Angles in Parallel Lines and Polygons - Higher

Match the angles in the same position at the two intersections.

1. Corresponding angles on parallel lines are equal.
2. Therefore x=68^\circ.

Look for the alternate angles between the parallel lines.

1. The marked angles are alternate angles.
2. Alternate angles are equal, so x=63^\circ.

Co-interior angles lie inside the parallel lines on the same side of the transversal.

1. Co-interior angles add to 180^\circ.
2. Thus x=180-113=67^\circ.

Find the angle sum, then share it equally between the vertices.

1. A pentagon has interior angle sum (5-2)\times180=540^\circ.
2. Each angle of a regular pentagon is 540\div5=108^\circ.

The exterior angles of a convex polygon add to a full turn.

1. The exterior angles add to 360^\circ.
2. There are 360\div40=9 equal angles, so the polygon has 9 sides.

First find the exterior angle beside the interior angle.

1. At each vertex the interior and exterior angles lie on a straight line, so they add to 180^\circ.
2. Each exterior angle is 180-156=24^\circ.
3. The exterior angles of a polygon add to 360^\circ, and in a regular polygon they are all equal, so the number of sides is 360\div24=15.

Subtract the known angles from the angle sum for six sides.

1. A hexagon has interior angle sum (6-2)\times180=720^\circ.
2. The five known angles total 650^\circ.
3. The sixth is 720-650=70^\circ.

Use the quadrilateral angle sum to find the value of one ratio part.

1. A quadrilateral has angle sum 360^\circ.
2. The ratio has 15 parts, so one part is 360\div15=24^\circ.
3. The largest angle is 6\times24=144^\circ.

Transfer the angle on the parallel line into the triangle before using its angle sum.

1. Alternate angles give \angle CED=54^\circ.
2. The angles in triangle CDE add to 180^\circ.
3. Therefore \angle DCE=180-54-67=59^\circ.

Use the interior angles of the two polygons and a full turn.

1. Each interior angle of a regular hexagon is (6-2)\times180\div6=120^\circ.
2. Each angle of a square is 90^\circ.
3. Angles around a point add to 360^\circ, so the gap is 360-120-90=150^\circ.

A whole number of its interior angles must fit into a full turn.

1. The angles meeting at the point must add to exactly 360^\circ, so the interior angle must divide exactly into 360^\circ.
2. Triangle: 360\div60=6. Square: 360\div90=4. Hexagon: 360\div120=3. All whole numbers, so these all fit.
3. A regular pentagon has interior angle (5-2)\times180\div5=108^\circ, and 360\div108=3.33\ldots, which is not a whole number.
4. Three pentagons leave a gap of 36^\circ and four would overlap, so the regular pentagon can't be used.

In a regular polygon all the exterior angles are equal.

1. In a regular polygon all the exterior angles are equal, so 2x+12=3x.
2. Subtract 2x from both sides: x=12.
3. Each exterior angle is 3\times12=36^\circ.
4. The exterior angles add to 360^\circ, so the number of sides is 360\div36=10.

Use the relationship between the number of triangles and the angle sum.

1. For n sides, (n-2)180=1620.
2. Thus n-2=9, so n=11.

Imagine a line through the bend parallel to the two given lines.

1. Draw a line through B parallel to the two given lines. It splits angle x into an upper part and a lower part.
2. The upper part and the 38^\circ angle at A are alternate angles between parallel lines (a Z shape), so the upper part is 38^\circ.
3. The lower part and the 57^\circ angle at C are also alternate angles, so the lower part is 57^\circ.
4. So x=38+57=95^\circ.

Subtract the different angle from the pentagon total before dividing.

1. The pentagon angle sum is 540^\circ.
2. The four equal angles total 540-100=440^\circ.
3. Each is 440\div4=110^\circ.

Use alternate angles and the isosceles triangle to write all three angles of triangle ABC in terms of x.

1. DE is parallel to BC, so angle ABC and angle DAB are alternate angles: angle ABC=2x^\circ.
2. AB=AC, so triangle ABC is isosceles and its base angles are equal: angle ACB=2x^\circ too.
3. Angles in a triangle add to 180^\circ: 2x+2x+(x+15)=180.
4. So 5x+15=180, which gives 5x=165 and x=33.
5. Angle BAC=x+15=33+15=48^\circ. (Check: 66+66+48=180.)

Use the equal base angles before transferring an angle along the parallel lines.

1. Triangle ABC is isosceles, so each base angle is (180-44)\div2=68^\circ.
2. Angles ABC and BCD are co-interior angles on parallel lines.
3. Thus \angle BCD=180-68=112^\circ.

A larger exterior angle means proportionally fewer sides.

1. Let A have n sides. Then B has n+6 sides.
2. The exterior angles of a regular polygon are 360^\circ divided by the number of sides: A's is \frac{360}{n} and B's is \frac{360}{n+6}.
3. A's exterior angle is twice B's: \frac{360}{n}=2\times\frac{360}{n+6}.
4. Divide both sides by 360 and cross-multiply: n+6=2n, so n=6.
5. B has 6+6=12 sides. (Check: 360\div6=60^\circ is twice 360\div12=30^\circ.)

Check whether the exterior angle leads to a whole number of sides.

1. The interior and exterior angles at a vertex add to 180^\circ, so the exterior angle would be 180-132=48^\circ.
2. The exterior angles of a regular polygon add to 360^\circ, so the number of sides would be 360\div48=7.5.
3. A polygon must have a whole number of sides, so no regular polygon has an interior angle of 132^\circ.
4. The first option is wrong because being less than 180^\circ is not enough. The last option is wrong because it is the exterior angle, not the interior angle, that must divide into 360^\circ (for example, a regular pentagon's 108^\circ does not divide 360^\circ).

Use the interior angle at the middle vertex of the isosceles triangle.

1. Each interior angle of a regular octagon is 180-360\div8=135^\circ.
2. In triangle ABC, AB=BC, so the angles at A and C are equal.
3. Each is (180-135)\div2=22.5^\circ.