geometry isosceles and equilateral triangles find the value of x in each diagram

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28-11-2018
Mathematics

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geometry isosceles and equilateral triangles find the value of x in each diagram

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calculista calculista · Answer 1 · 2018-12-07T20:23:56+01:00

Answer:

Part 9) [tex]x=12[/tex]

Part 10) [tex]x=12\°[/tex]

Part 11) [tex]x=10\°[/tex]

Part 12) [tex]x=11\°[/tex]

Part 13)

a) [tex]SU=4\ units[/tex]

b) m∠VWX=[tex]40\°[/tex]

c) m∠WVX=[tex]50\°[/tex]

d) m∠XTV=[tex]60\°[/tex]

e) m∠XVT=[tex]30\°[/tex]

Step-by-step explanation:

Part 9) we have that

[tex]12=2x-12[/tex] ----->by SSA

solve for x

[tex]2x=24[/tex]

[tex]x=12[/tex]

Part 10)

In the isosceles triangle of the left the vertex angle is equal to

[tex]180\°-68\°*2=44\°[/tex]

Find the measure of angle 2

m∠2=[tex]90\°-44\°=46\°[/tex]

m∠2=[tex]4x-2[/tex]

[tex]4x-2=46\°[/tex]

solve for x

[tex]4x=48\°[/tex]

[tex]x=12\°[/tex]

Part 11)

Find the base angle in the isosceles triangle of the top

[tex]180\°-118\°=62\°[/tex]

Find the vertex angle in the isosceles triangle of the top

[tex]180\°-2*62\°=56\°[/tex]

Find the vertex angle 2 in the isosceles triangle of the bottom

[tex]180\°-56\°=124\°[/tex] ------> this is the measure of angle 2

m∠2=[tex]124\°[/tex]

m∠2=[tex]12x+4[/tex]

[tex]12x+4=124\°[/tex]

[tex]12x=120\°[/tex]

[tex]x=10\°[/tex]

Part 12)

m∠2=[tex]146\°[/tex] ------> by corresponding angles

m∠2=[tex]13x+3[/tex]

[tex]13x+3=146\°[/tex]

[tex]13x=143\°[/tex]

[tex]x=11\°[/tex]

Part 13)

a) we have that

SU=UW ------> given problem

[tex]3x+1=x+3[/tex]

[tex]2x=2[/tex]

[tex]x=1[/tex]

therefore

[tex]SU=3x+1=3+1=4[/tex]

[tex]SU=4\ units[/tex]

b) we know that

m∠VWX=[tex]4y\°[/tex]

in the right triangle UVW find the value of y

The sum of the internal angles of a triangle is equal to [tex]180\°[/tex]

so

[tex]180\°=4y+90\°+50\°[/tex]

[tex]180\°=4y+140\°[/tex]

[tex]4y=40\°[/tex]

[tex]y=10\°[/tex]

so

m∠VWX=[tex]4y\°=40\°[/tex]

Part c) we know that

in the right triangle VWX

The sum of the internal angles of a triangle is equal to [tex]180\°[/tex]

so

m∠WVX=[tex]180\°-(4y+90\°)[/tex]

m∠WVX=[tex]180\°-(40\°+90\°)[/tex]

m∠WVX=[tex]50\°[/tex]

Part d) we know that

in the right triangle XTV

The sum of the internal angles of a triangle is equal to [tex]180\°[/tex]

so

m∠XTV=[tex]180\°-(40\°-y+90\°)[/tex]

m∠XTV=[tex]180\°-(40\°-10\°+90\°)[/tex]

m∠XTV=[tex]60\°[/tex]

Part e) we know that

m∠XVT=[tex](40\°-y)[/tex]

substitute the value of y

m∠XVT=[tex](40\°-10\°)[/tex]

m∠XVT=[tex]30\°[/tex]