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Solving SBT lesson 3 Two parallel lines (C4 Math 7 – Horizon)

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### Solution 1 page 83 SBT Math 7 Creative horizon episode 1 – CTST

Given a // b, find the measures of x in Figure 10.

## Detailed instructions for solving Lesson 1

**Solution method**

We use the property: If 2 lines intersect 2 parallel lines, they form a pair of staggered internal equal angles; pairs of isotopic angles are equal.

**Detailed explanation**

a) Because \(a // b \Rightarrow x=\widehat {ACD}\) (two isotopic angles).

Which \(\widehat {ACD}=135^0\)

\(\Rightarrow x= 135^0\)

b) Because \(a // b \Rightarrow x=\widehat {NFE}\) (two staggered interior angles).

Which \(\widehat {NFE}=90^0\)

\(\Rightarrow x= 90^0\)

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### Solve problem 2 page 83 SBT Math 7 Creative horizon episode 1 – CTST

Find pairs of parallel lines in Figure 11 and explain.

## Detailed instructions for solving Lesson 2

**Solution method**

Applying Signs to identify 2 parallel lines: If a line c intersects 2 lines a and b, forming a pair of staggered internal angles or equal isotopes, then a//b

**Detailed explanation**

a)

Set the angles A_{first} and BUT_{first} as shown above.

We have \(\widehat {{A_1}}\)=\(\widehat {{B_1}}\)=45°

Where \(\widehat {{A_1}}\) and \(\widehat {{B_1}}\) are staggered in.

Therefore a // b (Sign to identify 2 parallel lines).

So in Figure 11a there is a // b.

b)

Set the corners C_{first} and EASY_{first} as shown above.

We have \(\widehat {{C_1}}\) and \(\widehat {{D_1}}\) in staggered positions but these two angles are not equal in measure (\(\widehat {{C_1}) }\)=90°≠\(\widehat {{D_1}}\)=80°) so the two lines d and e are not parallel.

So in Figure 11b no two lines are parallel.

c)

Set the US angles_{first} and FEMALE_{first} as shown above.

We have \(\widehat {{M_1}}\)=\(\widehat {{N_1}}\)=60°

Where \(\widehat {{M_1}}\) and \(\widehat {{N_1}}\) are isotopic.

Therefore m // n (Signs to identify 2 parallel lines).

So in Figure 11c there are m // n.

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### Solve problem 3 page 83 SBT Math 7 Creative horizon episode 1 – CTST

a) Let ABC be a triangle. State how to draw a line a through vertex A and parallel to BC, how to draw a line b passing through B and parallel to AC.

b) How many a and b lines can be drawn? Why?

## Detailed instructions for solving Lesson 3

**Solution method**

We use the Eculid axiom and the properties of 2 staggered internal angles, isotopes to draw the figure.

**Detailed explanation**

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a) We draw the line xy through A such that \(\widehat {xAB}\)=\(\widehat {ABC}\)

Since \(\widehat {xAB}\)=\(\widehat {ABC}\) these two angles are in staggered positions in

So xy // BC.

So the line xy is the line to draw through A and parallel to BC.

We draw the line zt through B such that \(\widehat {tBC}\)=\(\widehat {BCA}\)

Since \(\widehat {tBC}\)=\(\widehat {BCA}\) these two angles are in staggered positions in

So zt // AC.

So the line zt is the line to be drawn that passes through B and is parallel to AC.

b) According to Euclidean axiom we have through a point outside a line only one line parallel to that line.

So we can only draw one line a and one line b.

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### Solve problem 4 page 83 SBT Math 7 Creative horizon episode 1 – CTST

Find the angles with equal measures of the triangles ABC and DEC in Figure 12.

## Detailed instructions for solving Lesson 4

**Solution method**

We rely on the properties of two staggered internal, isotope and opposite angles to determine a pair of congruent angles

**Detailed explanation**

We have a⫽b and BE, AD intersects 2 lines a,b

\(\widehat u = \widehat v\) (2 isotopic angles)

\(\widehat x = \widehat y\) (2 opposite angles)

\(\widehat z = \widehat t\) (2 staggered interior angles)

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### Solve problem 5 page 83 SBT Math 7 Creative horizon episode 1 – CTST

For Figure 13

a) Why m // n?

b) Calculate the measure x of angle ABD

## Detailed instructions for solving Lesson 5

**Solution method**

a) We use the relationship of 2 lines that are perpendicular to another line

b) We use the property of 2 staggered internal angles to calculate the adjacent angle complementary to angle x from which we can calculate the angle x

**Detailed explanation**

a) Since m CD and n CD

So m // n (same perpendicular to CD).

So m // n.

b) Set the angle BEFORE_{first} as shown below:

Since m // n (according to sentence a), so:

\(\widehat {{B_1}}\)=\(\widehat {CAB}\)=120° (two isotopic angles)

Again \(\widehat {ABD}\) and \(\widehat {{B_1}}\) are complementary adjacent angles, so:

\(\widehat {ABD}\)+\(\widehat {{B_1}}\)=180°

So \(\widehat {ABD}\)=180°−\(\widehat {{B_1}}\)=180°−120°=60°

So x = 60°.

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