ProphetesAI is thinking...
subtend
Answers
MindMap
Loading...
Sources
subtend
subtend, v. (səbˈtɛnd) [ad. L. subtendĕre, f. sub- sub- 2 + tendĕre to stretch, tend. Cf. Sp., Pg. subtender.] 1. trans. (Geom.) To stretch or extend under, or be opposite to: said esp. of a line or side of a figure opposite an angle; also, of a chord or angle opposite an arc.1570 Billingsley Euclid...
Oxford English Dictionary
prophetes.ai
subtend
subtend/səbˈtend; səb`tɛnd/ v[Tn](geometry 几) (of a chord2(1) or the side of a triangle) be opposite to (an arc(1) or angle) (指弦或三角形之边)对向(某弧或角) The chord AC subtends the arc ABC. The side XZ subtends the angle XYZ. AC 弦对向 ABC 弧. XZ 边对向 XYZ 角. =>illus 见插图.
牛津英汉双解词典
prophetes.ai
subtention
† subˈtention Obs. [f. L. subtent-, pa. ppl. stem of subtendĕre to subtend: see -tion.] = subtense n.1610 Hopton Baculum Geodæt. vii. ii. 297 Any right lines being applied to a circle is called a subtention, which may be Sines, Tangents, or Secants.
Oxford English Dictionary
prophetes.ai
subtendent
† subˈtendent, a. and n. Obs. Also 7 -ant. [ad. L. subtendens, -entem, pr. pple. of subtendĕre to subtend.] A. adj. That subtends.1571 Digges Pantom. i. vi. C iij b, In equiangle triangles, al their sides are proportional aswel such as conteyne the equall angles, as also their subtendente sides. Ibi...
Oxford English Dictionary
prophetes.ai
Subtend perpendicular at origin **Question:** > The locus of the foot of perpendicular, from the origin to chords of circle $x^2+y^2-4x-6y-3=0$, which subtend a right angle at the origin, is what? **My attempt:** A...
From the expressions you found for $x_0$ and $y_0$ one also gets: $$ x_0^2+y_0^2={c^2\over1+m^2}. $$ From ${c^2+2mc-3c\over1+m^2}={3\over2}$ we have then: $$ {c^2+2mc-3c\over1+m^2}=x_0^2+y_0^2-2x_0-3y_0={3\over2}. $$ That is the equation of a circle, centered at $\left(1,{3\over2}\right)$ and of rad...
prophetes.ai
$AB$ is any chord of the circle $x^2+y^2-6x-8y-11=0,$which subtend $90^\circ$ at $(1,2)$.If locus of mid-point of $AB$ is circle $x^2+y^2-2ax-2by-c=0$ $AB$ is any chord of the circle $x^2+y^2-6x-8y-11=0,$which subtend...
Let $C(3,4),D(1,2)$. Also, let $E(X,Y)$ be the midpoint of $AB$. $\qquad\qquad\qquad$^2+(Y-4)^2}$$ we have $$|AE|^2=|AC|^2-|CE|^2=36-(X-3)^2-(Y-4)^2\tag1$$ Also, since we can see that $D$ ...
prophetes.ai
conditions that the chord subtend a right angle Find the conditions that the chord of contact of tangents from the point $(x',y')$ to the circle $x^2+y^2=a^2$ should subtend a right angle at the centre. can someone p...
The equation of the circle is given as: $$x^2+y^2=a^2\tag 1 $$ and the point $P (x',y') $. The equation of the chord of contact of $P $ with respect to the equation of the circle will be: $$\frac {xx'}{a^2} + \frac {yy'}{a^2} = 0 \tag 2$$ To get the combined equation of the lines which join the orig...
prophetes.ai
If $PQ$ subtends right angle at the centre of ellipse then find $\frac{1}{OP^2}+\frac{1}{OQ^2}. $ $PQ$ is a variable chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ . If $PQ$ subtends right angle at the centr...
