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Solving SBT Lesson 1: Angles and sides of a triangle (C8 SBT Math 7 Horizons)
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Solution 1 page 41 SBT Math 7 Creative horizon episode 2 – CTST
Find the measure of the unknown angles of the triangles in figure 5
Detailed instructions for solving Lesson 1
Solution method
The measure of the remaining angle in a triangle is \({180^o}\) minus the sum of the two known angles.
Detailed explanation
Considering triangle ABC, we have:
\(\widehat {{A^{}}} + \widehat B + \widehat C = {180^o}\) so: \(\widehat {{A^{}}} = {180^o} – \widehat B – \widehat C = {180^o} – {70^o} – {35^o} = {75^o}\)
– Considering triangle EFD we have:
\(\widehat D + \widehat E + \widehat F = {180^o}\) deduce: \(\widehat D = {180^o} – \widehat E – \widehat F = {180^o} – {65^o} – {25^o} = {90^o}\)
Considering the triangle MNP, we have:
\(\widehat M + \widehat N + \widehat P = {180^o}\) derives: \(\widehat N = {180^o} – \widehat M – \widehat P = {180^o} – {131^o} – {21^o} = {28^o}\)
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Solution 2 page 41 SBT Math 7 Creative horizon episode 2 – CTST
Find the measure of x in figure 6
Detailed instructions for solving Lesson 2
Solution method
The measure of a right angle is \({90^o}\)
The measure of the remaining angle in a triangle is \({180^o}\) minus the sum of the two known angles
Detailed explanation
Picture A:
– Considering triangle AHC we have: \(\widehat C = {180^o} – \widehat {{A^{}}} – \widehat H = {180^o} – {61^o} – {90^ o} = {29^o}\)
Consider a triangle ABC with: \(\widehat {{A^{}}} + \widehat B + \widehat C = {180^o}\) so: \(x = \widehat B = {180^o} – \widehat A – \widehat C = {180^o} – {90^o} – {29^o} = {61^o}\)
– Consider triangle EKG with: \(\widehat {GEK} + \widehat {EKG} + \widehat G = {180^o}\)
Derive: \(\widehat {GEK} = {180^o} – \widehat {EKG} – \widehat G = {180^o} – {90^o} – {41^o} = {49^o} \)
Which: \(\widehat {GEK} + \widehat {{\rm{KEF}}} = {90^o}\) yields: \(x = \widehat {{\rm{KEF}}} = {90 ^o} – \widehat {GEK} = {90^o} – {49^o} = {41^o}\)
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Solution 3 page 42 SBT Math 7 Creative horizon episode 2 – CTST
Find the sum of the 4 angles in a rhombus ABCD
Detailed instructions for solving Lesson 3
Solution method
Divide the rhombus into two triangles.
Detailed explanation
The diagonal AC divides the rhombus into two triangles, we have the sum of 4 angles of a rhombus ABCD equal to the sum of the angles of two triangles ABC and ADC and equal to \({360^o}\).
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Solution 4 page 42 SBT Math 7 Creative horizon episode 2 – CTST
Which of the following triples can be the lengths of the three sides of a triangle:
a) 1 cm, 7 cm, 9 cm
b) 2 cm, 6 cm, 8 cm
c) 5 cm, 6 cm, 10 cm
Detailed instructions for solving Lesson 4
Solution method
Check if the triple of length satisfies the triangle inequality
Detailed explanation
a) The triplet of lengths 1 cm, 7 cm, 9 cm is not the length of the three sides of a triangle because it does not satisfy the triangle inequality: 9 > 1 + 7.
b) The triplet of lengths 2 cm, 6 cm, and 8 cm is not the length of the three sides of a triangle because the triangle inequality is not satisfied: 8 = 2 + 6.
c) The triplet of lengths 5 cm, 6 cm, 10 cm can be the lengths of the three sides of a triangle because the triangle inequality is satisfied: 6 – 5 < 10 < 6 + 5
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Solution 5 page 42 SBT Math 7 Creative horizon episode 2 – CTST
Let ABC be a triangle with BC = 9 cm, AB = 1 cm. Find the length of side AC, knowing this length is an integer.
Detailed instructions for solving Lesson 5
Solution method
Using triangle inequality: BC – AB < AC < BC + AB
Detailed explanation
We have: 9 – 1 = 8 < AC < 10 = 9 +1, AC is an inferred integer AC = 9 cm
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Solution 6 page 42 SBT Math 7 Creative horizon episode 2 – CTST
In a research station, people mark three areas M, N, P as the three vertices of a triangle, knowing the distances MN = 30 m, MP – 90 m.
a) If a broadcasting station is located in area P with an operating radius of 60 m, will the signal be received in area N? Why?
b) Same question as above with an operating radius of 120 m.
Detailed instructions for solving Lesson 6
Solution method
Use the triangle inequality: MP + MN > PN > MP – MN to check the conditions.
Detailed explanation
a) Applying the triangle inequality in triangle MNP, we get:
\(\begin{array}{l}MP + MM > PN > MP – MN\\90 + 30 > PN > 90 – 30\\120 > PN > 60\end{array}\)
Thus, with a broadcast radius of 60 m, area N cannot receive the signal.
b) With a broadcast radius of 120 m, area N receives the signal.
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