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2 years ago

Find the indicated angle measures:

Mathematics

2 answers:

Lera25 [3.4K]2 years ago

5 0

Answer:

Angle 1: 55 degrees
Angle 2: 55 degrees
Angle 3: 70 degrees

Step-by-step explanation:

Finding angle 1: We know that in a triangle, all three angles must add up to 180 degrees. In the triangle on the left, 2 of the angle measures are already given to us. Therefore, we can simply do 180 - 40 - 85, thus the measure of angle 1 is 55 degrees.
Finding angle 2: We know that opposite angles are congruent. Therefore, angle 2 and angle 1 have the same measure.
Finding angle 3: Using the same thought process as we used when finding the measure of angle 1, we can subtract the other 2 angles. 180 - 55 - 55 is equal to 70.

Firlakuza [10]2 years ago

4 0

Answer:

first, we take the first triangle( the right one), since a triangle is equal to 180 we add 85 and 40 which gives us 125, we then subtract 125 with 180 getting 55 so (1) =55

to find (2) we put 55 the same number as no. (1) because of the property VERTICALLY OPPOSITE ANGLES.

then to find (3) we do the same steps as (1), we add 55 and 55 and subtract by 180. 55+55=110

=110 - 180

=70

There u go, please mark me as brainless

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2 years ago

Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.

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Answer:

$T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

Step-by-step explanation:

Remember that for any curve $r(t)$

The tangent vector is given by

$T(t) = \frac{r'(t) }{| r'(t)| }$

And the normal vector is given by

$N(t) = \frac{T'(t)}{|T'(t)|}$

For this case, using the chain rule

$r'(t) = ( -10*2tsin(t^2) , 102t cos(t^2) )\\$

And also remember that

$|r'(t)| = \sqrt{(-10*2tsin(t^2))^2 + ( 10*2t cos(t^2) )^2} \\\\ = \sqrt{400 t^2*( sin(t^2)^2 + cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t$

Therefore

$T(t) = r'(t) / |r'(t) | = ( -10*2tsin(t^2) , 10*2t cos(t^2) )/ 20t\\\\ = ( -10*2tsin(t^2)/ 20t , 10*2t cos(t^2) / 20t )\\= ( -sin(t^2), cos(t^2) )$

Similarly, using the quotient rule and the chain rule

$T'(t) = ( -2t cos(t^2) , -2t sin(t^2))$

And also

$|T'(t)| = \sqrt{ ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t$

Therefore

$N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

Notice that

1. $|N(t)| = |T(t) | = \sqrt{ cos(t^2)^2 + sin(t^2)^2 } = \sqrt{1} = 1$

2. $N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0$

Simlarly

$r'(t) = (2t,-6,0) \\$

and

$|r'(t)| = \sqrt{(2t)^2 + 6^2} = \sqrt{4t^2 + 36}$

Therefore

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

Then

$T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)$

and also

$|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 + ( (3 t)/(9 + t^2)^{3/2})^2 + 0^2 }\\= 3/(t^2 + 9 )$

And since

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

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