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 The Cosine Formula

A B C 15 cm 10 cm A 12 cm B 13 cm C 40  18 cm Only three sides are given Two sides and their included angle are given We have another formula to solve these triangles. The sine formula cannot solve these triangles.

Consider △ ABC in the following two cases, and let h be the height of the triangle with base BC. Case 1: C is an acute angle. x D h b A B C c a  x Let CD = x, then BD = a  x. In △ ACD, b 2 – x 2 = h 2 (Pyth. theorem) In △ ABD, c 2  (a  x) 2 = h 2 (Pyth. theorem)

Case 1: C is an acute angle. (cont’d) In △ ACD, x = b cos C h D x b A B C c a  x ∴ c 2 = a 2 + b 2  2ab cos C ∴ b 2 – x 2 = c 2  (a  x) 2 b 2 – x 2 = c 2  (a 2  2ax + x 2 ) a 2 + b 2  2ax = c 2 b 2 – x 2 = c 2  a 2 + 2ax  x 2 Let’s check if this result also holds when C is an obtuse angle.

Case 2: C is an obtuse angle. x Let CD = x, In △ ABD, c 2 – (a + x) 2 = h 2 (Pyth. theorem) In △ ACD, b 2 – x 2 = h 2 (Pyth. theorem) h D b A B C c a then DB = a + x.

Case 2: C is an obtuse angle. (cont’d) In △ ACD, x = b cos (180  – C) ∴ c 2 = a 2 + b 2 + 2ab cos (180  – C) ◄  ACD = 180   C ∴ b 2  x 2 = c 2  (a + x) 2 b 2  x 2 = c 2  (a 2 + 2ax + x 2 ) b 2  x 2 = c 2  a 2  2ax  x 2 a 2 + b 2 + 2ax = c 2 i.e.c 2 = a 2 + b 2  2ab cos C ◄ cos(180   C)=  cos C x h D b A B C c a

The cosine formula The above results are known as the cosine formula. In △ ABC, In fact, for any △ ABC, we have c 2 = a 2 + b 2  2ab cos C Similarly, we can prove that b 2 = a 2 + c 2  2ac cos B and a 2 = b 2 + c 2  2bc cos A. 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. A C a b c B

In △ ABC, A C a b c B The cosine formula can also be written as follows: The cosine formula is also known as the cosine law or the cosine rule.

In general, the cosine formula is useful in solving a triangle in the following cases. Case 1: Given two sides and their non-included angle e.g. a B c S-A-S Problem Case 2: Given three sides e.g. c a b S-S-S Problem

By the cosine formula, x 2  a 2  b 2  2ab cos C Given two sides and their included angle A 14 cm B 15 cm C 42  Can you find the value of x in the figure correct to 1 decimal place? x cm

By the cosine formula, 5 cm A B C 7.5 cm 8 cm Given three sides  sum of △ C  180      Can you solve this triangle? (Give your answers correct to 1 decimal place.)

In the previous question, after finding the first angle, i.e. A, is it possible to find the second angle, i.e. B, by applying the sine formula? 5 cm A B C 7.5 cm 8 cm 76.7  Yes, but we need to check the two possible solutions obtained. Therefore, it is better to use the cosine formula in this situation.

By the cosine formula, Follow-up question Solve the triangle as shown on the right. (Give your answers correct to 1 decimal place.) 6 cm A B C 7 cm 100  BC 2  AB 2  AC 2  2(AB)(AC) cos A

Follow-up question  sum of △ C  180   100    Solve the triangle as shown on the right. (Give your answers correct to 1 decimal place.) 6 cm A B C 7 cm 100 

Sometimes, we need to use both the sine formula and the cosine formula to solve a problem.

A B C 5 cm 9 cm 8 cm D 6 cm 80  For example, can you find  ACD in the figure correct to 3 significant figures?

A B C 5 cm 9 cm 8 cm D 6 cm 80  In △ BDC, by the sine formula, In △ ABC, by the cosine formula,

Follow-up question Find x and  in the figure, correct to 3 significant figures. In △ ABC, by the cosine formula, x 2  AC 2 + AB 2  2(AC)(AB) cos  BAC In △ BDC, by the sine formula, A B C 4 cm7 cm 65  x cm D 8 cm  50 