1. 1515 15.1Area of Triangles 15.2Sine Formula 15.3Cosine Formula Chapter Summary Case Study Trigonometry (2) 15.4Heron’s Formula

2. P. 2 Today is the school’s Open Day. John helps his teacher to hang a banner outside the school building. In order to calculate the distance between the bottom of the banner and the ground, he should first find out the angles of elevation of the top and bottom of the banner. Case Study Then he can use the formula  height of the observer’s eyes to calculate the required distance h. I have measured the banner and its length is 16 m. How can I know the distance between the bottom of the banner and the ground? It is not difficult. Let me tell you how to find it out.

3. P. 3 15.1 Area of Triangles However, if we do not know the height of the triangle, then consider the following two cases: Substituting h  b sin C into (*), For a given triangle, if we know the base and the height of the triangle, then:

4. P. 4 15.1 Area of Triangles For a triangle with two sides and their included angle given, we can find the area of the triangle using the formula: Similarly, we can show that: Remark: If  C is a right angle, the area of  ABC becomes ab. In this case, b is the height of the triangle.

5. P. 5 Example 15.1T Solution: 15.1 Area of Triangles The figure shows a trapezium with AD // BC. If AD  12 cm, AC  16 cm, AB  9.3 cm,  DAC  35  and  ABC  80 , find the area of the trapezium. (Give the answer correct to 1 decimal place.) (  BAC  35  )  80   180  (int.  s, AD // BC)  BAC  65  Area of the trapezium  Area of  ABC  Area of  ACD (cor. to 1 d. p.)

6. P. 6 Example 15.2T 15.1 Area of Triangles  ABC  39.1494  or 180   39.1494  (cor. to 3 sig. fig.) The area of  ABC is cm 2. If and, find  ABC. (Give the answer correct to 3 significant figures.) Solution: Remarks: From this example, we find two possible values of  ABC and we can draw  ABC in two different ways.

7. P. 7 Example 15.3T 15.1 Area of Triangles  AOB  30.7971   41.9931cm 2 Area of the shaded region  (41.9931  40) cm 2 (cor. to 3 sig. fig.) The figure shows a sector with centre O and radius 12.5 cm. If the area of  OAB is 40 cm 2, find the area of the shaded region. (Give the answer correct to 3 significant figures.) Solution:

8. P. 8 15.2 Sine Formula In the last section, we learnt that the area of a triangle, such as  ABC in the figure, can be expressed in three different but equivalent forms: If we divide each of the above expressions by, we have: The above results are known as sine formula. In other words, Sine Formula For any triangle, the lengths of the sides are directly proportional to the sine ratios of the opposite angles. From the sine formula, we have a : b : c  sin A : sin B : sin C

9. P. 9 15.2 Sine Formula For any triangle, if we know the values of any two angles, then we can always find the remaining one by using the property of angle sum of triangle. Moreover, if we know the length of one side, then we can use the sine formula to find the lengths of the other two sides. A. Solving a Triangle with Two Angles and Any Side Given Any Side Given

10. P. 10 Example 15.4T 15.2 Sine Formula A  B  C  180  (  sum of  ) By sine formula, (cor. to 3 sig. fig.) In  ABC, A  40 , B  80  and c  10 cm. Solve the triangle and correct the answers to 3 significant figures if necessary. Solution: ‘Solve a triangle’ means to find the measures of all unknown angles and sides of a triangle. (cor. to 3 sig. fig.) A. Solving a Triangle with Two Angles and Any Side Given Any Side Given

11. P. 11 Example 15.5T 15.2 Sine Formula  18.0326 cm (cor. to 1 d. p.)  BAD   BCD  180  (opp.  s, cyclic quad.) In  BCD,  BDC  75   60  180  (  sum of  ) The figure shows a cyclic quadrilateral ABCD. AB  12 cm,  BAD  105 ,  ADB  40  and  DBC  60 . (a)Find the lengths of BD and BC. (b)Hence find the area of quadrilateral ABCD. (Give the answers correct to 1 decimal place.) Solution: (a)By sine formula, 105    BCD  180   BCD  75  18.0 cm ? ? ?  BDC  45  A. Solving a Triangle with Two Angles and Any Side Given Any Side Given

