Does trigonometry only work for right triangles?( sin, cos, and tan )?
If you are using ratios of the sides, then you must have a right triangle. However, you will learn the Law of Sines and the Law of Cosines, which can be applied to any triangle.
Why has the ratio of the sides of a right triangle been given so importance to make a separate branch in mathematics named trigonometry? Why is the ratio of other shapes like quadrilateral not defined, i.e., quadrometry can also be possible?
The trigonometric functions are commonly taught as a property of right triangles, but they are actually a property of the angles. You can see the same values in differently sized triangles if the angles are identical.As for quadrilaterals, you can divide them solely into triangles. So the trigonometric ratios extend to them, eliminating the need for any as you put it, “quadrometric” functions.So, the concept of quadrometry could make sense, but it would just be a derivative of trigonometry.
Solving a triangle without sin, cos, or tan (or any of those laws)?
There is not enough information. To solve for any triangle, there must be 3 things known, for example: 2 angles, 1 sides 2 sides, 1 angle 3 sides 3 angles is not enough either because there is not enough to figure out any side. So, we know 1 angle and 1 side. if we know 1 more angle we will be able to solve for everything, but we mustt use sin, cos and tan.. If we know one more side, then we will be able to solve for the third side with the pythagorean theory, but we cant solve for the angle winthout sin, cos and tan. peace
How can we define sin and cos functions in a non-90 degree triangle? Is this possible?
They can’t be defined for non-right triangles. But, the Law of Sines and Law of Cosines work for all triangles since we can add auxiliary line segments to any triangle to form right angles.Law of cosines - WikipediaLaw of sines - Wikipedia
How do you know when to use cos, sin, tan in math?
The most popular way to remember is is using "Soh Cah Toa" Soh = Sine Opposite/Hypotenuse Cah = Cosine Adjacent/Hypotenuse Toa = Tan Opposite/Adjacent Pick any angle in a right-triangle aside from the right-angle. Using "Soh Cah Toa", the SINE of that angle for example is equal to the OPPOSITE side of that angle divided by the HYPOTENUSE of the triangle (longest side, opposite the right-angle).
How do I calculate a right triangle's angles when I know all three side lengths?
I am building something, and have no idea how to calculate angles. I have a right triangle with all three lengths known. However, I need to know how to the two angles not known.
How can you prove that for a right angled isosceles triangle ABC, where AB hypotenuse and AD bisector of When I was first asked to answer this question, I had a lot of options. But as I looked at the answers already posted here, I wondered is there any other way I can do it. I have been out of touch with geometry for a long time. Took a long time, but I did manage to come out with an approach.Internal bisector theorem: The internal bisector of any angle of a triangle bisects the opposite side of that angle in a ratio that is equal to the ratio of the other two sides, in the correct order of correspondence.So we should have[math]\dfrac{AC}{AB}=\dfrac{CD}{DB} = \dfrac{\text{side of triangle ACD}}{\text{side of triangle ADB}}[/math][math]\sin 45^{\circ}=\dfrac{AC}{AB}=\dfrac{1}{\sqrt{2}}[/math][math]AC = BC = CD + DB[/math]Using Pythagoras[math]AB^2=AC^2+BC^2[/math][math]\implies AB = \sqrt{2}AC=\sqrt{2}BC[/math][math]\dfrac{CD}{DB}=\dfrac{1}{\sqrt{2}}[/math][math]\implies DB=\sqrt{2} CD[/math][math]\dfrac{AC}{AB}=\dfrac{CD}{DB}[/math][math]\implies \dfrac{BC}{AB}=\dfrac{CD}{DB}[/math][math]\implies \dfrac{CD+DB}{AB}=\dfrac{CD}{DB}[/math][math]\implies \dfrac{CD+DB}{CD}=\dfrac{AB}{DB}[/math][math]\implies 1+\dfrac{DB}{CD}=\dfrac{AB}{DB}[/math][math]\implies 1+\dfrac{\sqrt{2}CD}{CD}=\dfrac{AB}{\sqrt{2}CD}[/math][math]\implies 1+\sqrt{2}=\dfrac{AB}{\sqrt{2}CD}[/math][math]\implies AB=\sqrt{2}CD+2CD.........[i][/math][math]\implies \sqrt{2}AC=\sqrt{2}CD+2CD[/math][math]\implies AC = CD+\sqrt{2}CD[/math][math]\implies AC + CD = CD+\sqrt{2}CD + CD………[ii][/math]Comparing [math][i][/math] and [math][ii][/math][math]AB = AC+CD[/math]
Why is trigonometry applied only in right angled triangles?
Ok let me first say it, keep reading for having more clarity.Trigonometry is applicable to every possible triangle. You're thinking that it is applicable to only right angled triangles because that is what you have studied till now. As you move further in academics, you will study trigonometry in any general triangle.First of all, see the definition of trigonometry- it is a branch of mathematics that studies relationships involving lengths and angles of triangles. The definition doesn't say right triangle.As you'll study more, you'll know that formulas which you are currently using in right angle trigonometry are generated from trigonometric formula for general triangles. For example- see the link shown and carefully read the paragraph for FORMULAS FOR OBLIQUE TRIANGLES. The point which I wanted to make here is clearly written there.
Sine, cosine, tangent?
Remember this mnemonic: "Some People Have, Curly Brown Hair, Turned Permanently Black" (Where S = Sine, P = Perpendicular(adjacent), H = Hypotenuse, C = Cos , B = base, T = Tan.) For the triangle given, this is: Sine = 5/13 Cosine = 12/13 Tangent = 5/12