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## Hot answers tagged triangle

4

Call the side opposite $33^\circ$ as $a$. Therefore we have: $$\dfrac {x}{a}=\sin 25^\circ$$ and from the sine rule for the triangle we know that: $$\dfrac {20}{\sin 122^\circ}=\dfrac {a}{\sin 33^\circ}$$ Therefore from the above two equations we have $x=\dfrac{20*\sin 33^\circ * \sin 25^\circ}{\sin 122^\circ}$, or $$x\approx5.428336828982414$$

3

WOLOG assume that $a\geq b\geq c>0$. The constraints the problem imposes are $c=\frac{1-ab}{a+b}$, $b+c>a$. Equivalently, we have $\frac{1-ab}{a+b}+b > a\geq b$ and $b\geq \frac{1-ab}{a+b}$. These inequalities yield $$0\leq a^2-b^2<1-ab<1\quad (1)$$ and $$ab\geq \frac{1-b^2}{2}.\quad(2)$$ Note that $(1)$ implies $0<b\leq a<1$. Now, ...

3

Primitive pythagorean triples are generated by positive integers $p,q$, where $\gcd{(p,q)}=1$ and $p,q$ are of opposite parity. In particular, if $(a,b,c)$ form a primitive pythagorean triple, then we must have $a=2pq, b=p^2-q^2$ and $c=p^2+q^2$. For example, if $p=1, q=2$, then $a=2\cdot 2\cdot 1=4, b=2^2-1^2=3$ and $c^2=2^2+1^2=5$. Thus, you have your ...

3

What you mean by dist(P, ℓ1) = dist(P, ℓ2), is I guess the perpendicular or the shortest distance of ℓ1 and ℓ2 from $P$. So that is equivalent to the condition $PQ = PR$. So in triangles $AQP$ and $APR$: $\angle AQP= \angle ARP = 90^{\circ}$ $AP$ is common. $PQ = PR$. Hence, triangles $AQP$ and $ARP$ are congruent by the RHS criterion. So, $\angle ... 3 The terms "opposite" and "adjacent" are relative terms, which depend on a chosen one of the two non-right angles in a right triangle. So if the triangle is$ABC$with the right angle at vertex$C$, then if you are considering nonright angle/vertex$A$, its opposite is the side not containing that vertex, so is side$BC$, while its adjacent is the other ... 2 Note that $$\dfrac {h}{a}=\sin 38^\circ$$ and using the cosine rule for triangle$ABC$, we have $$c=\sqrt {a^2+b^2-2ab\cos C}$$ or $$a^2-2ab\cos C+b^2-c^2=0$$ Plugging in values and solving for$a$, we get $$a=67.02701552,3.89395228$$ which gives $$h=a\sin38^\circ \implies h=41.265951, 2.3973563$$ 2$\triangle ACD$has the same area as$\triangle ACP$since they both have the same base and altitude. Thus,$ABCD$has the same area as$\triangle ABP$. Thus, we need to find when the area of$\triangle ABP$has the same area as$\triangle APD$. This happens precisely when the distance from$D$to$\overline{AP}$is the same as the distance from$B$to ... 2 1 I'm a little confused about how you choose to use either sine or cosine or tangent over the others. Are they interchangeable given the same information you have about a right triangle? What are the circumstances that should dictate the use of one over the other? Or is it preference? I assume that as in part 2 you have a right angled triangle. In this ... 2 Sorry, I misunderstood your question the first time round. First, I'm getting $$9 \sec \theta \tan \theta + 9 \ln | \sec \theta + \tan \theta | + C$$ as my intermediate answer. From there, you need to take out all the thetas and put in$x$. From what you wrote, I'm assuming you see how we get$\tan \theta = x / \sqrt{6}$and$\sec \theta = \sqrt{(x^2 ...

2

The fact that $\overline{AS}$ is a median is irrelevant here. You're asking if, given lengths $AB$ and $AS$, you can determine (uniquely) the triangle $\triangle ABS$. There are infinitely many such triangles — imagine fixing $\overline{AS}$ and drawing a circle of radius $c=8$ centered at $A$. All but two of the points on that circle can be vertex $B$.

