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4 years ago

In ABC, BC=4 cm, angle b=angle c, and angle a=20 degrees, what is ac to two decimal places

Mathematics

2 answers:

zubka84 [21]4 years ago

8 0

Answer:

Therefore,

$AC=11.52\ cm$

Step-by-step explanation:

Given:

In ΔABC, BC=4 cm,

angle b=angle c, and

angle a=20°

To Find:;

AC = ?

Solution:

Triangle sum property:

In a Triangle sum of the measures of all the angles of a triangle is 180°.

$\angle a+\angle b+\angle c=180\\\therefore 2m\angle b =180-20=160\\m\angle b=\dfrac{160}{2}=80\°$

We know in a Triangle Sine Rule Says that,

In Δ ABC,

$\dfrac{a}{\sin A}= \dfrac{b}{\sin b}= \dfrac{c}{\sin C}$

substituting the given values we get

$\dfrac{BC}{\sin a}= \dfrac{AC}{\sin b}$

$\dfrac{4}{\sin 20}= \dfrac{AC}{\sin 80}\\\\AC=11.517=11.52\ cm$

Therefore,

$AC=11.52\ cm$

Montano1993 [528]4 years ago

7 0

Answer:

AC = 11.518 cm

Step-by-step explanation:

Given ,

BC = 4 cm

∠ b = ∠ c = x ( say )

∠ a = 20°

AC = ?

From the given data it is known that Δ ABC is isosceles triangle and we know that sum of angles of a triangle is 180° .

⇒ ∠ a + ∠ b + ∠ c = 180°

⇒ 20° + x + x = 180°

⇒ 2x = 160°

⇒ x = 80 °

∴ ∠ b = ∠ c = 80°

Now by applying sine rule,

$\frac{BC}{sin 20} = \frac{AB}{sin 80} = \frac{AC}{sin 80}$

$\frac{4}{0.342} = \frac{AC}{0.984}$ ( sin 20 = 0.342 & sin 80 = 0.984 )

AC = 11.518 cm

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Step-by-step explanation:

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3 years ago

A horse ran 800 meters in 40 seconds , 1200 meters in 60 seconds , and 480 meters in 24 seconds . Is this a proportional relatio

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Answer:

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Step-by-step explanation:

In order to find out whether this relationship is proportional, you need to see if the rate at which the horse runs is constant (the same). If you look at the three sets of data (24, 480), (40, 800) and (60, 1200) where the pair is (seconds, meters), you can see that for any two sets of data the change in meters divided by the change in seconds is consistently 20m/s. For example:

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4 years ago

Strain-displacement relationship) Consider a unit cube of a solid occupying the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 After loa

Anastasy [175]

Answer:

please see answers are as in the explanation.

Step-by-step explanation:

As from the data of complete question,

$0\leq x\leq 1\\0\leq y\leq 1\\0\leq z\leq 1\\u= \alpha x\\v=\beta y\\w=0$

The question also has 3 parts given as

Part a: Sketch the deformed shape for α=0.03, β=-0.01 .

Solution

As w is 0 so the deflection is only in the x and y plane and thus can be sketched in xy plane.

the new points are calculated as follows

Point A(x=0,y=0)

Point A'(x+αx,y+βy)

Point A'(0+(0.03)(0),0+(-0.01)(0))

Point A'(0,0)

Point B(x=1,y=0)

Point B'(x+αx,y+βy)

Point B'(1+(0.03)(1),0+(-0.01)(0))

Point B'(1.03,0)

Point C(x=1,y=1)

Point C'(x+αx,y+βy)

Point C'(1+(0.03)(1),1+(-0.01)(1))

Point C'(1.03,0.99)

Point D(x=0,y=1)

Point D'(x+αx,y+βy)

Point D'(0+(0.03)(0),1+(-0.01)(1))

Point D'(0,0.99)

So the new points are A'(0,0), B'(1.03,0), C'(1.03,0.99) and D'(0,0.99)

The plot is attached with the solution.

Part b: Calculate the six strain components.

Solution

Normal Strain Components

$\epsilon_{xx}=\frac{\partial u}{\partial x}=\frac{\partial (\alpha x)}{\partial x}=\alpha =0.03\\\epsilon_{yy}=\frac{\partial v}{\partial y}=\frac{\partial ( \beta y)}{\partial y}=\beta =-0.01\\\epsilon_{zz}=\frac{\partial w}{\partial z}=\frac{\partial (0)}{\partial z}=0\\$

Shear Strain Components

$\gamma_{xy}=\gamma_{yx}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=0\\\gamma_{xz}=\gamma_{zx}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}=0\\\gamma_{yz}=\gamma_{zy}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=0$

Part c: Find the volume change

$\Delta V=(1.03 \times 0.99 \times 1)-(1 \times 1 \times 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\$

Also the change in volume is 0.0197

For the unit cube, the change in terms of strains is given as

$\Delta V={V_0}[(1+\epsilon_{xx})]\times[(1+\epsilon_{yy})]\times [(1+\epsilon_{zz})]-[1 \times 1 \times 1]\\\Delta V={V_0}[1+\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}+\epsilon_{xx}\epsilon_{zz}+\epsilon_{yy}\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}\epsilon_{zz}-1]\\\Delta V={V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the strain values are small second and higher order values are ignored so

$\Delta V\approx {V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\ \Delta V\approx [\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the initial volume of cube is unitary so this result can be proved.

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3 years ago