Example Question - angle
Here are examples of questions we've helped users solve.
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>
Determining the Value of an Angle in a Diagram
<p>Unfortunately, the image is blurry and the specific details needed to solve the problem, such as angle measurements and the exact configuration of the diagram, are not clearly visible. Without these details, it's impossible to provide the steps to find the value of the angle in question. Please provide a clearer image or additional information for a proper solution.</p>
Solving for the Sine of an Angle in Similar Triangles
<p>Given that \(\sin(F) = \frac{308}{317}\) and triangles FGH and JKL are similar with angle F corresponding to angle J, and angles G and K are right angles, we have:</p> <p>\(\sin(J) = \sin(F) = \frac{308}{317}\)</p> <p>Therefore, the value of \(\sin(J)\) is \(\frac{308}{317}\).</p> <p>The correct option is C) \(\frac{308}{317}\).</p>
Calculating Work Done by a Force at an Angle
<p>The work \( W \) done by a force \( \vec{F} \) on an object through a displacement \( \vec{d} \) is given by the dot product of the force and displacement vectors:</p> <p>\( W = \vec{F} \cdot \vec{d} \)</p> <p>When the force is applied at an angle \( \theta \) to the direction of displacement, the work done is:</p> <p>\( W = F \cdot d \cdot \cos(\theta) \)</p> <p>Given the magnitude of the force \( F = 50 \, \text{N} \), the displacement \( d = 10 \, \text{m} \), and the angle \( \theta = 30^\circ \), we can calculate the work done as follows:</p> <p>\( W = 50 \, \text{N} \cdot 10 \, \text{m} \cdot \cos(30^\circ) \)</p> <p>\( W = 500 \, \text{N} \cdot \text{m} \cdot \frac{\sqrt{3}}{2} \)</p> <p>\( W = 250\sqrt{3} \, \text{J} \)</p> <p>Thus, the work done is \( 250\sqrt{3} \, \text{Joules} \).</p>
Calculating Work Done by a Force at an Angle
<p>The work \( W \) done by a force when moving an object through a displacement \( d \) at an angle \( \theta \) to the direction of the force is given by:</p> <p>\[ W = F \cdot d \cdot \cos(\theta) \]</p> <p>Given that the force \( F \) is \( 50 \text{N} \), the displacement \( d \) is \( 10 \text{m} \), and the angle \( \theta \) is \( 30^\circ \):</p> <p>\[ W = 50 \cdot 10 \cdot \cos(30^\circ) \]</p> <p>First, calculate \( \cos(30^\circ) \):</p> <p>\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]</p> <p>Then compute the work done \( W \):</p> <p>\[ W = 50 \cdot 10 \cdot \frac{\sqrt{3}}{2} \]</p> <p>\[ W = 500 \cdot \frac{\sqrt{3}}{2} \]</p> <p>\[ W = 250\sqrt{3} \text{J} \]</p> <p>Therefore, the work done by the force is \( 250\sqrt{3} \) joules.</p>
Work Done by a Force at an Angle
<p>The work done by a force when the force is applied at an angle to the direction of displacement can be calculated using the formula:</p> <p>\( W = F \cdot d \cdot \cos(\theta) \)</p> <p>where:</p> <p>\( W \) is the work done,</p> <p>\( F \) is the magnitude of the force,</p> <p>\( d \) is the displacement,</p> <p>\( \theta \) is the angle between the force and the displacement.</p> <p>Given that \( F = 50 \) N, \( d = 10 \) m, and \( \theta = 30^\circ \), the work done \( W \) is</p> <p>\( W = 50 \cdot 10 \cdot \cos(30^\circ) \)</p> <p>\( W = 500 \cdot \frac{\sqrt{3}}{2} \)</p> <p>\( W = 250\sqrt{3} \) J</p> <p>Therefore, \( 250\sqrt{3} \) joules of work is done by the force.</p>
Solving for the Sine of Angle R in a Right Triangle Given the Tangent of Angle S
Given that \(\tan(S) = \frac{1}{\sqrt{3}}\) and \(\angle T = 90\) degrees, so \(\angle S + \angle R = 90\) degrees. \(\tan(S) = \frac{1}{\sqrt{3}} = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sin(S)}{\cos(S)}\) \(\sin(S) = \cos(R)\) since \(\sin(\theta) = \cos(90 - \theta)\) Therefore, \(\sin(S) = \frac{1}{\sqrt{3}}\) Now, since \(\sin(S) = \frac{1}{\sqrt{3}}\), we have \(\sin(R) = \sin(90 - S)\) \(\sin(R) = \cos(S) = \sqrt{1 - \sin^2(S)}\) Calculating \(\cos(S)\) given \(\sin(S)\): \(\cos(S) = \sqrt{1 - \left(\frac{1}{\sqrt{3}}\right)^2}\) \(\cos(S) = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}}\) \(\sin(R) = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{2} \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}\) \(\sin(R) = \frac{\sqrt{6}}{3}\) Since this option is not provided in the multiple-choice answers, it must be simplified further: \(\sin(R) = \frac{\sqrt{6}}{3} = \frac{2 \cdot \sqrt{6}}{2 \cdot 3}\) \(\sin(R) = \frac{2\sqrt{6}}{6}\) \(\sin(R) = \frac{\sqrt{6}}{3}\) Hence, the correct answer is \(\sin(R) = \frac{\sqrt{6}}{3}\), which is not listed in the provided options, there might be an error in the provided options or question setup.
