Acceso
Prima
Hogar Banco de preguntas Matemáticas Trigonometría
Pregunta
upstudy study bank question image url

**\#1.) In right triangle \( A B C, m \angle A=90^{\circ} \). Which of the following is true? a) \( \sin B=\sin C \) b) \( \sin C=\cos C \) c) \( \sin B=\cos C \) d) \( \cos C=\tan B \)

Ask by Simpson Thornton. in the United States
Mar 11,2025

Solución de inteligencia artificial de Upstudy

Respuesta verificada por el tutor

Responder

\( \sin B = \cos C \)

Solución

Given that triangle \( ABC \) is right with \( m \angle A = 90^{\circ} \), the remaining angles satisfy \[ m \angle B + m \angle C = 90^{\circ}. \] This implies that \( \angle B \) and \( \angle C \) are complementary. Using the complementary angle relationships in trigonometry, we have: \[ \sin B = \cos (90^{\circ} - \angle B). \] Since \( 90^{\circ} - \angle B = \angle C \), it follows that \[ \sin B = \cos C. \] Thus, the correct statement is: c) \( \sin B = \cos C \).

Respondido por UpStudy AI y revisado por un tutor profesional

error msg
Explicar
Simplifique esta solución

Mind Expander

In right triangle \( A B C \), where \( m \angle A = 90^{\circ} \), there’s a nifty relationship between the angles and the sides! Specifically, the sine of an angle is the ratio of the opposite side to the hypotenuse. So, for angle \( B \), \( \sin B = \frac{\text{opposite side to B}}{\text{hypotenuse}} \), and for angle \( C \), it would be \( \sin C = \frac{\text{opposite side to C}}{\text{hypotenuse}} \). The true statement from the choices provided is \( c) \sin B = \cos C \). This relationship holds because \( B \) and \( C \) are complementary angles (they add up to 90 degrees), making the sine of one equal the cosine of the other. So keep that in your math toolbox!

preguntas relacionadas

шамданыз: \( \left(\operatorname{ctg} \alpha+\frac{\sin \alpha}{1+\cos \alpha}\right)^{2}-\operatorname{ctg}^{2} \alpha \)
Trigonometría Kazakhstan Mar 12, 2025
STION 6 Without using a calculator, simplify the following expression to a single trigonometric term: \( \frac{\sin 10^{\circ}}{\cos 440^{\circ}}+\tan \left(360^{\circ}-\theta\right) \cdot \sin 2 \theta \) Given: \( \sin \left(60^{\circ}+2 x\right)+\sin \left(60^{\circ}-2 x\right) \) 6.2 .1 \( 6.2 .2 \quad \) Calculate the value of \( k \) if \( \sin \left(60^{\circ}+2 x\right)+\sin \left(60^{\circ}-2 x\right)=k \cos 2 x \). If \( \cos x=\sqrt{t} \), without using a calculator, determine the value of \( \tan 60^{\circ}\left[\sin \left(60^{\circ}+2 x\right)+\sin \left(60^{\circ}-2 x\right)\right] \) in terms of \( t \).
Trigonometría South Africa Mar 12, 2025
6. Rezolvaṭi următoarele ecuatii trigonometrice elementare: \( \begin{array}{ll}\text { a) } \sin 5 x=\sin 3 x ; & \text { b) } \cos x=\cos 4 x \\ \text { c) } \cos 5 x=-\sin 3 x ; & \text { d) } \operatorname{tg} 4 x=\operatorname{ctg}\left(3 x+\frac{\pi}{4}\right) ; \\ \begin{array}{ll}\text { e) } \sin \left(\frac{3 \pi}{2}+x\right)+\cos 3 x=0 ; & \text { f) } \cos \left(\frac{3 x}{2}+\frac{\pi}{4}\right)+\sin \left(\frac{2 x}{3}-\frac{\pi}{4}\right)=0\end{array} \\ \text { 7. Rezolvați următoarele ecuații trigonometrice elementare: } \\ \begin{array}{lll}\text { a) } \sin x=\sin 7 x ; & \text { b) }-\sin 2 x=\sin x ; & \text { e) } \operatorname{tg} 4 x=\operatorname{tg} 2 x ;\end{array} & \text { c) } \cos 4 x=\cos 5 x \\ \text { d) } \cos 2 x=-\cos 6 x ; & \text { f) } \operatorname{ctg} 2 x=\operatorname{ctg} x\end{array} \)
Trigonometría Romania Mar 12, 2025
мданыз: \( \frac{\sin (2 \pi-\alpha)}{\operatorname{ctg}\left(\frac{\pi}{2}-\alpha\right)} \cdot \frac{\operatorname{tg}\left(\frac{\pi}{2}+\alpha\right)}{\cos \left(\frac{\pi}{2}+\alpha\right)}+\frac{\cos (2 \pi+\alpha)}{\sin \left(\frac{3 \pi}{2}+\alpha\right)} \)
Trigonometría Kazakhstan Mar 12, 2025
\( 12.3 \quad \) 12.3.1 \( \quad \) Prove that, for angles \( A \) and \( B \) \[ \frac{\sin A}{\sin B}-\frac{\cos A}{\cos B}=\frac{2 \sin (A-B)}{\sin 2 B} \] 12.3.2 Hence, or otherwise, without using a calculator, show that: (a) \( \frac{\sin 5 B}{\sin B}-\frac{\cos 5 B}{\cos B}=4 \cos 2 B \) (b) \( \quad \frac{1}{\sin 18^{\circ}}=4 \cos 36^{\circ} \)
Trigonometría South Africa Mar 12, 2025