12. P. 12 Example 15.5T 15.2 Sine Formula (b)In  ABD,  ABD  105   40  180  (  sum of  )  ABD  35  The figure shows a cyclic quadrilateral ABCD. AB  12 cm,  BAD  105 ,  ADB  40  and  DBC  60 . (a)Find the lengths of BD and BC. (b)Hence find the area of quadrilateral ABCD. (Give the answers correct to 1 decimal place.) Solution: 18.0 cm 45  75  By sine formula,  13.2008 cm (cor. to 1 d. p.) Area of quadrilateral ABCD  Area of  ABD  Area of  BCD A. Solving a Triangle with Two Angles and Any Side Given Any Side Given

13. P. 13 15.2 Sine Formula If two sides and one non-included angle are given, we can also use the sine formula to solve the triangle. However, there may be more than one set of solution. Construct a  ABC such that AB  3 cm,  ABC  30  and (a)AC  4 cm; (b)AC  2 cm; (c)AC  1.5 cm; (d)AC  1 cm. (Use a pair of compasses to construct each of the above lengths of AC.) B. Solving a Triangle with Two Sides and One Non-included Angle Given One Non-included Angle Given Consider the following cases: For example, in (a), since AC  AB, only one triangle can be constructed:

14. P. 14 15.2 Sine Formula (a)AC  4 cm In the above task, three cases can be concluded: Case 1:Only one triangle can be formed  (a) and (c) Case 2:Two triangles are formed  (b) Case 3:No triangle can be formed  (d) B. Solving a Triangle with Two Sides and One Non-included Angle Given One Non-included Angle Given The construction: (b)AC  2 cm (c)AC  1.5 cm(d)AC  1 cm C C

15. P. 15 Example 15.6T 15.2 Sine Formula (cor. to 1 d. p.) B. Solving a Triangle with Two Sides and One Non-included Angle Given One Non-included Angle Given In  ABC, a  15 cm, c  19 cm and C  65 , find the possible values of A. (Give the answer correct to 1 decimal place.) Solution: By sine formula, Since the angle sum of triangle is 180 , (180   45.7  ) is not a possible value of A.

16. P. 16 Example 15.7T 15.2 Sine Formula B. Solving a Triangle with Two Sides and One Non-included Angle Given One Non-included Angle Given In  ABC, a  16 cm, b  14 cm and B  48 . (a)Find the possible values of A. (Give the answers correct to 1 decimal place.) (b)How many triangles can be formed? Solution: (a)By sine formula, (cor. to 1 d. p.) (b)Two triangles can be formed.

17. P. 17 15.3 Cosine Formula In the last section, we learnt how to use the sine formula to solve a triangle. However, in some cases, such as the triangles in the following figures, we cannot solve the triangles using the sine formula. We are going to learn another formula to solve triangles with  two sides and the included angle given, or  three sides given.

18. P. 18 Case 1 : A is an acute angle.Case 2 : A is an obtuse angle. Let AP  x In  PBC, we have h 2  a 2  (c  x) 2 (Pyth. theorem)  a 2  c 2  2cx  x 2....... (2) 15.3 Cosine Formula Consider the following two cases, we draw a perpendicular line from C to meet AB (or its extension) at P. Let CP  h. In  APC, we have h 2  b 2  x 2 (Pyth. theorem)... (1) From (1) and (2), a 2  c 2  2cx  b 2  a 2  b 2  c 2  2bc cos A  b cos ALet AP  x In  PBC, we have h 2  a 2  (c  x) 2 (Pyth. theorem)  a 2  c 2  2cx  x 2....... (2) In  APC, we have h 2  b 2  x 2 (Pyth. theorem)... (1)  a 2  b 2  c 2  2bc cos A  b cos(180   A)   b cos A a 2  b 2  c 2  2cx From (1) and (2), a 2  c 2  2cx  b 2 a 2  b 2  c 2  2cx

