1. Starter
Construct accurately two different triangles with sides 3 cm and 5 cm and an angle of 30° opposite the 3 cm side
Use a ruler, protractor and compasses
What are the other two angles?

2. We are Learning to……
Use The Sine Law

3. The sine law Consider any triangle ABC,
If we drop a perpendicular line, h from C to AB, we can divide the triangle into two right-angled triangles, ACD and BDC. C b a h
a is the side opposite A and b is the side opposite B. A B D
We call the perpendicular h for height (not h for hypotenuse). h b h a
sin A =
sin B =
h = b sin A
h = a sin B
So,
b sin A = a sin B

4. The sine law b sin A = a sin B
Dividing both sides of the equation by sin A and then by sin B
we have: b
sin B = a
sin A
If we had dropped a perpendicular from A to BC we would have found that:
b sin C = c sin B
Rearranging: b
sin B = c
sin C

5. The sine law For any triangle ABC, C A B b c a a sin A = b sin B c
We can use the first form of the formula to find side lengths and the second form of the equation to find angles. a
sin A = b
sin B c
sin C
sin A
sin B
sin C a = b c or

6. Using the sine law to find side lengths
If we are given two angles in a triangle and the length of a side opposite one of the angles, we can use the sine law to find the length of the side opposite the other angle. For example,
Find the length of side a a
7 cm
118°
39° A B C
Using the sine law, a
sin 118° = 7
sin 39°
When trying to find a side length it is easier to use the formula in the form a/sin A = b/sin B.
Encourage pupils to wait until the last step in the equation to evaluate the sines of the required angles. This avoids errors in rounding. a =
7 sin 118°
sin 39°
a = (to 2 d.p.)

7. Using the sine law to find side lengths
Be aware that the lengths of the sides and the angles have been rounded. This means that, for example, the three angles in the triangle may not add up to exactly 180°.
Reveal an angle and the side opposite it. Reveal one more angle and ask pupils to find the side opposite it by using the sine rule.
Generate a new example by modifying the shape of the triangle.

8. Using the sine law to find angles
If we are given two side lengths in a triangle and the angle opposite one of the given sides, we can use the sine law to find the angle opposite the other given side. For example,
Find the angle at B
6 cm
46° B
8 cm A C
Using the sine rule,
sin B 8 = 6
sin 46°
When trying to find an angle it is easier to use the formula in the form sin A/a = sin B/b.
sin B =
8 sin 46° 6
sin–1
B =
8 sin 46° 6
B = 73.56° (to 2 d.p.)

9. Finding the second possible value
Suppose that in the last example we had not been given a diagram but had only been told that AC = 8 cm, CB = 6 cm and that the angle at A = 46°.
There is a second possible value for the angle at B.
Instead of this triangle …
… we could have this triangle.
6 cm
46° B
8 cm A C
Remember, sin θ = sin (180° – θ)
46°
6 cm B
So for every acute solution, there is a corresponding obtuse solution.
Remind pupils that the sine of angles in the second quadrant (that is angles between 90° and 180°) are positive. This means that for every angle between 0° and 90° there is another angle between 90° and 180° that has the same sine. This angle is found by subtracting the associated acute angle from 180°.
Ask pupils to imagine constructing the given triangle using a ruler and compasses (or ask them to do this as a practical). If the compass needle is placed at C and opened to 6 cm, there are two places that it can cross the line AB. These two points give the two possible triangles.
B = 73.56° (to 2 d.p.) or
B = 180° – 73.56°
= ° (to 2 d.p.)

10. Using the sine law to find angles
Be aware that the lengths of the sides and the angles have been rounded. This means that, for example, the three angles in the triangle may not add up to exactly 180°.
Reveal an angle and the side opposite it. Reveal one more side and ask pupils to find the angle opposite it by using the sine rule.
Generate a new example by modifying the shape of the triangle.

11. McGraw-Hill 12 Page 101 #s 1 – 6 BLM 2-7 #s 1 – 4
To succeed at this lesson today you need to…
1. Identify the angles and sides
2. Substitute the values into the formula
3. Calculate the missing value
McGraw-Hill 12 Page 101 #s 1 – 6 BLM 2-7 #s 1 – 4

12. Homework
McGraw-Hill 12 Page 101 #s 7 – 11
BLM 2.7 #s 5 – 8