# Search results

• (Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.) The proof using [[Routh's Theorem]] is extremely trivial, so we will not include it.
5 KB (842 words) - 02:35, 24 July 2020
• ...jacent side" is the side that is part of angle $\theta$, but is not the hypotenuse. ...ue to the way trig ratios are defined for non-acute angles, the value of a trig ratio could be positive or negative, or even 0.
4 KB (674 words) - 02:18, 29 December 2019
• ==== Non-Trig Approach ==== .../math> is $\frac{\pi}{3}$. We also know that the triangles will not be in terms of ${\pi}$. Looking at the answers, choices $\t 5 KB (802 words) - 23:20, 30 December 2020 • ...>b$, and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c$. ==Solution 2 (no trig)==
2 KB (405 words) - 00:00, 16 February 2021
• ...th> where $m$ and $n$ are positive integers and n is not divisible by the square of any prime. Find $m + n.$ ==Solution 4 (No Trig)==
8 KB (1,357 words) - 19:14, 31 January 2021
• ==Solution 2(no trig)== ==Solution 3(trig)==
4 KB (561 words) - 21:11, 30 October 2020
• ...th> and $r$ are [[relatively prime]], and $q$ is not [[divisibility | divisible]] by the [[perfect square | square]] of any prim == Solution 3: Trig ==
5 KB (915 words) - 17:56, 22 March 2021
• Note: Once $DY$ is found, there is no need to do the trig. Notice that the hexagon consists of two trapezoids, $ABPQ$ and ...= X, AP \cap CD = Y.[/itex] Then, angle chasing shows that $CQ$ not only bisects $BX,$ but is also perpendicular to it. This makes i
12 KB (2,006 words) - 13:47, 15 February 2021
• ...reas of the forums. You should also try to strengthen in the areas you are not as good at. This guide is intended to help you get started. ...n the right level for you. '''Alcumus''' is a good resource even if you do not own any of the AoPS books.
16 KB (2,332 words) - 15:36, 26 March 2021
• ...irc[/itex] and $\angle ADC = 60^\circ$ as you can see, let's do trig. Drop an altitude from $A$ to $BC$; call this point < ...75^\circ[/itex] or $105^\circ$. Since $105^\circ$ is not a choice, we know $\angle ACB = \boxed{75^\circ}$.
5 KB (853 words) - 11:56, 27 December 2020
• ...th>\sin^{-1}(\sin(6x)) = 180 - 6x[/itex]. As $\frac{180}{7}$ is not on the interval $30 \leq x \leq 45$, this yields no solution. .... Thus, $\sin^{-1}(\sin(6x)) = 6x - 360$. As $72$ is not on the interval $45 \leq x \leq 60$, this yields no solution.
6 KB (1,050 words) - 14:43, 20 December 2020
• ...egative. Continuing the pattern and accounting for doubled roots (which do not flip sign), we realize that there are $5$ negative intervals fro ...count it as two disjoint intervals. Note that this will be important as to not undercounting disjoint intervals. )
9 KB (1,423 words) - 13:17, 9 December 2020
• while word not in vowels: ...erything from position 1, including position 1, all the way to position 3, not including position 3. This is called a slice.
28 KB (4,756 words) - 16:56, 18 January 2021
• ==Solution 2 (with trig, not recommended)== ==Solution 3(Easier Trig)==
4 KB (562 words) - 23:15, 12 January 2021
• ...d \mathrm{(C) \ } 15 \qquad \mathrm{(D) \ }18 \qquad \mathrm{(E) \ } \text{Not uniquely determined} [/itex] .../math> to get that $BD=\frac{c\cos{\beta}}{2}$. We can also use trig manipulation on $BCE$ to get that $CE=a\tan{\beta}$.
2 KB (332 words) - 20:37, 12 February 2017
• ...latively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ ...ector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig)
8 KB (1,392 words) - 20:33, 24 January 2021
• ...[/itex], where $a,b,c$ are positive integers, $b$ is not divisible by the square of any prime, and $a$ and $c$ == Solution 2 (Trig) ==
5 KB (831 words) - 17:55, 21 July 2018
• ...h>, where $a$ and $d$ are relatively prime, and c is not divisible by the square of any prime. Find $a+b+c+d$. ...ath>H[/itex], and note that $QP = 1 + \sqrt{3}$ after using the trig functions for $75$ degrees.
7 KB (1,070 words) - 20:11, 24 January 2021
• ...enominator a difference (or rather a sum) of squares. The denominator does not matter. Only the numerator, because we are trying to find $\frac{P}{Q} ...theta}-2\sin{2\theta}}{8+2\cos{2\theta}-3\sin{\theta}}$. We use some trig identities and we get $\frac{\cos{\theta}(5-4\sin{\theta})}{10-4\sin^2 10 KB (1,678 words) - 20:13, 24 January 2021 • ...th>m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$. ...\cos C[/itex], so $\cos C = \frac{1}{8}$. Using the Pythagorean trig identity $\sin^2 + \cos^2 = 1$, [itex]\sin^2 C = 1 - \frac{1}{64
9 KB (1,465 words) - 10:19, 20 December 2020

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