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viva [34]
3 years ago
11

In an obtuse isosceles triangle the angle measurements are, x∘, x∘, and (10x−2)=128∘. Find the measurement of one of the acute a

ngles.
Mathematics
1 answer:
klemol [59]3 years ago
7 0

Answer:

26°

Step-by-step explanation:

An obtuse triangle is a triangle that has one obtuse angle. Obtuse angle is an angle that is greater than 90 degrees but less than 180 degrees.

An isosceles triangle is a triangle that has two equal sides and angles.  

Therefore, an obtuse isosceles triangle is a triangle with an obtuse angle and two equal sides that have two equal acute angles (angle less than 90° ).

Given:

The three angles of the triangle are given to be x°, x°  and (10x−2) = 128°. The obtuse angle is 128°, the two x° are acute angles. We are not using equation 10x − 2 since the value of the obtuse angle has been given as 128°

The sum of angles in a triangle is 180°

∴  x° + x° + 128° = 180°

2x° = 180° - 128°

2x° = 52°

x° = 52° / 2

x° = 26°

The measurement of one of the acute angles is 26°

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Answer:

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in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)
Nimfa-mama [501]

Answer:

The component of \vec u orthogonal to \vec a is \vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right).

Step-by-step explanation:

Let \vec u and \vec a, from Linear Algebra we get that component of \vec u parallel to \vec a by using this formula:

\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a (Eq. 1)

Where \|\vec a\| is the norm of \vec a, which is equal to \|\vec a\| = \sqrt{\vec a\bullet \vec a}. (Eq. 2)

If we know that \vec u =(2,1,1,2) and \vec a=(4,-4,2,-2), then we get that vector component of \vec u parallel to \vec a is:

\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)

\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)

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Lastly, we find the vector component of \vec u orthogonal to \vec a by applying this vector sum identity:

\vec  u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a} (Eq. 3)

If we get that \vec u =(2,1,1,2) and \vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right), the vector component of \vec u is:

\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10}    \right)

\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)

The component of \vec u orthogonal to \vec a is \vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right).

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3 years ago
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Step-by-step explanation:


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Answer:

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