Yes if it's a right angle triangle.
If side is given too, then you can find area with one diagonal. As diagonals bisect each other in a rhombus at 90°, Using Pythogoras Theorem: (Half d1)² = (side)² - (Half d2)²
Sine= Opposite/ Hypotenuse Cosine= Adjacent/ Hypotenuse
Remember SOHCAHTOA which means, the Sin of an angle is equal to the Opposite side divided by the Hypotenuse, the Cos of an angle is equal to the Adjacent side divided by the hypotenuse, and the Tangent of an angle is equal to the Opposite side divided by the Adjacent side. So as long as you have two sides of a right triangle, then you can find the angles and the length of the third side.
Take the inverse tangent -- tan-1(opposite side/adjacent side)
Third side = sqrt(hyp2 - opp2)
Yes
you can use pythagorean theorem formula:c² = a² + b²you must have the values of the opposite side and adjacent side.you may try below link for easy calculation
If side is given too, then you can find area with one diagonal. As diagonals bisect each other in a rhombus at 90°, Using Pythogoras Theorem: (Half d1)² = (side)² - (Half d2)²
In a right-angled triangle, the hypotenuse is the longest side, opposite the right-angle. There are two ways of finding the length of the hypotenuse using mathematics: Pythagoras' theorem or trigonometry, but for both you need either two other lengths or an angle. For Pythagoras' theorem, you need the other two lengths. The theorem is a2+b2=c2, or the square root of the sum of two angles squared, where c=the hypotenuse. Let's say that one length is 4.8cm and the other 4cm. 4.82+42=6.22. So, the answer is 6.2cm. If you have one side and one angle, use trigonometry. You will need a calculator for this. Each side of the right-angled triangle has a name corresponding to the positioning of the angle given. The opposite is the side opposite the given angle, the adjacent is the side with the right-angle and the given angle on it, and the hypotenuse is the longest side or the side opposite the right-angle. There are three formulas in trigonometry: sin, cos and tan. Sin is the opposite/hypotenuse; cos is the adjacent/hypotenuse; and tan is the opposite/adjacent. As we are trying to find the hypotenuse, we already have either the opposite or the adjacent, and one angle. Let's say that our angle is 50o and we have the adjacent side, and that is 4cm. So, we have the adjacent and want to know the hypotenuse. The formula with both the adjacent and the hypotenuse in is cos. So, Cos(50o)=4/x where x=hypotenuse. We can single out the x by swapping it with the Cos(50o), so x=4/Cos(50o) -> x=6.22289530744164. This is the length of the hypotenuse, and is more accurate that Pythagoras' theorem.
The ratio of the opposite side over the adjacent side is called the tangent.Expressing the fraction (opposite/adjacent) as a decimal, you can find the angle by looking in a table of values for the tangents of various angles.
You would have to find the adjacent and the hypotenuse. For example, if the adjacent was 80 degrees and the hypotenuse 29 you would do this: cos(29)/29 which wiuld give you 0.03. When finding the cosine its adjacent over hypotenuse. For sin its opposite over hhypotenuse and for tan its opposite over adjacent.
Sine= Opposite/ Hypotenuse Cosine= Adjacent/ Hypotenuse
you need a calculator to do Sin-1 Opposite/hypotenuse OR Cos-1 Adjacent/Hypotenuse OR Tan-1 Opposite/Adjacent
Usually the two familiar components are opposite and adjacent. For opposite sine function and for adjacent cosine function have to used. Hence as R is to be resolved then the components are R sin@ and R cos@, where @ is the angle of R with its adjacent.
You would have to use its opposite tangent, tan-1on your scientific calculator. It would be tan-1(opposite side/adjacent side), and you must have the opposite and adjacent sides of the angle you are trying to solve.
Sin is opposite over hypotenuse. Just think of this to help you: SOH-CAH-TOA Sine = opposite over hypotenuse Cosine = adjacent over hypotenuse Tangent = opposite over adjacent Hope that helps you.
No. Adjacent sides, yes. (Twice the sum)