19. P. 19 15.3 Cosine Formula In the two cases, we obtain the same result: a 2  b 2  c 2  2bc cos A Using the same method as shown above, we can also show that b 2  a 2  c 2 – 2ac cos B; c 2  a 2  b 2 – 2ab cos C. Case 1 : A is an acute angle.Case 2 : A is an obtuse angle. The above results are known as cosine formulas. Cosine Formulas a 2  b 2  c 2  2bc cos A b 2  a 2  c 2  2ac cos B c 2  a 2  b 2  2ab cos C

20. P. 20 15.3 Cosine Formula Thus, Pythagoras’ theorem is a special case of cosine formula for right-angled triangles. Notes: When A  90 , a 2  b 2  c 2  2bc cos A  b 2  c 2 ( ∵ cos 90   0)

21. P. 21 Example 15.8T 15.3 Cosine Formula  13.1424 cm (cor. to 3 sig. fig.) In the figure, ABCD is a cyclic quadrilateral with  BAD  94 , BC  8 cm and CD  11 cm. (a) Find  BCD. (b) Find the length of BD. (Give the answer correct to 3 significant figures.) Solution: (a)  BCD  94   180  (opp.  s, cyclic quad.) (b)By cosine formula,

22. P. 22 15.3 Cosine Formula Alternate Forms of Cosine Formula Consider the cosine formula: a 2  b 2  c 2  2bc cos A 2bc cos A  b 2  c 2  a 2 Similarly, we can express cos B and cos C in terms of a, b and c. Thus, the alternate forms of the cosine formula are: These formulas are useful in finding the unknown angles when three sides of a triangle are given.

23. P. 23 Example 15.9T 15.3 Cosine Formula  BAC  44.3640  (cor. to 1 d. p.) In  ABC, a  14 cm, b  18 cm and c  19 cm. (a)Find  BAC. (b)Hence find the area of  ABC. (Give the answers correct to 1 decimal place.) Solution: (a)By cosine formula, (b) Area of  ABC

24. P. 24 The formula is credited to the ancient Greek mathematician Heron of Alexandria (about 75 AD). Another important formula for calculating the area of a triangle is Heron’s formula. 15.4 Heron’s Formula Heron’s Formula For a triangle with known sides as shown in the figure, area of triangle where The formula is stated as follows.

25. P. 25 15.4 Heron’s Formula By cosine formula,. Consider Area of  ABC

26. P. 26 Example 15.10T 15.4 Heron’s Formula  10x cm By Heron’s formula, Find the area of a triangle with sides 5x cm, 7x cm and 8x cm. (Express the answer in surd form.) Solution:

27. P. 27 Example 15.11T 15.4 Heron’s Formula  11 cm (cor. to 3 sig. fig.) In  ABC, a  (k  5) cm, b  2k cm and c  (2k  3) cm. If the perimeter of the triangle is 22 cm, find (a)the value of k; (b)the area of the triangle correct to 3 significant figures. Solution: (a) ∵ Perimeter  22 cm ∴ (k  5)  2k  (2k  3)  22 k  4 (b)The sides of  ABC are 9 cm, 8 cm and 5 cm respectively. Area of  ABC

28. P. 28 15.1 Area of Triangles In  ABC, Chapter Summary

29. P. 29 15.2 Sine Formula Chapter Summary In  ABC, the length of a side is directly proportional to the sine of its opposite angle.

30. P. 30 15.3 Cosine Formula Chapter Summary In  ABC, a 2  b 2  c 2  2bc cos A b 2  a 2  c 2  2ac cos B c 2  a 2  b 2  2ab cos C

31. P. 31 15.4 Heron’s Formula Chapter Summary In  ABC, area of  ABC

32. Follow-up 15.1 15.1 Area of Triangles The figure shows a trapezium ABCD with AD // BC, AD  15 cm, BC  20 cm, BD  18 cm and  DBC  30 . Find the area of the trapezium. Solution:  ADB  30  (alt.  s, AD // BC) Area of the trapezium  Area of  ABD  Area of  BCD

