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3 years ago

Find the measure of the angle x to the nearest tenth. Please help ty!!!

Mathematics

1 answer:

den301095 [7]3 years ago

5 0

Answer:

Option A) 33.5 degrees

Step-by-step explanation:

The above is a right angled triangle, thus we can apply trigonometry.

Above here the hypotenous is 58 units and opposite side to angle x is 32 units.

$sin(x) = \frac{opposite}{hypotenous} = \frac{32}{58} = 0.552$

Thus,

$x = sin^{-1}(0.552) = 33.5 degrees$

Thus, the measure of angle x is 33.5 degrees, which means Option A is correct.

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15. If x=a Sin2t (I+Cos2t) and y=b Cos 2t (1-Cos2t) then find dy/dx at =22/7*4

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By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$

It looks like we're given

$\begin{cases}x=a\sin(2t)(1+\cos(2t))\\y=b\cos(2t)(1-\cos(2t))\end{cases}$

where a and b are presumably constant.

Recall that

$\cos^2t=\dfrac{1+\cos(2t)}2$

$\sin^2t=\dfrac{1-\cos(2t)}2$

so that

$\begin{cases}x=2a\sin(2t)\cos^2t\\y=2b\cos(2t)\sin^2t\end{cases}$

Then we have

$\dfrac{\mathrm dx}{\mathrm dt}=4a\cos(2t)\cos^2t-4a\sin(2t)\cos t\sin t$

$\dfrac{\mathrm dy}{\mathrm dt}=-4b\sin(2t)\sin^2t+4b\cos(2t)\sin t\cos t$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4b\cos(2t)\sin t\cos t-4b\sin(2t)\sin^2}{4a\cos(2t)\cos^2t-4a\sin(2t)\cos t\sin t}$

$\implies\boxed{\dfrac{\mathrm dy}{\mathrm dx}=\dfrac ba\tan t}$

where the last reduction follows from dividing through everything by $\cos(2t)\cos^2t$ and simplifying.

I'm not sure at which point you're supposed to evaluate the derivative (22/7*4, as in 88/7? or something else?), so I'll leave that to you.

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3 years ago