A rhombus is related to a “Parallelogram”
Why? - because it fulfills the requirements of a parallelogram: a quadrilateral with two pairs of parallel sides. It goes above and beyond that to also have four equal-length sides, but it is still a type of parallelogram.
hope this helped <3
Answer: 0.4
Step-by-step explanation:
i think
Let ABC be a triangle in the 3rd quadrant, right-angled at B.
So, AB-> Perpendicular BC -> Base AC -> Hypotenuse.
Given: sinθ=-3/5 cosecθ=-5/3
According to Pythagorean theorem, square of the hypotenuse is equal to the sum of square of the other two sides.
Therefore in triangle ABC, 〖AC〗^2=〖AB〗^2+〖BC〗^2 ------
--(1)
Since sinθ=Perpendicular/Hypotenuse ,
AC=5 and AB=3
Substituting these values in equation (1)
〖BC〗^2=〖AC〗^2-〖AB〗^2
〖BC〗^2=5^2-3^2
〖BC〗^2=25-9
〖BC〗^2=16
BC=4 units
Since the triangle is in the 3rd quadrant, all trigonometric ratios, except tan
and cot are negative.
So,cosθ=Base/Hypotenuse Cosθ=-4/5
secθ=Hypotnuse/Base secθ=-5/4
tanθ=Perpendicular/Base tanθ=3/4
cotθ=Base/Perpendicular cotθ=4/3
Hi there there's several ways this could be proven one way us to consider the allied angle theory where two angles formed between parallel lines are supplementary which in this case can be proven by
2(45)+90=180⁰ ✔
or 3(45)+45=180⁰✔
this would not be the case if it wasn't parallel
Consequently, you can also use the alternate angle theory where you essentially extend one of the lines and you'll see two equal alternate angles
A = 5 + B
A + 5 = 2B
Let's get one of the variables on one side.
A = 5 + B
A = 2B - 5
By the transitive property
5 + B = 2B - 5
Solve for B.
5 = B - 5
10 = B
Use 10 = B in an earlier equation to find A.
A = 5 + B
A = 5 + 10
A = 15
Abe has 15$ and Ben has 10$.
