Example Question - triangle
Here are examples of questions we've helped users solve.
Determining Angles in a Triangle
<p>Given triangle ABC with angle C = 108° and angle E = 36°, we can find angle A.</p> <p>Using the angle sum property of triangles:</p> <p>Angle A + Angle B + Angle C = 180°</p> <p>Let Angle B = 180° - 108° - 36° = 36°</p> <p>Thus, Angle A = 180° - (108° + 36°) = 36°.</p>
Finding the Value of a Variable in a Triangle
<p>Para encontrar el valor de \( x \), utilizamos la propiedad de que la suma de los ángulos en un triángulo es igual a 180 grados.</p> <p>Los ángulos dados son \( 37^\circ \) y \( x \). Si el tercer ángulo se denota como \( 120^\circ \), entonces podemos plantear la ecuación:</p> <p> \( 37^\circ + x + 120^\circ = 180^\circ \)</p> <p>Resolviendo, tenemos:</p> <p> \( x = 180^\circ - 37^\circ - 120^\circ \)</p> <p> \( x = 23^\circ \)</p> <p>En conclusión, el valor de \( x \) es \( 23^\circ \).</p>
Finding Distance in a Triangle
<p>Дано треугольник ABC со сторонами AB = 5 см, BC = 8 см, AC = 9 см.</p> <p>Сначала находим полупериметр треугольника: </p> <p>s = (AB + BC + AC)/2 = (5 + 8 + 9)/2 = 11 см.</p> <p>Находим радиус вписанной окружности r: </p> <p>r = A / s, где A — площадь треугольника. Используем формулу Герона для нахождения A: </p> <p>A = √(s(s - AB)(s - BC)(s - AC)) = √(11(11 - 5)(11 - 8)(11 - 9)) = √(11 * 6 * 3 * 2) = √(396) = 6√11.</p> <p>Следовательно, r = (6√11) / 11.</p> <p>Теперь находим расстояние от точки K до точки M биссектрисы BM. Сначала находим длину BM: </p> <p>BM = (AC * AB) / (AB + AC) = (9 * 5) / (5 + 9) = 45/14.</p> <p>Теперь, зная BM и KL = r, можем найти KM: </p> <p>KM = BM - r.</p>
Triangle Geometry Problem
<p>Given triangle ABC with angle A = 32° and sides AC = 8 and BC = 5.5, we will use the Law of Sines to find side AB.</p> <p>According to the Law of Sines:</p> <p> \[ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \] </p> <p>Let AB = c, then:</p> <p> \[ \frac{c}{\sin(32°)} = \frac{8}{\sin(B)} \] \end{p> <p>To find angle B, we can use the sine rule again:</p> <p> \[ \frac{5.5}{\sin(B)} = \frac{8}{\sin(32°)} \end{p> <p>Rearranging gives:</p> <p> \[ \sin(B) = \frac{5.5 \cdot \sin(32°)}{8} \end{p> <p>Calculating sin(B) and then angle B, we can find angle C = 180° - A - B.</p> <p>Finally, using angle C, apply the Law of Sines again to find side c:</p> <p> \[ \frac{c}{\sin(C)} = \frac{5.5}{\sin(B)} \end{p>
Geometry Problem Involving Tangents and Circles
<p>Let $\angle ABC = \angle BAD = y$ since they are angles subtended by the same arc AD. Similarly, let $\angle ACB = \angle ADB = z$ since they are subtended by the same arc AB.</p> <p>Since AE and AX are tangents to the circle at points B and C, and tangents from a point outside a circle are equal, we have AE = AB and AX = AC.</p> <p>Triangle ABE is isosceles with AE = AB, so $\angle ABE = \angle AEB = y$. Similarly, triangle ACX is isosceles with AC = AX, so $\angle ACX = \angle AX = z$.</p> <p>Angles around point A add up to 360°, thus $\angle EAX + 2y + 2z = 360°$.</p> <p>Substitute $\angle EAX = 80°$ to find $2y + 2z = 280°$.</p> <p>Divide by 2 to find $y + z = 140°$.</p> <p>The angle sum in triangle ABC is 180°, thus $y + z + \angle BAC = 180°$.</p> <p>Substitute $y + z = 140°$ into the previous equation to find $\angle BAC = 40°$.</p> <p>Now, focusing on quadrilateral ADBC, notice that $\angle ABC + \angle BAD + \angle BAC + \angle BCD = 360°$.</p> <p>Since $\angle BAC = 40°$, $\angle ABC + \angle BAD = 2y = 140°$, and $\angle BCD = 90°$ (angle in a semicircle), we substitute these into the quadrilateral angle sum to find $2y + 40° + 90° = 360°$.</p> <p>Combine like terms to get $2y + 130° = 360°$.</p> <p>Subtract 130° from both sides to solve for $2y$, yielding $2y = 230°$.</p> <p>Divide by 2 to find $y = 115°$.</p> <p>Finally, $\angle AFX = 115°$ because it is equal to $\angle ABC$ which is equal to $y$.</p>
Geometry Problem with Intersecting Circles and Triangle
Unfortunately, the image does not provide a specific question to be solved. It simply shows two intersecting circles and a triangle, with some points labeled. To solve a problem, specific measurements, or a question regarding the relationships between the points, lines, and circles, would need to be provided. As such, without additional context or a specified problem, I am unable to provide a solution. Please provide a clear question or the measurements necessary to solve a particular problem related to this diagram.