2

Converting comment to answer with a few more details, as requested. Draw the circumcircle of $\triangle BCD$; let its center be $K$, and let $D^\prime$ be a point on the major arc $\stackrel{\frown}{BC}$. Note that $\angle BDC$ and $\angle BD^\prime C$ are supplementary. By the Inscribed Angle Theorem, point $K$ is such that $$\angle BKC = 2\;\angle ... 2 We know that$$ \left(\frac{\sin\alpha}{\sin\beta}\right)^{100}=\frac{1-\cos\alpha}{1-\cos\beta} $$Then$$ \frac{(\sin\alpha)^{100}}{1-\cos\alpha}=\frac{(\sin\beta)^{100}}{1-\cos\beta} $$Let$$ f(x):=\frac{(\sin x)^{100}}{1-\cos x} $$Then$$ f'(x)=\frac{(\sin x)^{99}}{(1-\cos x)^2}\left( 100\cos x - 99 \cos^2 x -1 \right) $$Let g(x):=100y - 99 ... 1 This is nothing more than an elaboration of Hayden's answer, but I'll post it anyway since I spent 10 minutes typing it on my phone while standing in a crowded train. You can't tell anything about the angles of a right triangle if you only know its hypotenuse. Imagine that the hypotenuse is a (fixed length) ladder leaning against a vertical wall, so that ... 1 This is not really different from previous answers, but slightly more general. If you know the length of one side of a triangle and the size of the opposite angle, you know that the third angle lies on an arc of a circle. With a right-angle you know this is the full circle. If R is the circumradius, a is the length of the side and A the opposite ... 1 Although lab bhattacharjee has already said, we have to use the Law of Sines. If you aren't familiar with it or its proof, see the link. I will tell you how to proceed in a detailed manner. Here we have our \triangle ABC and its circumscribed circle with center O. We now construct a diameter BOD. So, \angle BAC=\angle BDC and \angle ... 1 In a triangle, the bisector of an angle divides the opposite side in the same ratio as the ratio of the leg lengths of the bisected angle. The cosine of angle ABC, which we know to be 60°, is \frac12, so \overline{BD} is 6cm. \overline{AE} is twice the size of \overline {DE}. For a triangle having an angle of 60°, the chances are simply ... 1 According to this Wikipedia article, the diameter of the circumscribed circle of a triangle can be found in terms of the length of its three sides, a,b,c :$$\frac{abc}{2\sqrt{s(s-a)(s-b)(s-c)}}$$where$$s = \frac{a + b + c}{2}$$. 1 A triangle \{e_1,e_2,e_3\}⊆V(LG) is said to be an odd triangle if there exists a vertex e∈V(G) incident to exactly one or all of \{e_1,e_2,e_3\}, and it is said to be even otherwise. [source] 1 The data you give does not allow the calculation of the bearings. You can get the three angles A,B,C using the law of cosines. From the data you have, the triangle could be oriented any way on the earth's surface. If I look on the map, it looks like the line from Bacoor to Sto. Tomas is at an angle of about 60 degrees South of East, and San Pablo is at ... 1 Using Law of Sines we have:$$\frac b{\sin B} = \frac c{\sin C} \iff \sin B = \frac{b \cdot \sin C}{c}$$Now solving this you'll get \angle B \approx 41,27^{\circ} or \angle B \approx 138.72^{\circ} Now check both cases. Use A+B+C = 180^{\circ} and Law of Sines to get 2 solutions. 1 Are they interchangeable, given the same information you have about a right triangle? \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad Yes. If \sin x is known, then \cos x=\sqrt{1-\sin^2x} and \tan x=\dfrac{\sin x}{\cos x}=\dfrac{\sin x}{\sqrt{1-\sin^2x}} . Similarly for \cos x. If \tan x is known, then \cos ... 1 Another answer without any nasty mathematical symbols or equations (and no pictures, either, except the ones in your mind) :-) Imagine that the triangle is drawn on a square piece of rubber. Now stretch the rubber square so that it's length and width get doubled. Then the sides of the triangle will be doubled, too. But, what happens to the area of the ... 1 the easy way is to have BP extended to Y, now prove AQ=YC AQ // YC,if we can prove AY// CQ, the problem is solved. check \angle YAC and \angle ACN \angle ACN +\angle CAN=? \angle YAC +\angle CAN=? you should get answer now. 1 We observe that \angle AIB = 90^{\circ} + (C/2). Extend CA to D such that AD = AI. Then, CD = CB by the hypothesis. Hence, \angle CDB = \angle CBD = 90^{\circ} - (C/2). Thus \angle AIB + \angle ADB = 180^{\circ}. Hence ADBI is a cyclic quadrilateral. This implies that \angle ADI = \angle ABI = B/2. But ADI is isosceles, since AD = ... 1 Use the angle bisector theorem to get that:$$\frac 45 = \frac {AB}{BC}$$(Note that BC > AB WHY???) where BC is hypotenuse. Now we have$$AB = \frac 45 BC$$Now since we know that AC = 9, use the Pyhtagorean Theorem and we have:$$\left(\frac {4BC}5\right)^2 + 9^2 = BC^2\frac{16BC^2}{25} + 81 = BC^2\frac{9BC^2}{25} = 81BC^2 = ...

1

You need to de-noise the surface in some way. One technique is to use Chambolle's algorithm, which is a computer vision technique based on total variation regularization. It is very fast, and it preserves planar dimensionality. It is fairly easy to implement, but your data do not appear to be in a grid, so it will be annoying to implement. Another method ...

1

The mistake on the last step: $$\frac{6}{\sqrt{2\pi}} \left[ \frac {\sin^2\frac{3\omega}2}{\frac{3\omega^2}2} \right] = \frac{6}{\sqrt{2\pi}} \left[ \frac {\sin^2\frac{3\omega}2}{\frac{2}{3}\left(\frac{3\omega}2\right)^2} \right] = \frac{9}{\sqrt{2\pi}} \left[ \frac {\sin\frac{3\omega}2}{\frac{3\omega}2} \right]^2 = \frac{9}{\sqrt{2\pi}} {{\rm ... 1 Solving ab+bc+ca=1 for c gives$$ c=\frac{1-ab}{a+b}\tag{1} $$The triangle inequality says that for non-degenerate triangles$$ |a-b|\lt c\lt(a+b)\tag{2} $$Multiply (2) by a+b to get$$ |a^2-b^2|\lt1-ab\lt(a+b)^2\tag{3} $$By (3), we have (a+b)^2-1+ab\gt0; therefore,$$ \begin{align} (a+b+1)(a+b+ab-1) &=\left[(a+b)^2-1+ab\right]+(a+b)ab\\ ...

1

Hint: label the points on the polygon $A,B,C,D,E,F$ such that $\angle ABC = 36^{\circ}$ and $\angle BCD = (360-84)^{\circ} = 276^{\circ}$. For any polygon with $n$ sides, the sum of the internal angles must be $(180(n-2))^{\circ} = (36 + 276 + 4a)^{\circ}$. Can you figure out why, and can you take it from here?

1

There is no single explicit equation that describes any triangle. What you can do is find the equations of each of the three lines joining the vertices. The points of intersection of those lines, of course, are the given vertices. Recall, given two points $(x_1, y_1), (x_2,y_2)$, the equation of the line that joins them is given by: (y - ...

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