Triangle lurlurl, u, r has a uniform thickness of 2.52.52, point, 5, as shown at left. the length of \overline{ud} ​ud ​ ​​ start overline, u, d, end overline, which is the height of the triangle, is 25\sqrt325√ ​3 ​ ​​ 25, square root of, 3, end square root. the triangle has a right angle at vertex uuu and an angle of \dfrac\pi6 ​6 ​ ​π ​​ start fraction, pi, divided by, 6, end fraction at vertex rrr. given that the width, www, of the shaded triangle is about 50\sqrt{3}50√ ​3 ​ ​​ 50, square root of, 3, end square root, what is the height, hhh, of the shaded triangle to the nearest tenth?

37.5

Step-by-step explanation:

The line which marks the height 25\sqrt325  

3

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25, square root of, 3, end square root of the large triangle is vertical, and the base of the triangle is horizontal, so their intersection is a right angle. The angles in the large triangle add up to \piπpi, so the remaining angle is:

\qquad \begin{aligned} \pi - \dfrac\pi2 - \dfrac\pi6 =& \dfrac\pi3 \end{aligned}  

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The original triangle L U R is shown, but now the length of segment U D is shown to be '25 times the square root of 3', and angle U L R is shown to have a measure of ' pi divided by 3' radians.

The original triangle L U R is shown, but now the length of segment U D is shown to be '25 times the square root of 3', and angle U L R is shown to have a measure of ' pi divided by 3' radians.

Hint #2

Looking at the left triangle ULDULDU, L, D, the hypotenuse length LULUL, U is:

\qquad \begin{aligned} \dfrac{25\sqrt3}{\sin\!\left({\Large\frac\pi3}\right)} \end{aligned}  

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Looking at the big triangle LURLURL, U, R, we see that the base length LRLRL, R is:

\qquad \begin{aligned} &\left(LU\right)\dfrac{1}{\sin\!\left({\Large\frac\pi6}\right)} \\ \\ =&\left(\dfrac{25\sqrt3}{\sin\!\left({\Large\frac\pi3}\right)}\right)\dfrac{1}{\sin\!\left({\Large\frac\pi6}\right)} \\ \\ =&\left(\dfrac{25\sqrt3}{ \quad\large\frac{\sqrt{3}}{2}\quad }\right)\dfrac{1}{ \quad\large\frac{1}{2}\quad } \\ \\ =&\left(\dfrac{25}{ \quad\large\frac{1}{2}\quad }\right)\dfrac{1}{ \quad\large\frac{1}{2}\quad } \\ \\ =&100 \\ \end{aligned}  

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100

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Hint #3

The large triangle has uniform thickness, so the inner triangle's sides are parallel to the outer triangle's sides. Therefore, the inner triangle's angles are congruent to the outer triangle's angles. The two triangles are similar.

Hint #4

Similar triangles have proportional lengths, so we can set up the equation:

\qquad \begin{aligned} \dfrac{h}{w} =& \dfrac{H}{W} \\ \\ \dfrac{h}{50\sqrt{3}} \approx& \dfrac{25\sqrt3}{100} \end{aligned}  

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100

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Hint #5

\qquad\qquad\, \begin{aligned} \dfrac{h}{50\sqrt{3}} \approx& \dfrac{\sqrt3}{4} \\ \\ {h} \approx& \dfrac{{50\sqrt{3}}\sqrt3}{4} \\ \\ {h} \approx& \dfrac{25\cdot3}{2} \\ \\ h \approx& 37.5 \\ \\ \end{aligned}  

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37.5

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Hint #6

To the nearest tenth, the height of the inner triangle is:

\qquad\qquad\, \begin{aligned} h \approx& 37.5 \end{aligned}  

h≈

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37.5

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*note: The answers 37.337.337, point, 3 and 37.437.437, point, 4 may have been obtained by other methods of solving for hhh since this is an approximation. Those answers are also accepted as correct.

stop using this to cheat on home work

step-by-step explanation:

dfbvfvdf

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)