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2 years ago

What is the measure of ∠EGF? What is the measure of ∠CGF?

Mathematics

2 answers:

andrey2020 [161]2 years ago

4 0

measure of ∠EGF = 1/2( 180 - 50)
= 1/2(130)
= 65

the measure of ∠CGF = 180 - 65
= 115

Leni [432]2 years ago

3 0

Hello!

This is an isosceles triangle, which means its base angles are congruent. Interior angles of triangle add to equal 180 degrees, so you can make an equation to solve for the measure of the base angles. I'll call the base angles x.

180 = 50 + 2x
130 = 2x
65 = x

That gives you m∠EGF and m∠FGE (65°).

To find m∠CGF, you must know that it and ∠EGF are supplementary (they add to equal 180 degrees). Make another equation setting them to equal 180 using m∠EGF (65°):

180 = 65 + x
115 = x

Answer:
m∠EGF = 65°
m∠CGF = 115°

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Find the value of x and BC if B is between C and D CB=4x-9 BD=3x+5 and CD= 17

snow_tiger [21]

X = 3 and CB = 19. To find x you would put 4x-9 + 3x+5 = 17. Remember to get all the x’s on one side and the other numbers on the other aide of the equal sign. Then plug x into the length of BC.

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1 year ago

Two forces, F1 and F2, are represented by vectors with initial points that are at the origin. The first force has a magnitude of

Strike441 [17]

Answer:

Step-by-step explanation:

The two force F1 and F2 are represented by vectors with initial points that are at the origin.

the terminal point of the vector is point P(1, 1, 0)

Therefore, the direction of the vector force is

$v_1=(1-0)\hat i+(1-0)\hat j +(0-0)\hat k\\\\=1\hat i+1\hat j+0\hat k\\\\=\hat i + \hat j$

The unit vector in the direction of force will be

$\frac{v_1}{|v_1|} =\frac{\hat i+ \hat j}{\sqrt{1^2+1^2+0^2} } \\\\=\frac{1}{\sqrt{2} (\hat i +\hat j)}$

The magnitude of the force is 40lb, so the force will be

$F_1=40\times \frac{1}{\sqrt{2} } (\hat i+\hat j)\\\\=20\sqrt{2} (\hat i+\hat j)$

The terminal point of its vector is point Q(0, 1, 1)

Therefore, the direction of the vector force is

$v_2=(0-0)\hat i+(1-0)\hat j +(1-0)\hat k\\\\=0\hat i+1\hat j+1\hat k\\\\=\hat j + \hat k$

$\frac{v_2}{|v_2|} =\frac{\hat j+ \hat k}{\sqrt{0^2+1^2+1^2} } \\\\=\frac{1}{\sqrt{2} (\hat j +\hat k)}$

The magnitude of the force is 60lb, so the force will be

$F_2=60\times \frac{1}{\sqrt{2} } (\hat j+\hat k)\\\\=30\sqrt{2} (\hat j+\hat k)$

The resultant of the two forces is

$F=F_1+F_2\\\\=[20\sqrt{2} (\hat i+\hat j)]+[30\sqrt{2} (\hat j +\hat k)]\\\\=20\sqrt{2} \hat i+20\sqrt{2} \hat j +30\sqrt{2} \hat j+30\sqrt{2} \hat k\\\\=20\sqrt{2} \hat i+50\sqrt{2} \hat j+30\sqrt{2} \hat k$

The magnitude force will be

$|F|=\sqrt{(20\sqrt{2} )^2+(50\sqrt{2} )^2+(30\sqrt{2} )^2} \\\\=\sqrt{800+5000+1800} \\\\=\sqrt{3100} \\\\=55.68$

to (1 decimal place)=55.7lb

b) The direction angle of force F

The angle formed by F and x axis

$\alpha=\cos^{-1}(\frac{20\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(0.5080)\\\\=59.469$

The angle formed by F and y axis

$\alpha=\cos^{-1}(\frac{50\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(1.270)\\\\=$

The angle formed by F and z axis

$\alpha=\cos^{-1}(\frac{30\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(0.7620)\\\\=40.359$

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