A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry. A triangle with vertices A, B, and C is denoted
△
A
B
C
{\displaystyle \triangle ABC}
.In Euclidean geometry, any three points, when non-collinear, determine a unique triangle and simultaneously, a unique plane (i.e. a two-dimensional Euclidean space). In other words, there is only one plane that contains that triangle, and every triangle is contained in some plane. If the entire geometry is only the Euclidean plane, there is only one plane and all triangles are contained in it; however, in higher-dimensional Euclidean spaces, this is no longer true. This article is about triangles in Euclidean geometry, and in particular, the Euclidean plane, except where otherwise noted.
Let $ABC$ be a triangle with $\angle A= 60^{\circ},$ and $AD,BE$ are bisectors of $A,B$ respectively where $D\in BC, E\in AC.$ Find the measure of $B$ if $AB+BD=AE+BE.$
Hello MHB, I saw one question that really tickles my intellectual fancy and because of the limited spare time that I have, I could not say I have solved it already! But, I will most definitely give the question more thought and will post back if I find a good solution to it.
Here goes the...
I'm confused about it is not clearly given in task that all the little changes Δ are approaching 0. Especially that Feynman does not mention limits in chapter exercise is for. He is using relatively big values as a little changes (like 4cm). Let's assume that Δ means value is approaching 0...
From @fresh_42's Insight
https://www.physicsforums.com/insights/10-math-things-we-all-learnt-wrong-at-school/
Please discuss!
We all live on a globe, a giant ball. The angles of a triangle on this ball add up to a number greater than ##180°##.
And the amount by which the sum extends...
Triangle $ABC$ is inscribed in a circle of radius 2 with $\angle B\ge 90^{\circ}$ and $x$ is a real number satisfying the equation $x^4+ax^3+bx^2+cx+1=0$ where $a=BC,\,b=CA$ and $c=AB$. Find all possible values of $x$.
Here is my attempt to draw a diagram for this problem:
I'm confused about the "the perpendicular bisector of ##BC## cuts ##BA##, ##CA## produced at ##P, \ Q##" part of the problem.
How does perpendicular bisector of ##BC## cut the side ##CA##?
Let $ABC$ be a triangle with $\angle ACB=2\alpha,\, \angle ABC=3\alpha$, $AD$ is an altitude and $AE$ is a median such that $\angle DAE=\alpha$. If $BC=a,\,CA=b$ and $AB=c$, prove that $\dfrac{a}{b}=1+\sqrt{2\left(\dfrac{c}{b}\right)^2-1}$.
Hi,
I'm new to programming in python [total beginner in programming] and I would like to ask you for your help.
Here is what I got so far:
import numpy as np
import random
from math import sqrt
p = np.array([(0, 0), (1, 0), (1, (1/sqrt(2)))], dtype=float)
t = np.array((0, 0)...
Please, I need help! I need to calculate the moment of inertia of a triangle relatively OY. I have an idea to split my triangle into rods and use Huygens-Steiner theorem, but after discussed this exercise with my friend, I have a question: which of these splits are right (picture 1 and 2)? Or...
By considering a vector triangle at any point on its circular path, at angle theta from the x -axis,
We can obtain that:
(rw)^2 + (kV)^2 - 2(rw)(kV)cos(90 + theta) = V^2
This can be rearranged to get:
(r thetadot)^2 + (kV)^2 + 2 (r* thetadot)(kV)sin theta = V^2.
I know that I must somehow...
I am looking for the name of the triangle centre point from which the vertices subtend 120°.
I am concerned really with deviant equilateral triangles, thru to right angle triangles. Flat triangles, that have an internal angle greater than 120°, do not have such a point within the triangle...
Hello everyone. I am having trouble finding the area of the shaded region using the determinant area formula. I know where to plug in the numbers into the formula. My problem here is finding the needed points in the form (x, y) from the given picture for question 21.
Let $a,\,b,\,c$ be the sides of a triangle. Prove that
$\dfrac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\dfrac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\dfrac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\le 3$.
Consider a square with the side of length n and $(n+1)^2$ points inside it. Show that we can choose 3 of them to determine a triangle (possibly degenerate) of area at most $\frac{1}{2}$.
I think that I know how to solve the problem for the cases $n=1$ and $n=2$:
For $n=1$ we can easily prove...
There are 3 lights in the form of a triangle...
A, B, and C are lights and are stationary with respect to each other. S1, S2, S3 are spaceships.
B
S1 S2
A S3...
$\tiny{act.ge.5}$
ok you have 2 seconds to figure this one out:unsure:
This question has live answer choices. Select all the answer choices that apply. The correct answer to a question of this type could consist of as few as one, or as many as all five of the answer choices.
\item In triangle...
In triangle ABC, ∠C = 90 degrees, ∠A = 30 degrees and BC = 1. Find the minimum length of the longest side of a triangle inscribed in triangle ABC (that is, one such that each side of ABC contains a different vertex of the triangle).