Finding the General Coordinates of a Point in a Triangle
<p>Let the coordinates of A be $(x_1, y_1, z_1)$ and B be $(x_2, y_2, z_2)$, and let C be $(x_3, y_3, z_3)$. The centroid G of triangle ABC has coordinates given by:</p> <p>$G \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right)$</p> <p>Given that G is at (1, 1, 1), we have:</p> <p>$1 = \frac{x_1+x_2+x_3}{3}$</p> <p>$1 = \frac{y_1+y_2+y_3}{3}$</p> <p>$1 = \frac{z_1+z_2+z_3}{3}$</p> <p>Given the coordinates of A $(x_1, y_1, z_1) = (3, -5, 7)$ and B $(x_2, y_2, z_2) = (-1, 7, 6)$, we can substitute them into the equations:</p> <p>$1 = \frac{3 - 1 + x_3}{3}$</p> <p>$1 = \frac{-5 + 7 + y_3}{3}$</p> <p>$1 = \frac{7 + 6 + z_3}{3}$</p> <p>Now solve for $x_3, y_3, z_3$:</p> <p>$1 = \frac{2 + x_3}{3}$ -> Multiply both sides by 3 -> $3 = 2 + x_3$ -> $x_3 = 1$</p> <p>$1 = \frac{2 + y_3}{3}$ -> Multiply both sides by 3 -> $3 = 2 + y_3$ -> $y_3 = 1$</p> <p>$1 = \frac{13 + z_3}{3}$ -> Multiply both sides by 3 -> $3 = 13 + z_3$ -> $z_3 = -10$</p> <p>Thus, the coordinates of point C are $(x_3, y_3, z_3) = (1, 1, -10)$.</p>
Triangle Angle Calculation Problem
<p>Let the angles of the triangle be A, B, and C.</p> <p>Given: A = 50 degrees, B = 60 degrees.</p> <p>The sum of the angles in a triangle is always 180 degrees.</p> <p>A + B + C = 180 degrees</p> <p>50 degrees + 60 degrees + C = 180 degrees</p> <p>110 degrees + C = 180 degrees</p> <p>C = 180 degrees - 110 degrees</p> <p>C = 70 degrees</p>
Triangle Side Relationships
Para resolver el problema, necesitamos usar el teorema de Pitágoras para verificar la validez de las longitudes de los lados del triángulo. El teorema de Pitágoras se expresa como \( c^2 = a^2 + b^2 \), donde \( c \) es la longitud de la hipotenusa y \( a \) y \( b \) son las longitudes de los catetos. <p>Tomamos los valores dados en la imagen:</p> <p>\( a = 9 \) cm (cateto opuesto)</p> <p>\( b = 12 \) cm (cateto adyacente)</p> <p>\( c = 16 \) cm (hipotenusa)</p> <p>Aplicamos el teorema de Pitágoras:</p> \[ c^2 = a^2 + b^2 \] \[ 16^2 = 9^2 + 12^2 \] \[ 256 = 81 + 144 \] \[ 256 = 225 \] <p>Como \( 256 \neq 225 \), la relación dada no cumple con el teorema de Pitágoras. Por lo tanto, las longitudes dadas para los lados del triángulo son incorrectas.</p>
Trigonometric Ratio Calculation from Triangle Dimensions
Para el inciso a: \[ \sin(\alpha) = \frac{opuesto}{hipotenusa} = \frac{BC}{AB} = \frac{2 \text{ mm}}{4 \text{ mm}} = \frac{1}{2} \] Para el inciso b: \[ \sin(\alpha) = \frac{opuesto}{hipotenusa} = \frac{BC}{AB} = \frac{4 \text{ cm}}{\sqrt{10} \text{ cm}} = \frac{4}{\sqrt{10}} = \frac{4\sqrt{10}}{10} = \frac{2\sqrt{10}}{5} \]
Solving for a Triangle Side Using Trigonometric Ratios
<p>De acuerdo con la información proporcionada y utilizando la relación trigonométrica del seno, podemos encontrar el valor de \( x \). La fórmula para el seno de un ángulo en un triángulo rectángulo es:</p> <p>\[ \sin(\theta) = \frac{\text{opuesto}}{\text{hipotenusa}} \]</p> <p>Dado que \( \sin(\alpha) = 0.28 \) y la hipotenusa del triángulo es 8 unidades, podemos establecer la siguiente ecuación:</p> <p>\[ \sin(\alpha) = \frac{x}{8} \]</p> <p>Entonces:</p> <p>\[ 0.28 = \frac{x}{8} \]</p> <p>\[ x = 0.28 \cdot 8 \]</p> <p>\[ x = 2.24 \]</p> <p>Por lo tanto, el valor de \( x \) es 2.24 unidades.</p>
Calculation of a Triangle Side
