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HKDSE MATHEMATICS Ronald Hui Tak Sun Secondary School MISSING HOMEWORK RE2 Kelvin SHW2-R1 Kelvin, Charles SHW2-P1 Daniel, Kelvin, Sam L, Charles 22 October 2015 Ronald HUI MISSING HOMEWORK SHW3-01 Kelvin, Charles SHW3-A1 Charles SHW3-B1 Charles SHW3-C1 Daniel, Kelvin(RD), Sam L, Charles 22 October 2015 Ronald HUI MISSING HOMEWORK SHW3-E1 Kelvin, Charles SHW3-R1 Daniel, Charles SHW3-P1 Kelvin, Charles, Isaac, Macro S (RD) 22 October 2015 Ronald HUI MISSING HOMEWORK Chapter 4 HW SHW4-01 SHW4-A1 SHW4-B1 SHW4-C1 SHW4-D1 SHW4-R1 SHW4-P1 Deadline: 26 Nov (Thursday) 22 October 2015 Ronald HUI Book 5A Chapter 5 Area of a Triangle In trigonometry, we usually represent the angles and sides of a triangle, say △ABC, as follows: b A C c a B (a) The interior angles are denoted by capital letters A, B and C respectively. (b) The lengths of the sides opposite to angles A, B and C are denoted by the corresponding small letters a, b and c respectively. In trigonometry, we usually represent the angles and sides of a triangle, say △ABC, as follows: b A C c a B Example: A 62 5 cm B 7 cm C A= 62 a= 7 cm b= 5 cm There is a formula to calculate the area of a triangle when only two of its sides and their included angle are known. For example: 7 cm 140 40 9 cm 9 cm 7 cm Given two sides and their included angle Suppose only two sides a and b of △ABC and their included angle C are given. Consider the following two cases. Case 1: C is an acute angle. Case 2: C is an obtuse angle. A b h B C D a Consider △ADC. h sin C b h b sin C A To find the area of △ABC, we have to find its height first. B a b h C D Consider △ADC. h ◄ ACD = 180 sin (180 C ) b C h b sin (180 C ) h b sin C ◄ sin (180 C) = sin C Case 1: C is an acute angle. Case 2: C is an obtuse angle. A A b h B C D B a b h C D a h b sin C In both cases, the height of the triangle with base a is equal to b sin C. h b sin C Case 1: C is an acute angle. Case 2: C is an obtuse angle. A A b h B B C D a b h C D a h b sin C h b sin C Thus, we have: 1 Area of △ABC base height 2 1 a b sinC 2 1 ab sin C 2 Similarly, for any △ABC, we have: 1 ab sin C 2 1 = bc sin A 2 1 = ac sin B 2 the included angle of a and b. Area of △ABC = A b c B a C the included angle the included angle of b valid and cfor . right-angled triangles. Note: The above formula is also of a and c. When C = 90, 1 1 1 area of △ABC ab sin C ab sin 90 ab 2 2 2 A 10 cm B 105 Find the area of △ABC correct to 1 decimal place. 6 cm C b = 6 cm, c = 10 cm, A = 105 Area of △ABC 1 6 10 sin105 cm2 2 29.0 cm2 (cor. to 1 d. p.) Follow-up question The figure shows △ABC with area 10 cm2. If AC = 7 cm and BAC = 75, find the y cm value of y correct to 3 significant figures. B A 75 7 cm area = 10 cm2 C 1 Area of △ABC AB AC sin A 2 1 10 y 7 sin 75 2 y 2.96 (cor. to 3 sig. fig.) When only the lengths of the three sides of a triangle are known, could we find the area of the triangle? Yes, we can use Heron’s formula to find the area of the triangle. Heron’s formula Area of △ABC s(s a)(s b)(s c) , where s abc . 2 s is called the semi-perimeter. B A b c a C Let me show you how to find the area of △ABC correct to 3 significant figures. A abc Let s . 2 875 ∴ s cm 10 cm 2 5 cm 7 cm B 8 cm Area of △ABC s(s a )(s b )(s c ) 10 (10 – 8) (10 – 7) (10 – 5) cm2 17.3 cm2 (cor. to 3 sig. fig.) C Follow-up question A 9 cm There figure shows △ABC, where B AB = 9 cm, BC = 12 cm and AC = 13 cm. (a) Find the area of △ABC. 12 cm (b) Find the length of BD. (Give your answer correct to 3 significant figures.) abc . (a) Let s 2 12 13 9 s cm 17 cm ∴ 2 D 13 cm C Area of △ABC s(s a)(s b)(s c ) 17 (17 12) (17 13) (17 9) cm2 2720 cm2 52.2 cm2 (cor. to 3 sig. fig.) Follow-up question A 9 cm There figure shows △ABC, where B AB = 9 cm, BC = 12 cm and AC = 13 cm. (a) Find the area of △ABC. 12 cm (b) Find the length of BD. (Give your answer correct to 3 significant figures.) (b) ∵ ∴ 1 Area of △ABC AC BD 2 1 AC BD 2720 cm2 2 1 13 BD 2720 cm 2 BD 8.02 cm (cor. to 3 sig. fig.) D 13 cm C