Finding the Value of an Angle in a Kite-like Figure
\begin{align*} m\angle EAD + m\angle ADC + m\angle DAB &= 180^\circ \quad \text{(Sum of interior angles of $\triangle ADB$)} \\ x + 118^\circ + 30^\circ &= 180^\circ \\ x + 148^\circ &= 180^\circ \\ x &= 180^\circ - 148^\circ \\ x &= 32^\circ \end{align*}
Rate of Change of Triangle Area with Changing Sides and Angle
To solve the given problem, we must determine the rate of change of the area of a triangle with sides A and B and angle B between them, given how A and B are changing with time. We are given: - The area of a triangle is given by the formula \( A = \frac{1}{2} ab \sin{\theta} \), where \( a \) and \( b \) are the sides of the triangle and \( \theta \) is the angle between them. For our problem, we can rewrite the given area formula as \( Area = \frac{1}{2} AB \sin{B} \). - Side \( A \) is increasing at 0.1 units/s. - Side \( B \) is decreasing at 0.1 units/s. - Angle \( B \) is increasing at 0.2 radians/s. We want to find the rate of change of the triangle's area, which can be found by differentiating the area formula with respect to time \( t \). This is an application of the chain rule in calculus. So we differentiate \( Area = \frac{1}{2} AB \sin{B} \) with respect to \( t \): \( \frac{d(Area)}{dt} = \frac{1}{2} \left[ B \sin{B} \frac{dA}{dt} + A \sin{B} \frac{dB}{dt} + AB \cos{B} \frac{dB}{dt} \right] \) Plugging in the rates given: \( \frac{dA}{dt} = 0.1 \, \text{units/s} \) \( \frac{dB}{dt} = -0.1 \, \text{units/s} \) \( \frac{dB}{dt} = 0.2 \, \text{radians/s} \) Substitute these into the derivative: \( \frac{d(Area)}{dt} = \frac{1}{2} \left[ B \cdot \sin{B} \cdot 0.1 + A \cdot \sin{B} \cdot (-0.1) + AB \cdot \cos{B} \cdot 0.2 \right] \) Unfortunately, the image doesn't provide the exact values of \( A \), \( B \), and angle \( B \) at the moment of calculation, so we cannot give a numerical answer. You would need to substitute the current values of \( A \), \( B \), and angle \( B \) into the derivative to find the rate of change at that particular moment.
Understanding Inverse Trigonometric Functions
The question in the image asks to evaluate the expression: \[ \cos(\cos^{-1}(-0.9)) \] By definition, the inverse cosine function, \(\cos^{-1}(x)\), returns the angle whose cosine is \(x\). Therefore, when we apply the cosine function to the result of an inverse cosine function, we get back the original value inside the inverse cosine. For the given problem: \[ \cos(\cos^{-1}(-0.9)) = -0.9 \] This is because the inverse cosine of \(-0.9\) gives us the angle whose cosine is \(-0.9\), and taking the cosine of that angle returns us back to the original value of \(-0.9\). So the correct answer is: -0.9
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