<p>Из условия задачи, \( LN = 3 \) и \( LM = 4 \). Так как \( L \) – прямой угол, треугольник \( MNL \) является прямоугольным и по теореме Пифагора:</p> <p>\[ LK^2 = LM^2 + LN^2 \]</p> <p>\[ LK^2 = 4^2 + 3^2 \]</p> <p>\[ LK^2 = 16 + 9 \]</p> <p>\[ LK^2 = 25 \]</p> <p>\[ LK = \sqrt{25} \]</p> <p>\[ LK = 5 \]</p> <p>Таким образом, длина стороны \( LK \) этого треугольника равна 5.</p>
Geometry Problem on Calculating the Median Length in a Triangle
<p>Обозначим длину медианы \(MK\) как \(x\).</p> <p>Так как \(MK\) является медианой треугольника \(MNL\), медиана делит сторону \(NL\) на две равные части, следовательно \(NK = KL = \frac{LM}{2} = 2\).</p> <p>По теореме Аполлония для медианы \(MK\) выполняется равенство: \(2(MK^2 + KN^2) = ML^2 + MN^2\).</p> <p>\(2(x^2 + 2^2) = 3^2 + 4^2\)</p> <p>\(2(x^2 + 4) = 9 + 16\)</p> <p>\(2x^2 + 8 = 25\)</p> <p>\(2x^2 = 25 - 8\)</p> <p>\(2x^2 = 17\)</p> <p>\(x^2 = \frac{17}{2}\)</p> <p>\(x = \sqrt{\frac{17}{2}}\)</p> <p>Длина медианы \(MK\) равна \(\sqrt{\frac{17}{2}}\).</p>
Finding Algebraic Expressions for the Areas of Geometric Shapes
Para el rectángulo de la izquierda, el área \( A \) es el producto de la longitud y la altura: <p>\( A_{\text{rectángulo izquierdo}} = 3z \cdot z \)</p> <p>\( A_{\text{rectángulo izquierdo}} = 3z^2 \)</p> Para el triángulo en el medio, el área \( A \) es un medio del producto de la base por la altura: <p>\( A_{\text{triángulo}} = \frac{1}{2} \cdot x \cdot h \)</p> Para el rectángulo de la derecha, el área \( A \) es el producto de la longitud y la altura: <p>\( A_{\text{rectángulo derecho}} = x \cdot (2x + 1) \)</p> <p>\( A_{\text{rectángulo derecho}} = 2x^2 + x \)</p>
Finding the Area of a Shaded Region in a Geometric Diagram
The problem is to find the area of region EGFB within rectangle ABCD. We're given that the area of rectangle ABCD is 112 cm². First, let's express the lengths BE, EG, and BF in terms of a single variable. Considering that BE is one-third of EG, let's say that EG = 3x, so BE = x. Furthermore, given that BF is half of 5FC and FC is equal to EG (that is, 3x), we have BF = \(\frac{5}{2}\)x. Next, we can find the side lengths of ABCD by exploiting the fact that the area of the rectangle is the product of its length and width: - The width of the rectangle is BE + EG, which is x + 3x = 4x. - The length of the rectangle is BF, which is \(\frac{5}{2}\)x. Area of ABCD is then: \[ 4x \times \frac{5}{2}x = 112 \] \[ 10x^2 = 112 \] \[ x^2 = \frac{112}{10} \] \[ x^2 = 11.2 \] \[ x = \sqrt{11.2} \] \[ x \approx 3.35 \] Now that we have x, we can calculate the length of EG (which is 3x): \[ EG = 3 \times 3.35 \approx 10.05 \] We also calculate BF: \[ BF = \frac{5}{2} \times 3.35 \approx 8.38 \] The area of triangle BEG is half of the rectangle ABCD's area because BEG cuts the rectangle diagonally in half: \[ \text{Area of BEG} = \frac{\text{Area of ABCD}}{2} \] \[ \text{Area of BEG} = \frac{112}{2} \] \[ \text{Area of BEG} = 56 \text{ cm}^2 \] Finally, we find the area of triangle BFG using its base BF and height x (remember that height is BE): \[ \text{Area of BFG} = \frac{1}{2} \times BF \times BE \] \[ \text{Area of BFG} = \frac{1}{2} \times 8.38 \times 3.35 \] \[ \text{Area of BFG} \approx \frac{1}{2} \times 28.07 \] \[ \text{Area of BFG} \approx 14.04 \text{ cm}^2 \] To find the area of the shaded region EGFB, we subtract the area of triangle BFG from the area of triangle BEG: \[ \text{Area of EGFB} = \text{Area of BEG} - \text{Area of BFG} \] \[ \text{Area of EGFB} = 56 - 14.04 \] \[ \text{Area of EGFB} \approx 41.96 \text{ cm}^2 \] So, the area of EGFB is approximately 41.96 cm².
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