Example Question - trigonometric ratios
Here are examples of questions we've helped users solve.
Solving a Right Triangle
<p>Paso 1: Determinar la hipotenusa \( C \)</p> <p>\( C = \sqrt{A^2+B^2} \)</p> <p>\( C = \sqrt{3^2+5^2} \)</p> <p>\( C = \sqrt{9+25} \)</p> <p>\( C = \sqrt{34} \)</p> <p>\( C \approx 5.83 \) (2 decimales)</p> <p>Paso 2: Encontrar las medidas de los ángulos \( A \) y \( B \)</p> <p>Usando las funciones trigonométricas:</p> <p>Para \( A \):</p> <p>\( \tan(A) = \frac{\text{Opuesto}}{\text{Adyacente}} = \frac{B}{A} \)</p> <p>\( \tan(A) = \frac{5}{3} \)</p> <p>\( A = \arctan\left(\frac{5}{3}\right) \)</p> <p>\( A \approx 59.04^\circ \) (2 decimales)</p> <p>Para \( B \) (sabemos que en un triángulo rectángulo \( A + B = 90^\circ \)):</p> <p>\( B = 90^\circ - A \)</p> <p>\( B \approx 90^\circ - 59.04^\circ \)</p> <p>\( B \approx 30.96^\circ \) (2 decimales)</p> <p>Paso 3: Determinar las 6 razones trigonométricas</p> <p>Sen \( A = \frac{\text{Opuesto a } A}{\text{Hipotenusa}} = \frac{B}{C} = \frac{5}{\sqrt{34}} \approx 0.857 \) (3 decimales)</p> <p>Cos \( A = \frac{\text{Adyacente a } A}{\text{Hipotenusa}} = \frac{A}{C} = \frac{3}{\sqrt{34}} \approx 0.515 \) (3 decimales)</p> <p>Tan \( A = \frac{\text{Opuesto a } A}{\text{Adyacente a } A} = \frac{B}{A} = \frac{5}{3} \approx 1.667 \) (3 decimales)</p> <p>Sen \( B = \frac{\text{Opuesto a } B}{\text{Hipotenusa}} = \frac{A}{C} = \frac{3}{\sqrt{34}} \approx 0.515 \) (3 decimales)</p> <p>Cos \( B = \frac{\text{Adyacente a } B}{\text{Hipotenusa}} = \frac{B}{C} = \frac{5}{\sqrt{34}} \approx 0.857 \) (3 decimales)</p> <p>Tan \( B = \frac{\text{Opuesto a } B}{\text{Adyacente a } B} = \frac{A}{B} = \frac{3}{5} \approx 0.600 \) (3 decimales)</p>
Solving a Triangle Problem Using Trigonometric Ratios
Para resolver este problema, primero identificamos que se trata de un triángulo rectángulo y que debemos usar las razones trigonométricas para encontrar la altura del objeto, representada por la línea BC en la figura 3.27. La información dada es el ángulo de elevación (40°) y la distancia horizontal a la base del objeto (20 cm), y necesitamos encontrar la longitud de la línea BC, es decir, la altura del objeto. Usaremos la tangente del ángulo de elevación para relacionar los lados opuesto y adyacente del triángulo rectángulo. La fórmula de la tangente de un ángulo en un triángulo rectángulo es: \[ \tan(\theta) = \frac{\text{Lado opuesto}}{\text{Lado adyacente}} \] Donde \( \theta \) es el ángulo de elevación, el lado opuesto es BC y el lado adyacente es AC. Usando la información dada en la figura 3.27 y el ángulo de elevación de 40°: \[ \tan(40^{\circ}) = \frac{BC}{20\text{ cm}} \] Ahora calculamos el valor numérico de \( \tan(40^{\circ}) \) y despejamos para BC: \[ BC = 20\text{ cm} \cdot \tan(40^{\circ}) \] Luego, al calcular el valor de \( \tan(40^{\circ}) \), obtenemos aproximadamente 0.8391 (este valor puede variar ligeramente dependiendo de la precisión de la calculadora utilizada). Finalmente, calculamos BC como: \[ BC = 20\text{ cm} \cdot 0.8391 \] \[ BC \approx 16.782\text{ cm} \] Por lo tanto, la altura del objeto, que es la línea BC, es aproximadamente 16.78 cm.
