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2 years ago

Use right triangle DCB to find the trig values for angle D. sinD = cosD = tanD =

Mathematics

2 answers:

Sever21 [200]2 years ago

4 0

Sin D=35/37
cos D=12/37
tan D= 35/12
Use SOHCAHTOA

Natali5045456 [20]2 years ago

4 0

Answer:

$sin(D)=\frac{35}{37}$

$cos(D)=\frac{12}{37}$

$tan(D)=\frac{35}{12}$

Step-by-step explanation:

Part a) Find $sin(D)$

we know that

In a right triangle the value of sine of an angle is equal to the opposite side to the angle divided by the hypotenuse

$sin(D)=\frac{CB}{DB}$

substitute the values

$sin(D)=\frac{35}{37}$

Part b) Find $cos(D)$

we know that

In a right triangle the value of cosine of an angle is equal to the adjacent side to the angle divided by the hypotenuse

$cos(D)=\frac{DC}{DB}$

substitute the values

$cos(D)=\frac{12}{37}$

Part c) Find $tan(D)$

we know that

In a right triangle the value of tangent of an angle is equal to the opposite side to the angle divided by the adjacent side to the angle

$tan(D)=\frac{CB}{DC}$

substitute the values

$tan(D)=\frac{35}{12}$

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Looks like we have

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$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV$

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$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

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$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\iint_D\vec F\cdot\mathrm d\vec S$

Parameterize $D$ by

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$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-u\,\vec k$

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Finally,

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\left(-\frac\pi4\right)=\boxed{\frac{13\pi}{20}}$

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