Latest Trigonometry Questions

6. Rezolvaṭi următoarele ecuatii trigonometrice elementare: \( \begin{array}{ll}\text { a) } \sin 5 x=\sin 3 x ; & \text { b) } \cos x=\cos 4 x \\ \text { c) } \cos 5 x=-\sin 3 x ; & \text { d) } \operatorname{tg} 4 x=\operatorname{ctg}\left(3 x+\frac{\pi}{4}\right) ; \\ \begin{array}{ll}\text { e) } \sin \left(\frac{3 \pi}{2}+x\right)+\cos 3 x=0 ; & \text { f) } \cos \left(\frac{3 x}{2}+\frac{\pi}{4}\right)+\sin \left(\frac{2 x}{3}-\frac{\pi}{4}\right)=0\end{array} \\ \text { 7. Rezolvați următoarele ecuații trigonometrice elementare: } \\ \begin{array}{lll}\text { a) } \sin x=\sin 7 x ; & \text { b) }-\sin 2 x=\sin x ; & \text { e) } \operatorname{tg} 4 x=\operatorname{tg} 2 x ;\end{array} & \text { c) } \cos 4 x=\cos 5 x \\ \text { d) } \cos 2 x=-\cos 6 x ; & \text { f) } \operatorname{ctg} 2 x=\operatorname{ctg} x\end{array} \)
Trigonometría Romania Mar 12, 2025
STION 6 Without using a calculator, simplify the following expression to a single trigonometric term: \( \frac{\sin 10^{\circ}}{\cos 440^{\circ}}+\tan \left(360^{\circ}-\theta\right) \cdot \sin 2 \theta \) Given: \( \sin \left(60^{\circ}+2 x\right)+\sin \left(60^{\circ}-2 x\right) \) 6.2 .1 \( 6.2 .2 \quad \) Calculate the value of \( k \) if \( \sin \left(60^{\circ}+2 x\right)+\sin \left(60^{\circ}-2 x\right)=k \cos 2 x \). If \( \cos x=\sqrt{t} \), without using a calculator, determine the value of \( \tan 60^{\circ}\left[\sin \left(60^{\circ}+2 x\right)+\sin \left(60^{\circ}-2 x\right)\right] \) in terms of \( t \).
Trigonometría South Africa Mar 12, 2025
мданыз: \( \frac{\sin (2 \pi-\alpha)}{\operatorname{ctg}\left(\frac{\pi}{2}-\alpha\right)} \cdot \frac{\operatorname{tg}\left(\frac{\pi}{2}+\alpha\right)}{\cos \left(\frac{\pi}{2}+\alpha\right)}+\frac{\cos (2 \pi+\alpha)}{\sin \left(\frac{3 \pi}{2}+\alpha\right)} \)
Trigonometría Kazakhstan Mar 12, 2025
prove the following identities: \( \begin{array}{ll}\text { (1) } \cos x \sin x+\sin ^{2} x \tan x=\tan x & \text { (2) } \frac{\tan A \cdot \sin 2 A}{1-\cos A}=2+2 \cos A \\ \text { (3) } \frac{\cos \alpha+\sin \alpha}{\cos 2 \alpha}=\frac{\cos ^{2} \alpha+\sin ^{2} \alpha}{\cos \alpha-\sin \alpha} & \text { (4) } \frac{\sin 2 x-2 \sin x}{\cos 2 x-1}=\frac{1}{\sin x}-\frac{1}{\tan x} \\ \text { (5) } \frac{\sin \theta-\sin 2 \theta}{\cos \theta-\cos 2 \theta-1}=\tan \theta & \text { (6) } \frac{\cos 2 \alpha+\cos \alpha}{\sin 2 \alpha-\sin \alpha}=\frac{\cos \alpha+1}{\sin \alpha}\end{array} \)
Trigonometría South Africa Mar 12, 2025
\( { }^{\circ} \cos 50^{\circ}+\sin 80^{\circ} \sin 50^{\circ} \) даныз: \( \left(\frac{\sin \alpha}{1+\cos \alpha}+\frac{\sin \alpha}{1-\cos \alpha}\right) \cdot \sin 2 \alpha \) \( \frac{\cos \left(\frac{2 \pi+\alpha)}{3 \pi+\alpha)}\right.}{\left.\frac{3 \pi}{\pi}+\alpha\right)} \)
Trigonometría Kazakhstan Mar 12, 2025
Anterior Próximo
¡Prueba Premium ahora!
¡Prueba Premium y hazle a Thoth AI preguntas de matemáticas ilimitadas ahora!
Quizas mas tarde Hazte Premium
Estudiar puede ser una verdadera lucha
¿Por qué no estudiarlo en UpStudy?
Seleccione su plan a continuación
Prima

Puedes disfrutar

Empieza ahora
  • Explicaciones paso a paso
  • Tutores expertos en vivo 24/7
  • Número ilimitado de preguntas
  • Sin interrupciones
  • Acceso completo a Respuesta y Solución
  • Acceso completo al chat de PDF, al chat de UpStudy y al chat de navegación
Básico

Totalmente gratis pero limitado

  • Solución limitada
Bienvenido a ¡Estudia ahora!
Inicie sesión para continuar con el recorrido de Thoth AI Chat
Continuar con correo electrónico
O continuar con
Al hacer clic en "Iniciar sesión", acepta nuestros términos y condiciones. Términos de Uso & Política de privacidad