Triangles that aren’t Right Angled To find unknown sides and angles in non-right angled triangles we can use one or both of 2 rules: the sine rule the cosine rule The next few slides prove the sine rule. The cosine rule is on the next presentation. You do not need to learn the proof.

a, b and c are the sides opposite angles A, B and C The Sine Rule ABC is a scalene triangle a, b and c are the sides opposite angles A, B and C A B C b a c

A B C b a h N c ABC is a scalene triangle Draw the perpendicular, h, from C to BA. A B C In b a h N c

ABC is a scalene triangle In b b a a h N c c B A

ABC is a scalene triangle In b a In h N c B A

ABC is a scalene triangle h b a c C In In

ABC is a scalene triangle h b a c C In In

so, and A B C b a c

so, and A B C b a c

A C b a c h C A b a c The triangle ABC . . . . . . can be turned so that BC is the base. We would then get A B C b a c h C A B b a c

So, We now have and So,

Q q p The sine rule can be used in a triangle when we know One side and its opposite angle, plus One more side or angle e.g. Suppose we know p, q and angle Q in triangle PQR Q q p Tip: We need one complete “pair” to use the sine rule. The angle or side that we can find is the one that completes another pair.

e.g. 1 In the triangle ABC, find the size of angles A and C. 12 10 B b a We don’t need the 3rd part of the rule Solution: Use B a b (3 s.f.)

y Z Y Z y Y e.g. 2 In the triangle XYZ, find the length XY. Y Z X 13 Solution: As the unknown is a side, we “flip” the sine rule over. The unknown side is then at the “top”. y Z Y Z y Y ( 3 s.f.)

A a SUMMARY The sine rule can be used in a triangle when we know One side and its opposite angle, plus One more side or angle We write the sine rule so that the unknown angle or side is on the left of the equation If 2 sides and 1 angle are known we use: A If 1 side and 2 angles are known we use: a

Exercises In triangle ABC, b = cm, c = cm and angle C = . Find the size of angles A and B. 2. In triangle PQR, PQ = 23 cm, angle R = and angle P = . Find the size of side QR. B A C a Solution: C c b

Exercises 2. In triangle PQR, PQ = 23 cm, angle R = and angle P = . Find the size of QR. P Q R p Solution: r P R ( 3 s.f.)

If an unknown angle is opposite the longest side, 2 triangles may be possible: one will have an angle greater than e.g. In a triangle PQR, p = 5 cm, r = 7.2 cm and angle P = . Drawing side r and angle P, we have: R1 5 This is one possible complete triangle. P Q 7.2

e.g. In a triangle PQR, p = 5 cm, r = 7.2 cm and angle P = . This is the other. P Q R1 7.2 5 5 R2

e.g. In a triangle PQR, p = 5 cm, r = 7.2 cm and angle P = . This is the other. The 2 possible values of R are connected since R1 Triangle is isosceles 5 5 R2 Q P 7.2

e.g. In a triangle PQR, p = 5 cm, r = 7.2 cm and angle P = . This is the other. The 2 possible values of R are connected since P Q R1 7.2 5 R2 Triangle is isosceles The calculator will give the acute angle ( < ). We subtract from to find the other possibility.

P p r e.g. In a triangle PQR, p = 5 cm, r = 7.2 cm and angle P = . Find 2 possible values of angle R and the corresponding values of angle Q. Give the answers correct to the nearest degree. Solution: P p r or

and We have either: and or: P Q1 R1 7.2 5 P Q2 7.2 5 R2

SUMMARY If the sine rule is used to find the angle opposite the longest side of a triangle, 2 values may be possible. Use the sine rule and a calculator to find 1 value. This will be an acute angle ( less than ). Subtract from to find the other possibility. Use each value to find 2 possible values for the 3rd angle.

Exercise Find 2 possible values of angle ACB in triangle ABC if AB = 15 cm, AC = 12 cm and angle B = . Sketch the triangles obtained. Solution: B b c

AB = 15 cm, AC = 12 cm, angle B = A Exercise (i) (ii) C C 12 12 B 15 A