these formulae can be changed
to get the 1/2 angles of B and
C

i hope that this helps

Answer 9

Use the law of cosines.

For a triangle with sides a, b, and c, and opposite angles α, β, and γ (respectively),

α = arccos(b² + c² - a²)/(2bc)
β = arccos(a² + c² - b²)/(2ac)
γ = arccos(a² + b² - c²)/(2ab)

Answer 10

right triangle?