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Semenov [28]
3 years ago
15

In right △ABC, the altitude CH to the hypotenuse AB intersects angle bisector AL in point D. Find the sides of △ABC if AD = 8 cm

and DH = 4 cm.

Mathematics
1 answer:
tangare [24]3 years ago
3 0

Answer:

AC=8\sqrt{3}\ cm\\ \\AB=16\sqrt{3}\ cm\\ \\BC=24\ cm

Step-by-step explanation:

Consider right triangle ADH ( it is right triangle, because CH is the altitude). In this triangle, the hypotenuse AD = 8 cm and the leg DH = 4 cm. If the leg is half of the hypotenuse, then the opposite to this leg angle is equal to 30°.

By the Pythagorean theorem,

AD^2=AH^2+DH^2\\ \\8^2=AH^2+4^2\\ \\AH^2=64-16=48\\ \\AH=\sqrt{48}=4\sqrt{3}\ cm

AL is angle A bisector, then angle A is 60°. Use the angle's bisector property:

\dfrac{CA}{CD}=\dfrac{AH}{HD}\\ \\\dfrac{CA}{CD}=\dfrac{4\sqrt{3}}{4}=\sqrt{3}\Rightarrow CA=\sqrt{3}CD

Consider right triangle CAH.By the Pythagorean theorem,

CA^2=CH^2+AH^2\\ \\(\sqrt{3}CD)^2=(CD+4)^2+(4\sqrt{3})^2\\ \\3CD^2=CD^2+8CD+16+48\\ \\2CD^2-8CD-64=0\\ \\CD^2-4CD-32=0\\ \\D=(-4)^2-4\cdot 1\cdot (-32)=16+128=144\\ \\CD_{1,2}=\dfrac{-(-4)\pm\sqrt{144}}{2\cdot 1}=\dfrac{4\pm 12}{2}=-4,\ 8

The length cannot be negative, so CD=8 cm and

CA=\sqrt{3}CD=8\sqrt{3}\ cm

In right triangle ABC, angle B = 90° - 60° = 30°, leg AC is opposite to 30°, and the hypotenuse AB is twice the leg AC. Hence,

AB=2CA=16\sqrt{3}\ cm

By the Pythagorean theorem,

BC^2=AB^2-AC^2\\ \\BC^2=(16\sqrt{3})^2-(8\sqrt{3})^2=256\cdot 3-64\cdot 3=576\\ \\BC=24\ cm

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and we have 72 because 15 plus 57 is 72

Now we have:

8w=72

Divide by 8 on both sides

8w cancels out to just w

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And you're left with:

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Step 1:
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Hope this helps! :)


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An optical inspection system is used to distinguish among different part types. The probability of a correct classification of a
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Answer:

Probability Mass Function:

   x:          0                         1                            2                          3

P(x):          0.000064          0.004608             0.115902             0.884736

Step-by-step explanation:

We are given the following information:

We treat correct classification  as a success.

P(correct classification) = 0.96

Then the number of classification follows a binomial distribution, where

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 3 and x = 0, 1, 2, 3

We have to evaluate:

P(x = 0)\\= \binom{3}{0}(0.96)^0(1-0.96)^3\\=0.000064

P(x = 1)\\= \binom{3}{1}(0.96)^1(1-0.96)^2\\=0.004608

P(x = 2)\\= \binom{3}{2}(0.96)^2(1-0.96)^1\\=0.115902

P(x = 3)\\= \binom{3}{3}(0.96)^3(1-0.96)^0\\=0.884736

PMF:

   x:          0                         1                            2                          3

P(x):          0.000064          0.004608             0.115902             0.884736

8 0
3 years ago
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