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Anna35 [415]
4 years ago
5

Find the area of a triangle base is 3 height is 11

Mathematics
2 answers:
Virty [35]4 years ago
6 0

we know that

The formula to calculate the area of a triangle is equal to

A=\frac{1}{2}bh

where

b is the base of the triangle

h is the height of the triangle

in this problem we have

b=3\ units\\h=11\ units

substitute the values in the formula

A=\frac{1}{2}*3*11

A=16.5\ units^{2}

therefore

<u>the answer is</u>

the area of the triangle is equal to 16.5\ units^{2}

lukranit [14]4 years ago
5 0
A=1/2*b*h

A=1/2*3*11

A=1/2*33

A=16.5 

I hope this helps!
~kaikers
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6 over 25

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Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) 5, 1,
Dahasolnce [82]

Answer:

The direction cosines are:

\frac{5}{\sqrt{42} }, \frac{1}{\sqrt{42} }  and  \frac{4}{\sqrt{42} }  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

Step-by-step explanation:

For a given vector a = ai + aj + ak, its direction cosines are the cosines of the angles which it makes with the x, y and z axes.

If a makes angles α, β, and γ (which are the direction angles) with the x, y and z axes respectively, then its direction cosines are: cos α, cos β and cos γ in the x, y and z axes respectively.

Where;

cos α = \frac{a . i}{|a| . |i|}               ---------------------(i)

cos β = \frac{a.j}{|a||j|}               ---------------------(ii)

cos γ = \frac{a.k}{|a|.|k|}             ----------------------(iii)

<em>And from these we can get the direction angles as follows;</em>

α =  cos⁻¹ ( \frac{a . i}{|a| . |i|} )

β = cos⁻¹ ( \frac{a.j}{|a||j|} )

γ = cos⁻¹ ( \frac{a.k}{|a|.|k|} )

Now to the question:

Let the given vector be

a = 5i + j + 4k

a . i =  (5i + j + 4k) . (i)

a . i = 5         [a.i <em>is just the x component of the vector</em>]

a . j = 1            [<em>the y component of the vector</em>]

a . k = 4          [<em>the z component of the vector</em>]

<em>Also</em>

|a|. |i| = |a|. |j| = |a|. |k| = |a|           [since |i| = |j| = |k| = 1]

|a| = \sqrt{5^2 + 1^2 + 4^2}

|a| = \sqrt{25 + 1 + 16}

|a| = \sqrt{42}

Now substitute these values into equations (i) - (iii) to get the direction cosines. i.e

cos α = \frac{5}{\sqrt{42} }

cos β =  \frac{1}{\sqrt{42} }              

cos γ =  \frac{4}{\sqrt{42} }

From the value, now find the direction angles as follows;

α =  cos⁻¹ ( \frac{a . i}{|a| . |i|} )

α =  cos⁻¹ ( \frac{5}{\sqrt{42} } )

α =  cos⁻¹ (\frac{5}{6.481} )

α =  cos⁻¹ (0.7715)

α = 39.51

α = 40°

β = cos⁻¹ ( \frac{a.j}{|a||j|} )

β = cos⁻¹ ( \frac{1}{\sqrt{42} } )

β = cos⁻¹ ( \frac{1}{6.481 } )

β = cos⁻¹ ( 0.1543 )

β = 81.12

β = 81°

γ = cos⁻¹ ( \frac{a.k}{|a|.|k|} )

γ = cos⁻¹ (\frac{4}{\sqrt{42} })

γ = cos⁻¹ (\frac{4}{6.481})

γ = cos⁻¹ (0.6172)

γ = 51.89

γ = 52°

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The direction cosines are:

\frac{5}{\sqrt{42} }, \frac{1}{\sqrt{42} }  and  \frac{4}{\sqrt{42} }  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

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y = -2x + 5

Compared to y = 2x + 5, the slopes are different, so this system won't be inconsistent. Not a good choice.

B: y = 2x + 5

Compared to y = 2x + 5, the slopes are the same and the y intercepts are the same. This system has infinitely many solutions. Not a good choice.

C: 2x - 4y = 10

-4y = 10 - 2x

-4y = -2x + 10

y = 2/4x -10/4

Here the slopes are different, so, like A this is not a good choice.

D: 2y - 4x = -10

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y = 2x - 5

Compared to y = 2x + 5 we have the same slopes and different y intercepts.  The lines will be parallel and the system is inconsistent.


Thus, D is the best choice.

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