Rotate the figure 90 deg clockwise. It's asking for all like the prime points as well.

Mathematics

1 answer:

ss7ja [257]2 years ago

3 0

Answer:

you need to use tracing paper for this

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Find sin D sin E cos D and cos E

Andrew [12]

9514 1404 393

Answer:

sin(D) = cos(E) = (√3)/2

cos(D) = sin(E) = 1/2

Step-by-step explanation:

The mnemonic SOH CAH TOA is intended to remind you of the relationships between trig functions and right triangle sides.

Sin = Opposite/Hypotenuse

Cos = Adjacent/Hypotenuse

For this diagram, this means ...

sin(D) = cos(E) = (13√3)/26 = (√3)/2

cos(D) = sin(E) = 13/26 = 1/2

5 0

2 years ago

Please help me worth 35 points!!

mixer [17]

Answer:

Solution given:

Volume of rectangular prism=length*breadth*height

for

1st one 4in,3in,6in

Volume=4*3*6=72cu.in

For second one 6in,3in,2in

Volume=6*3*2=36cu.in

For third one 4in ,5in,4in

Volume=4*5*4=80cu.in

For forth one 2in,9in,3in

Volume=2*9*3=54cu.in

7 0

2 years ago

Read 2 more answers

Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.

vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

$cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}$

$\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}$