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Answers To Coursebook Exercises: 5 Angles

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Answers to Coursebook exercises

5 Angles

✦ Exercise 5.1 Parallel lines


1 a p and t, q and u, s and w, r and v b q and w, r and t
2 a b b d
3 a q, r, u b p, s, t
4 a corresponding b alternate c CQX d BPY
5 a E b H, N, W
6 If they were parallel, then the angles XSA and XTC would be equal. This is not the case.
7 a b, f, j b, c c and e; c and i
8 a i and q; i and k b o and j; o and t
9 a neither b corresponding c corresponding d alternate e neither

✦ Exercise 5.2 Explaining angle properties


Alternative explanations are possible for some questions.
1 a 125° b 40° c 48°
2 a 72° and 73° b 145° and 107°
3 a Draw a line from R parallel to PQ; x = p, corresponding angles; y = q, S
alternate angles; the exterior angle is angle SRQ = x + y = p + q; this is the R x°
required result. y°

b x + y + r = 180, angles on a straight line; hence p + q + r = 180, which is the
p° q°
required result. Q
P
4 a alternate angles b alternate angles
c angle XAB + angle BAC + angle YAC = 180°, angles on a straight line;
angle ABC + angle BAC + angle ACB = 180°. This proves the result.
5 Draw HF to divide the quadrilateral into two triangles. Show that the six triangle angles are the four
quadrilateral angles.
6 a alternate angles b corresponding angles c x=a+y=a+c
7 a x is the exterior angle of triangle PQR.
b y=d+e
c x + y + c + f = 360, angles at a point; hence a + b + d + e + c + f = 360. These are the angles of the quadrilateral.
8 a alternate angles b corresponding angles
c angle CBD = angle XDY, corresponding angles; angle BCD = angle CDX, alternate angles. The six angles
round D add up to 360°. The result follows from this.

✦ Exercise 5.3 Solving angle problems


Alternative explanations are possible in some questions.
1 Because 30° and 20° are opposite angles and should be equal. Similarly, 150° and 160° are opposite angles and
should be equal.
2 a = 136°, alternate angles; b = 136°, corresponding angles; c = 180° − 136° = 44°, angles on a straight line;
d = 44°, alternate angles.
3 a d + b = 180°, angles on a straight line and b + a + c = 180°, angle sum of a triangle, so d = a + c
b e = a + b; f = b + c
c d + e + f = 2(a + b + c) = 360

Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 8 1


Unit 5 Answers to Coursebook exercises

4 angle BAC = 180 – (2 × 68) = 44°, isosceles triangle; angle EDC = 44°, corresponding angle
5 S how that the angles of the triangle and the quadrilateral together make the angles of the pentagon. The sum
of the angles is 180° + 360°.
6 T
 he angles at A and D are equal (corresponding angles); the angles at B and E are equal (corresponding angles);
the angle at C is common to both triangles.
7 Angle BAC = q, alternate angles; r = angle BAC + p, exterior angles. The result follows.
8 a w = a + c, exterior angle of a triangle; y = b + d, exterior angle of a triangle. The result follows.
b w + y = the sum of two angles of the quadrilateral; x + z = the sum of the other two angles of the quadrilateral;
w + x + y + z = the angle sum of the quadrilateral = 360°.
9 a exterior angle of a triangle
b exterior angle of a triangle
c a + x + y = 180°, angle sum of a triangle; hence a + (b + d) + (c + e) = a + b + c + d + e = 180°.

End-of-unit review
1 a e b f c c d d, f, b or h
2 a = 45°, corresponding angles; b = 45°, vertically opposite angles or alternate angles; c = 45°, vertically opposite
angles; d = 135°, angles on a straight line.
3 a and b, or f and g
4 82° + 27° = 109° so the angle between 82° and 27° is 180° – 109° = 71°; hence a = 71°, alternate angles.
b = 27°, corresponding angles.
5 a = 125° − 41° = 84°, external angle. b = 84° − 35° = 49°, external angle.
6 a corresponding angles b alternate angles c corresponding angles d alternate angles
7 A
 ngle ADB = angle ABD, isosceles triangles; angle CDB = angle CBD, isosceles;
Angle B = ABD + CBD = ADB + CDB = angle D.

2 Cambridge Checkpoint Mathematics 8 Copyright Cambridge University Press 2013

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