33. Follow-up 15.2 15.1 Area of Triangles 55  68.265sin  ABC  ABC  53.6763  or 180   53.6763  (cor. to 1 d. p.) The area of  ABC is 55 cm 2. If AB  12.3 cm and BC  11.1 cm, find  ABC. (Give the answers correct to 1 decimal place.) Solution:

34. Follow-up 15.3 15.1 Area of Triangles (cor. to 3 sig. fig.) The figure shows a sector with centre O and radius cm. If  AOB  67 , find the area of the shaded region. (Give the answer correct to 3 significant figures.) Solution: Area of the shaded region  (7.6009  5.9833) cm 2

35. Follow-up 15.4 15.2 Sine Formula A  B  C  180  (  sum of  ) By sine formula, Solve  ABC with A  85 , C  56  and b  9 cm. (Give the answers correct to 3 significant figures if necessary.) (cor. to 3 sig. fig.) Solution: A. Solving a Triangle with Two Angles and Any Side Given Any Side Given

36. Follow-up 15.5 15.2 Sine Formula  7.2579 cm (cor. to 1 d. p.) (b)70    BAC  65  180  (int.  s, AD // BC) (cor. to 1 d. p.) The figure shows a parallelogram ABCD with AB  7 cm,  ABC  70  and  CAD  65 . (a)Find the length of AC. (b)Hence find the area of the parallelogram. (Give the answers correct to 1 decimal place.) (a)  ACB  65  (alt.  s, AD // BC) In  ABC, Solution: by sine formula,  BAC  45  Area of the parallelogram  Area of  ABC  2 A. Solving a Triangle with Two Angles and Any Side Given Any Side Given

37. Follow-up 15.6 15.2 Sine Formula (cor. to 1 d. p.) B. Solving a Triangle with Two Sides and One Non-included Angle Given One Non-included Angle Given In  PQR, q  16 cm, r  20 cm and Q  42 , find the possible values of R. (Give the answers correct to 1 decimal place.) Solution: By sine formula,

38. Follow-up 15.7 15.2 Sine Formula  R  51.8  or 128.2   Two triangles can be formed. B. Solving a Triangle with Two Sides and One Non-included Angle Given One Non-included Angle Given In  PQR, p  9 cm, r  11 cm and P  40 . Find the number of triangles that can be formed with the above conditions. Solution: By sine formula,

39. Follow-up 15.8 15.3 Cosine Formula  13.1565 cm (cor. to 3 sig. fig.) In  PQR, p  9 cm, q  7 cm and R  110 . (a)Find the length of PQ. (b)Find  RPQ and  PQR. (Give the answers correct to 3 significant figures.) Solution: (a)By cosine formula, (b)By sine formula, (cor. to 3 sig. fig.)

40. Follow-up 15.9 15.3 Cosine Formula  Q  102.0515  (cor. to 2 d. p.)  R  37.2579   By angle sum of triangle, In  PQR, p  14 cm, q  21 cm and r  13 cm. (a)Find the largest angle in  PQR. (b)Find the other two angles in  PQR. (Give the answers correct to 2 decimal places.) Solution: (a)The largest angle in  PQR is Q. By cosine formula, (b)By cosine formula, (cor. to 2 d. p.)

41. Follow-up 15.10 15.4 Heron’s Formula  43 cm (cor. to 1 d. p.)  23.0  The height of the triangle is 23.0 cm. In the figure, the sides of the triangle are 23 cm, 27 cm and 36 cm respectively. Find (a)the area of the triangle; (b)the height, h cm, of the triangle. (Give the answers correct to 1 decimal place.) Solution: By Heron’s formula, (cor. to 1 d. p.)

42. Follow-up 15.11 15.4 Heron’s Formula The sides of the triangle are 16 cm, 24 cm and 20 cm respectively.  30 cm (cor. to 3 sig. fig.) In  ABC, a : b : c  4 : 6 : 5 and the perimeter of  ABC is 60 cm. (a)Find the sides of  ABC. (b)Hence find the area of  ABC. (Give the answer correct to 3 significant figures.) Solution: (a)Let a  4k cm, then b  6k cm and c  5k cm ∵ Perimeter  60 cm ∴ 4k  6k  5k  60 k  4 Area of  ABC