You are on the right track. You know that $\tan(\alpha)\tan(\beta)=-\frac{a^2}{b^2}$, and: $$\begin{eqnarray*} \frac{1}{OP^2}+\frac{1}{OQ^2} &=& \frac{\frac{1}{\cos^2\alpha}}{b^2\tan^2\alpha+a^2}+\frac{\frac{1}{\cos^2 \beta}}{b^2\tan^2\beta+a^2}\\\\[0.2cm]&=&\frac{1+\tan^2\alpha}{b^2\tan^2\alpha+a^2...
prophetes.ai
Nth roots subtending right angle at origin Let $\ z_1$ and $\ z_2$ be nth roots of unity which subtend a right angle at the origin. Then n must be of the form(A) 4k + 1(B) 4k + 2(C) 4k + 3(D) 4k My approach Let the ...
Let nth root be $z=\exp({2\pi i\frac{k}{n}})$. Now you can say other root of unity is $z'=\exp(2\pi i\frac{k+l}{n})$. Now a number subtending angle $90^\circ$ anticlockwise from $z$ is $\exp(\frac{\pi i}{2})\cdot z$. So $z' = \exp(\frac{\pi i}{2})\cdot z$ $$\exp\left(2\pi i\frac{k+l}{n}\right) =\exp...
prophetes.ai
Do Equal Angles at the Circumference Subtend Equal Chords? Do Equal Angles at the Circumference Subtend Equal Chords? This may seem like a pretty basic question but I can't seem to find a definite answer on the inter...
It is true that equal angles at circumference subtend equal chords.
prophetes.ai
At what distance does a man, whose height is 2m subtend an angle of 10' ? I tried using the angle in radian = length of arc/radius of circle and got the answer as 974.0286 m. But the answer given in the answer's secti...
Since no work was shown, it is hard to answer the question "What am I doing wrong?" However, providing a possible solution may give a clue. $10'=\frac\pi{1080}$ is a small enough angle that $$ \frac{\tan\left(\frac\pi{1080}\right)}{\frac\pi{1080}}=1.00000282 $$ So we need only compute $$ \frac{2\tex...
prophetes.ai
Euclidean Distance on a Sphere I have that the Euclidean distance on the surface of a sphere in terms of the angle they subtend at the centre is $(\sqrt{2})R\sqrt{1-\cos(\theta_{12})}$ (Where $\theta_{12}$ is the an...
Consider the diagram: $\hspace{4cm}$!enter image description here Using the identity $\cos(\theta)=1-2\sin^2(\theta/2)$, the distance is $$ 2r\sin(\theta/2)=r\sqrt{2-2\cos(\theta)} $$
prophetes.ai
If F is the Fermat point of triangle ABC, then what is the algebraic formula that gives minimum sum distance, i.e., FA+FB+FC=? Fermat point (F) of a triangle is at the least distance from triangle vertices. If one of ...
If $ABC$ is a triangle with every angle being $\leq 120^\circ$ and $F$ is its Fermat-Torricelli point, we may introduce the points $A',B',C'$ given by the vertices of the equilateral triangles externally built on the sides of $ABC$. Then we have that $(A,F,A'),(B,F,B'),(C,F,C')$ are collinear and $$...
prophetes.ai
Two circles touch internally at X and a straight line cuts them at A, B, C, D in order. Prove that AB, CD subtend equal angles at X. Two circles touch internally at X and a straight line cuts them at A, B, C, D in ord...
![enter image description here]( Let $AD$ intersect the smaller circle at $E\ne X$ and $h$ the common tangent of the two circles at $X$. It follows that $\angle ABX=\angle CEX=180^\circ - \angle CBX$. On the other hand, $\angle XCE=\angle XAD$ because both of them are equal to $\angle \alpha$.
prophetes.ai
Providing a figure for given problem The Question is from Challenge and Thrill of Pre College Mathematics, Ex. 6.10 ques no. 8 I've been trying this ques for a lot of time but I've been facing the following problems ...
Problem #4 solution Let midpoint of $AB$ be $O$. In right triangle $PON$ we have $ON = \dfrac{PN}{\tan\phi}$ In right triangle $PAN$ we have $AN = \dfrac{PN}{\tan\theta}$ Finally in right triangle $AON$ apply pythagoras : $$\begin{align} &OA^2+ON^2 = AN^2\\\ &a^2+\dfrac{PN^2}{\tan^2\phi} = \dfrac{PN...
prophetes.ai