Finding Length of Side in Right Triangle
Para resolver la pregunta necesitamos aplicar relaciones trigonométricas en el triángulo rectángulo ABC. La medida del lado AB es la hipotenusa de este triángulo rectángulo y es de 11 unidades. Queremos encontrar la longitud del lado BD. Primero determinamos la longitud del lado BC usando la razón trigonométrica del seno: \[ \sin(45^\circ) = \frac{BC}{AB} \rightarrow \sin(45^\circ) = \frac{BC}{11} \] Como \(\sin(45^\circ)\) es \( \frac{\sqrt{2}}{2} \), entonces: \[ \frac{\sqrt{2}}{2} = \frac{BC}{11} \] Despejamos BC: \[ BC = 11 \cdot \frac{\sqrt{2}}{2} = \frac{11\sqrt{2}}{2} \] Ahora bien, BD es el lado adyacente al ángulo de 30° en el triángulo BCD, donde CD es la hipotenusa. Usamos la razón trigonométrica del coseno: \[ \cos(30^\circ) = \frac{BD}{CD} \] Como \( \cos(30^\circ) \) es \( \frac{\sqrt{3}}{2} \) y CD es igual a BC, entonces sustituímos los valores: \[ \frac{\sqrt{3}}{2} = \frac{BD}{\frac{11\sqrt{2}}{2}} \] Despejamos BD: \[ BD = \frac{11\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{11\sqrt{6}}{4} \] Entonces, la longitud de BD es \( \frac{11\sqrt{6}}{4} \) unidades.
Calculating Length in a Right Triangle
To find the length of \(x\) in the right triangle as depicted in the image, we can use trigonometric ratios. The triangle has an angle of 35 degrees, and we're given the length of the side opposite to this angle, which is 16 units. Since we have the angle and the opposite side, we can use the tangent function (tan) to find the length of the adjacent side (\(x\)). The tangent of an angle in a right triangle is equal to the opposite side divided by the adjacent side. So, we have: \[ \tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}} \] Substituting the known values, we get: \[ \tan(35^\circ) = \frac{16}{x} \] To isolate \(x\), we'll multiply both sides by \(x\) and then divide both sides by \(\tan(35^\circ)\): \[ x \cdot \tan(35^\circ) = 16 \] \[ x = \frac{16}{\tan(35^\circ)} \] Using a calculator and making sure it's set to degree mode, we can calculate the value of \(x\): \[ x \approx \frac{16}{0.70020753820971} \] \[ x \approx 22.849 \] So, the length of \(x\) rounded to three decimal places is approximately 22.849 units.
Trigonometric Ratios in a Right-Angled Triangle
The image shows a right-angled triangle, ΔABC, with angle A = 60°, angle B = 90°, and the hypotenuse AC = 20 ft. We are required to solve a series of questions related to this triangle. a) Using trigonometric ratios to find the perpendicular (opposite side to angle A, which is BC) and base (adjacent side to angle A, which is AB): We use the sine and cosine functions, which are defined as follows for a right-angled triangle: - Sine of an angle is the ratio of the length of the opposite side to the hypotenuse. - Cosine of an angle is the ratio of the length of the adjacent side to the hypotenuse. For angle A, sin(60°) = opposite/hypotenuse => sin(60°) = BC/AC BC = AC * sin(60°) BC = 20 * (√3/2) (since sin(60°) is √3/2) BC = 10√3 ft For the base (AB), cos(60°) = adjacent/hypotenuse => cos(60°) = AB/AC AB = AC * cos(60°) AB = 20 * (1/2) (since cos(60°) is 1/2) AB = 10 ft b) Using the perpendicular (BC) and base (AB) obtained above, find the value of tan(60°): Tangent of an angle is the ratio of the length of the opposite side to the adjacent side. tan(60°) = opposite/adjacent tan(60°) = BC/AB tan(60°) = (10√3)/10 tan(60°) = √3 c) Find the values of Sin C and Cos C: In a right triangle, the sine of one non-right angle is the cosine of the other, and vice versa. Since angle C is the 90-angle B (90° - 60° = 30°), we can find the sine and cosine of angle C by using their known values at 30°. sin(C) = sin(30°) = 1/2 cos(C) = cos(30°) = √3/2 d) Using the values of Sin C and Cos C, prove that Sin² C + Cos² C = 1: This is a well-known trigonometric identity known as the Pythagorean identity. Now let's substitute the values obtained for Sin C and Cos C: Sin² C + Cos² C = (1/2)² + (√3/2)² Sin² C + Cos² C = 1/4 + 3/4 Sin² C + Cos² C = 4/4 Sin² C + Cos² C = 1 This proves the Pythagorean identity for angle C.