Given two moving points $A(x_1,\,y_1)$ and $B(x_2,\,y_2)$ on parabola curve $y^2=6x$ with $x_1+x_2=4$ and $x_1\ne x_2$ and the perpendicular bisector of segment $AB$ intersects $x$-axis at point $C$. Find the maximum area of $\triangle ABC$.
I first computed the operator ##\hat{T}## in the ##a,b,c## basis (assuming ##a = (1 \ 0 \ 0 )^{T} , b = (0 \ 1 \ 0)^{T}## and ##c = (0 \ 0 \ 1)^{T}##) and found
$$ \hat{T} = \begin{pmatrix} 0&0&1 \\ 1&0&0 \\ 0&1&0 \end{pmatrix}.$$
The eigenvalues and eigenvectors corresponding to this matrix...
Summary:: I think we are still in the earlier parts of Physics and I am confused at how "values" work for a velocity-time graph. We are using the formulas to solve an area of a triangle and rectangle to find the total displacement. If a diagonal line begins from above and continue to go down...
Could I please ask for advice with the following:
ABC is a right-angled triangle in which AB = 4a; BC = 3a. Forces of magnitudes P, Q and R act along the directed sides AB, BC and CA respectively.
a) Find the ratios P:Q:R if their resultant is a couple.
b) If the force along the directed line...
Could I please ask for help with the following:
ABC is a right-angled triangle in which AB = 4a; BC = 3a. Forces of magnitudes P, Q and R act along the directed sides AB, BC and CA respectively. Find the ratios P:Q:R if their resultant is a couple.
Book answer is 4 : 3 : 5
Here's my diagram...
At first I had no idea of how to solve this problem, but checking online I found out that there is a formula linking the radius of the circumference and the side of the triangle... the formula is:
side=radius√3
The thing is that I can't understand why is this working... which deduction have been...
It is given that the solution is ideal, i.e. that we can take ##\gamma_A = 1##.
I wondered what that small triangle signifies in the second definition? Thanks!
I'm not really sure where to start with this problem, but I wanted to ask a few questions about the approach I should use.
Is it reasonable to say that a gradient could be set up that could describe the force on the fourth ion at any point?
The way I'm thinking of this problem is, I want to...
Actually we find the two position in the axis very easily, but, what am trying to find is if exist such position (Being the charge of the ions equal) away from the symmetry axis, but i really don't want to try find it numerically, it would be a disaster.
The only conclusion i got is, if such...
I did in this way:
## I = \int dm \rho^2 ##
Dividing the triangle in small rectangles with ##dA = dy x(y) ## where ##x(y) = 2 ctg( \alpha ) (h - y) ##
we have : ## dm = \sigma 2 ctg( \alpha ) (h - y) ##
Now i have ## \rho^2 = x^2 + (h-y)^2 ##
Now I don't know what I can do because it would be...
Hi all, I happened to see this primary 6 math geometry problem and thought it was a fun (not straightforward but not too hard) problem. Try it and post your solution if you are interested. (Cool)
In the figure, not drawn to scale, $UX=XY=YT$ and $UV=VS$. Given that the area of triangle $XVU$ is...
How would I have to calculate this question for an answer, a friend of mine told me he could get the answer without knowing that the bottom line was 16 meters, I can't seem to find a way that would work, I am not sure if I am missing something or he is lying.
Let $a,\,b$ and $c$ be the side lengths of a triangle. Prove that $\dfrac{a}{\sqrt[3]{4b^3+4c^3}}+\dfrac{c}{\sqrt[3]{4a^3+4b^3}}+\dfrac{a}{\sqrt[3]{4b^3+4c^3}}<2$.
In isosceles triangle ABC, the sides are of length AC = BC = 8 and AB = 4. Find the angles of the triangle . Express the answers both in radians, rounded to two decimal places, and in degrees, rounded to one decimal place.
Solution:
I think dropping a perpendicular line from the vertex at C...
Find the area of a triangle with angle 70° in between sides 6 cm and 4 cm.
Solution:
From the SOH-CAH-TOA mnemonic, I want the ratio of the opposite side (CD) to the hypotenuse (AC). I should be using the *sine* function, not cosine. Yes?
SOH leads to sin = opp/hyp
sin(70°) = CD/4
CD = 4...
A right triangle has an acute angle measure of 22°. Which two numbers could represent the lengths of the legs of this triangle?
OPTIONS
a. 2 and 5
b. 1 and 5
c. 3 and 5
d. 4 and 5
I know that each leg represents the sides of the right triangle opposite the hypotenuse. I think the tangent...
This is the question. The following is the solutions I found:
I understand that the first line was derived by setting one vertex on origin and taking the transpose of the matrix. However, I cannot understand where the extra row and column came from in the second line. Can anyone explain how...
Suppose the lengths of the three sides of $\triangle ABC$ are integers and the inradius of the triangle is 1. Prove that the triangle is a right triangle.