Example Question - opposite angles
Here are examples of questions we've helped users solve.
Vertical Angles and Angle Equality
The image shows two lines that intersect each other forming an angle of 113°. The other angle labeled "x°" is the vertical angle to the 113° angle. Vertical angles, also known as vertically opposite angles, are pairs of opposite angles made by two intersecting lines. Vertical angles are always equal to one another. Thus, the angle marked "x°" is also equal to 113°.
Solving Angles in a Rhombus
Claro, voy a ayudarte a resolver el problema en la imagen. La ecuación que se proporciona es: \[ x + \beta = 70^\circ \] Además, hay un dibujo de un rombo con ángulos marcados. Los ángulos opuestos de un rombo son iguales, y los ángulos en los vértices donde se encuentran los lados iguales son los ángulos agudos \( \alpha \) y \( \beta \). Los ángulos obtusos están marcados como \( 80^\circ \) en el dibujo, lo que significa que los ángulos \( \alpha \) y \( \beta \) son iguales a \( 80^\circ \) ya que son opuestos a estos. Dado que la suma de los ángulos en cualquier cuadrilátero es \( 360^\circ \), y sabemos que dos ángulos son \( 80^\circ \) y los otros dos son \( \alpha \) y \( \beta \), podemos establecer la siguiente ecuación: \[ 2 \cdot 80^\circ + \alpha + \beta = 360^\circ \] Si reemplazamos \( \beta \) por \( 70^\circ - x \) (usando la ecuación dada \( x + \beta = 70^\circ \)) en la ecuación anterior, obtenemos: \[ 2 \cdot 80^\circ + \alpha + (70^\circ - x) = 360^\circ \] Ahora simplificamos y resolvemos para \( x \): \[ 160^\circ + \alpha + 70^\circ - x = 360^\circ \] \[ 230^\circ + \alpha - x = 360^\circ \] \[ \alpha - x = 360^\circ - 230^\circ \] \[ \alpha - x = 130^\circ \] Sabiendo que \( \alpha = 80^\circ \) (ya que es opuesto al otro ángulo marcado en el dibujo), remplazamos para encontrar \( x \): \[ 80^\circ - x = 130^\circ \] \[ - x = 130^\circ - 80^\circ \] \[ - x = 50^\circ \] \[ x = -50^\circ \] Sin embargo, una respuesta negativa no tiene sentido en esta situación, ya que estamos hablando de ángulos de un rombo. Podría ser que haya un error en el dibujo o en la interpretación de los datos proporcionados. En el contexto de geometría euclidiana, los ángulos de un rombo no pueden ser negativos, por lo que es posible que necesites revisar las condiciones iniciales o el planteamiento del problema.
Finding the Measure of Angle WYX in a Rhombus
The image shows a geometric figure labeled WXYZ, which is identified as a rhombus. There is an angle at vertex Y that is marked as 75 degrees. The question asks for the measure of the angle ∠WYX. In a rhombus, opposite angles are equal and adjacent angles are supplementary (meaning they add up to 180 degrees). Since the angle at vertex Y is given as 75 degrees, the angle at vertex W (which is adjacent to angle Y) would be supplementary to it. Therefore, to find the measure of angle W (∠WYX), we need to subtract the angle at vertex Y from 180 degrees: ∠WYX = 180 degrees - ∠Y = 180 degrees - 75 degrees = 105 degrees So the measure of angle ∠WYX is 105 degrees.
Vertical Angles and Straight Line Angles Calculation
The image depicts two angles formed by a straight line and a vertically opposite angle, which are always equal. The two angles given are: 1. \((4x + 2)^\circ\) 2. \((7x - 6)^\circ\) Since they are vertically opposite angles, they are equal. Therefore, we can set up the following equation: \(4x + 2 = 7x - 6\) Now, let's solve for \(x\): \(4x + 2 = 7x - 6\) \(6 + 2 = 7x - 4x\) \(8 = 3x\) \(x = \frac{8}{3} \approx 2.67\) Now we can use the value of \(x\) to find \(y\). Looking at the lines, we can see that they form a straight line and the sum of angles on a straight line is \(180^\circ\). This information can be used to form an equation using either of the two expressions for the angles given: Using the angle \((4x + 2)^\circ\), we get: \(4x + 2 + y = 180\) Now substituting the value of \(x\) we found: \(4(\frac{8}{3}) + 2 + y = 180\) \( \frac{32}{3} + 2 + y = 180\) \( \frac{38}{3} + y = 180\) Multiply both sides by 3 to rid the fraction: \(38 + 3y = 540\) Now subtract 38 from both sides: \(3y = 540 - 38\) \(3y = 502\) Divide both sides by 3 to solve for \(y\): \(y = \frac{502}{3}\) \(y = 167.\overline{333}\) or \(167\frac{1}{3}^\circ\) Hence, the values are: \(x \approx 2.67\) \(y \approx 167.33^\circ\) or \(167\frac{1}{3}^\circ\)
Solving for a Side Length in a Right Triangle Using Law of Sines
To solve for \( w \), we'll use the Law of Sines, which states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is the same for all three sides. Given triangle has angles 30°, 60°, and because the sum of angles in a triangle is always 180°, the remaining angle is 90°, making it a right triangle. The side opposite the 30° angle is half the hypotenuse in a right triangle, while the side opposite the 60° angle is equal to \( \sqrt{3}/2 \) times the hypotenuse. Since we have: \[ \frac{w}{\sin(90°)} = \frac{10}{\sin(60°)} \] We know that \( \sin(90°) = 1 \) and \( \sin(60°) = \sqrt{3}/2 \), so we have: \[ w = 10 \times \frac{1}{\sqrt{3}/2} = 10 \times \frac{2}{\sqrt{3}} = \frac{20}{\sqrt{3}} \] To express in simplest radical form, we rationalize the denominator: \[ w = \frac{20}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{20\sqrt{3}}{3} \] Thus, \( w = \frac{20\sqrt{3}}{3} \) meters in simplest radical form.
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