Trigonometric Ratios in the Second Quadrant
The question gives us the cosine of an angle (θ) and the range for the angle, which is the second quadrant (\(\frac{\pi}{2} < \theta < \pi\)), and asks us to find the values of the remaining trigonometric ratios. Given: \[ \cos \theta = -\frac{\sqrt{3}}{2} \] Since the cosine of θ is negative and the angle is in the second quadrant, sine will be positive (as sine is positive in the second quadrant). Let's use the Pythagorean identity to find sine: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substitute the given value of cos θ: \[ \sin^2 \theta + \left(-\frac{\sqrt{3}}{2}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{3}{4} = 1 \] Solving for \(\sin^2 \theta\): \[ \sin^2 \theta = 1 - \frac{3}{4} = \frac{1}{4} \] Taking the square root to find sin θ yields two possible values, +1/2 and -1/2. Since we are in the second quadrant and sine is positive: \[ \sin \theta = \frac{1}{2} \] The tangent of θ is the ratio of sine to cosine: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] \[ \tan \theta = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} \] \[ \tan \theta = -\frac{1}{\sqrt{3}} \] \[ \tan \theta = -\frac{\sqrt{3}}{3} \] For the reciprocals: \[ \csc \theta = \frac{1}{\sin \theta} = 2 \] \[ \sec \theta = \frac{1}{\cos \theta} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \] \[ \cot \theta = \frac{1}{\tan \theta} = -\sqrt{3} \] So, the six trigonometric ratios for the given angle θ in the second quadrant are: \[ \sin \theta = \frac{1}{2}, \cos \theta = -\frac{\sqrt{3}}{2}, \tan \theta = -\frac{\sqrt{3}}{3} \] \[ \csc \theta = 2, \sec \theta = -\frac{2\sqrt{3}}{3}, \cot \theta = -\sqrt{3} \]
Solving for Length of Hypotenuse in Right-Angled Triangle
You have presented a right-angled triangle ABC with an altitude BD from the right angle B to the hypotenuse AC, creating two smaller right-angled triangles, ABD and BDC. You have provided several pieces of information: 1. The angle \( \angle ABC = 15^\circ \) 2. The angle \( \angle BAD = 45^\circ \) 3. The length of BD is equal to that of DC, i.e., \( BD = DC \) You've asked to solve for the value of y, where y is the length of AC. But there's no direct indication of how the lengths on the triangle relate to y. The given angles and conditions suggest that we can calculate the length of AC using trigonometric ratios. Since triangle ABD and BDC are both isosceles and right-angled, we know several relationships: 1. \( AB = BD \) (since \( \angle BAD = 45^\circ \) it is an isosceles right-angled triangle, so the two non-hypotenuse sides are equal) 2. \( BD = DC \) (given) 3. Therefore, \( AB = BD = BC \), and those lengths are all equal to each other. To find y, we need to use these relationships and the fact that \( \angle ABC = 15^\circ \). We'll use the trigonometric identity for the sine of a 15-degree angle, as we're given \( \angle ABC \) and we have a right-angled triangle. In right-angled triangle ABC: \[ \sin(15^\circ) = \frac{AB}{AC} = \frac{BD}{AC} \] We know that \( \sin(15^\circ) \) can be expressed using the half-angle identity: \[ \sin(15^\circ) = \sin(30^\circ / 2) = \sqrt{\frac{1 - \cos(30^\circ)}{2}} = \sqrt{\frac{1 - \sqrt{3}/2}{2}} \] Simplifying the expression under the square root gives us: \[ \sin(15^\circ) = \sqrt{\frac{2 - \sqrt{3}}{4}} \] Using this identity, we can relate the length of AC to BD: \[ \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{BD}{AC} \] Since \( BD = BC \) and \( BC = AB \), let's denote the length of BD (also AB and BC) as x. Then the above equation becomes: \[ \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{x}{y} \] We can solve for y by multiplying both sides by y and dividing by \( \sqrt{\frac{2 - \sqrt{3}}{4}} \): \[ y = \frac{x}{\sqrt{\frac{2 - \sqrt{3}}{4}}} \] However, we have not been provided the actual numerical length for x (BD, AB, or BC), so we cannot provide a numerical solution for y without additional information. If you have a length for BD, AB, or BC, you can substitute that value into the equation to solve for y.
