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3 years ago

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What is the value for y? Enter your answer in the box. y = An isosceles triangle A B C with horizontal base B C and vertex A is

above the base. Side A C and C B are labeled with single tick mark. All the three angles are labeled. Base angles C A B is labeled as 50 degrees and angle C B A is labeled as 2x degrees. The angle A C B is labeled left parenthesis 5 y plus 10 right parenthesis degrees.

1 answer:

vfiekz [6]3 years ago

7 0

<u><em>Answer:</em></u>

x = 25

y = 14

<u><em>Explanation:</em></u>

The described scenario can be represented using the attached triangle.

<u>1- getting the value of x:</u>

We know that ΔABC is an isosceles triangle with AC = BC

<u>This means that:</u>

∠CAB = ∠CBA

We know that ∠CAB = 50° and ∠CBA = 2x°

<u>Equating the two angles, we get:</u>

50 = 2x .................> Divide both sides by 2

x = 25

<u>2- getting the value of y:</u>

We know that the sum of the internal angles of a triangle is 180°

<u>This means that:</u>

∠ABC + ∠CAB + ∠ACB = 180°

<u>We have:</u>

∠ABC = 2x = 50°

∠ACB = 5y + 10

∠CAB = 50°

<u>Now, we substitute to get the value of y as follows:</u>

50 + 50 + 5y + 10 = 180

110 + 5y = 180

5y = 180 - 110

5y = 70 .............> Divide both sides by 5

y = 14

Hope this helps :)

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3 years ago

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SashulF [63]

Answer:

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And the anfle is approximately $\theta \approx 48$

Step-by-step explanation:

For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

a=[1,2,-2], b=[4,0,-3,]

The dot product on this case is:

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Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

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$|b| =\sqrt{(4)^2 +(0)^2 +(-3)^2}=\sqrt{25}= 5$

And finally we can calculate the angle between the vectors like this:

$cos \theta = \frac{ab}{|a| |b|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{ab}{|a| |b|})$

If we replace we got:

$\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190$

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately $\theta \approx 48$

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