Calculating Total Length of Timber for Symmetrical Roof Truss
The image provides a diagram of a symmetrical triangular roof truss with a pitch of 35°. It's given that \( AC = 6 \) meters, and we need to calculate the total length of timber required to make the truss, considering all the lengths \( AD, FB, \) and \( BG \) are equal, as well as \( GC \). Given the symmetry of the problem and the 35° pitch, we can break down the truss into different right-angled triangles and use trigonometric ratios to find the lengths of the sides. For the main triangle \( ABC \): - \( AC \) is the base and is given to be \( 6 \) meters. - \( AB \) can be found using the given pitch angle. The pitch of \( 35° \) tells us that angle \( BAC \) is \( 35° \). Since \( ABC \) is an isosceles triangle (because \( AB = BC \)), angle \( ABC = 35° \) as well. In right-angled triangle \( ABD \): - \( AD \) (which is half of \( AB \)) is the opposite side to angle \( BAC \) of \( 35° \). - \( AC/2 \) or \( 3 \) meters is the adjacent side to angle \( BAC \) of \( 35° \). Using the tangent function: \[ \tan(35°) = \frac{AD}{3} \] Solving for \( AD \): \[ AD = 3 \cdot \tan(35°) \] Since your truss is symmetrical and comprised of similar triangles, you can use this approach to calculate the total length of timber: - Length of \( AD \) (same as \( FB \), \( BG \), \( GC \)) = \( 3 \cdot \tan(35°) \) meters - Total length for all of these 4 members is \( 4 \cdot AD \). In the smaller triangles \( AFD \), \( BFE \), and \( BGC \), the angle at \( F \) and \( G \) will still be \( 35° \). Given that each smaller triangle is half the size of the larger one, we can apply the same approach to find the length of \( DF \) (which will also equal \( FE \) and \( GB \)): - \( DF \) (same as \( FE \), \( GB \); is half the size of \( AD \), so its length can be calculated as \( \frac{AD}{2} \)) = \( \frac{3 \cdot \tan(35°)}{2} \) meters. - Total length for all of these 3 members is \( 3 \cdot DF \). Now calculate the actual values for \( AD \) and \( DF \) using a calculator for tangent of \( 35° \), and then add them together to find the total length of timber needed: \[ \text{Total timber length} = 4 \cdot AD + 3 \cdot DF \] Let's calculate the values using a calculator: \[ AD = 3 \cdot \tan(35°) \approx 3 \cdot 0.7002 \approx 2.1006 \text{ meters} \] \[ DF = \frac{3 \cdot \tan(35°)}{2} \approx \frac{3 \cdot 0.7002}{2} \approx 1.0503 \text{ meters} \] Now, let's find the total length of timber: \[ \text{Total timber length} = 4 \cdot (2.1006) + 3 \cdot (1.0503) \approx 4 \cdot 2.1006 + 3 \cdot 1.0503 \approx 8.4024 + 3.1509 = 11.5533 \text{ meters} \] So, the total length of timber required to make the truss is approximately \( 11.5533 \) meters, depending on how accurate the approximation or calculator is.
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