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Home , Circumscribed circle, Law of cosines, Law of sines

Course I

Lesson １２ Law of and

１2A • Fundamental Properties of • The

1 Fundamental Properties of Triangles

Fundamental Properties A (1) A + B + C = 180° (2) a b c, b c a, c a b b < + < + < + c Determination of a triangle (1) Three sides B a C (2) Two sides and the included . (3) Two internal and the included side

Similar and Congruent Triangles Similar Congruent

2 The Law of Sines Oblique triangles A A

c b c h h b B B a C B a C C is acute C is obtuse

The Law of Sines The ratio of the length of a side of a to the of the angle opposite that side is the same for all sides and angles. a b c = = sin A sin B sinC 3 Proof of the Law of Sines ［Case 1]

[Case that all angles are acute] The passes three vertices. Let the of this circle be BD. From the , we have D = A , ∠ BCD = 90° Therefore,

a = 2Rsin D = 2Rsin A. Similarly b = 2Rsin B, c = 2Rsin C. Therefore, we have a b c = = = 2R sin A sin B sinC 4 Proof of the Law of Sines [ Case 2 ]

[ Example 12.1] Prove a = 2 R sin A . for the case A > 90 ° . Ans. Draw the line BD which passes the center. ∠BCD = 90° Since quadrangle has contact with the circle, the following relationship holds. [Note] A + D=180°

Therefore a = 2Rsin D= 2Rsin(180° − A) = 2Rsin A [ Note ] From the inscribed angle theorem, the angles with the same symbol has the same magnitude. Therefore A + D= α + β + γ + δ =180° 5 How Can We Use Sine Law ?

When two angles and one side of an acute triangle is given, we can know the other sides.

[ Example 12.2] In the triangle ABC, A = 45 ° , B = 60 ° , a = 10 are given. Find the lengths b and c .

Ans. C =180° − 45° − 60° = 75° From the Law of Sines 10 b c = = sin 45° sin 60° sin 75° Value of each sine is 2 3 6 + 2 sin 45° = , sin 60° = , sin 75° = sin(30° + 45°) = 2 2 4 Therefore sin60° sin 75° b =10 = 5 6, c =10 = 5( 3 +1) 6 sin 45° sin 45° Ambiguous Case

Even if we know “ two side and an angle not between “, we cannot determine the last side.

C Angle A and sides a and b are given. b But a a • Triangle ABC • Triangle AB′C are possible. A BB B′ You can swing side a to left and right.

Huh? 7 Exercise

[Ex.12.1] In triangle ABC, b = 12 , C = 135 ° and the R = 12 of its are given. Find angle B and side c .

Pause the video and solve the problem by yourself.

8 Answer to the Exercise

[Ex.12.1] In triangle ABC, b = 12 , C = 135 ° and the radius R = 12 of its circumscribed circle are given. Find B and c .

Ans. From the law of sine and its relation to the radius of circumscribed circle a 12 c = = = 2×12 sin A sin B sin135°

Therefore 12 1 sin B = = ∴ B = 30° (∵ B <180° −135°) 2×12 2

c = 2×12×sin135° = 2×12×sin135° = 24×sin45° =12 2

9 Course I

Lesson １２ The Law of Sines and the Law of Cosines

１２B • Law of Cosines

10 The Law of Cosines

If we know two sides and the included angle, we can find the side which is opposite to this angle. A The Law of Cosines b a2 = b2 + c2 − 2bc cos A c b2 = c2 + a2 − 2ca cos B c2 = a2 + b2 − 2ab cosC B a C

Another Expression

b2 + c2 − a2 c2 + a2 −b2 a2 + b2 − c2 cos A = , cos B = , cosC = 2bc 2ca 2ab We can find angles from three sides. 11 Proof of the Law of Cosines

Take projections of point A to side BC and name this point H. a = CH + BH = b cos C + c cos B (1) Similarly b = c cos A + a cos C (2) c = a cos B + b cos A (3)

From (2)×b + (3)× c − (1)× a b2 + c2 − a2 = 2bc cos A a2 b2 c2 2bc cos A ∴ = + − 12 Example

[ Example 12.3] In triangle ABC, b = 2 , c = 1 + 3 and A = 60 ° are given. Solve for a , B , C .

Ans. From the law of cosines a2 = 22 + (1+ 3)2 − 2⋅2(1+ 3)cos60° = 6 From the law of sines 6 2 2 3 1 = ∴ sin B = ⋅ = sin60° sin B 6 2 2 ∴ B = 45° or 135° Since a = 6 > b = 2 , the angle B < A = 60° Therefore, B = 45° ∴ C =180° − (A + B) = 75°

13 Exercise

[ Ex 12.2] When we see the top P of the mountain from two points A and B which are separated by 2km, the angles are ∠ PAB = 75 ° and ∠ PBA = 60 ° . In addition, the angle of elevation of the mountain top P from the point A is 30°. What is the height of the mountain ?

Ans.

Pause the video and solve the problem by yourself.

14 Answer to the Exercise

[ Ex 12.2] When we see the top P of the mountain from two points A and B which are separated by 2km, the angles are ∠ PAB = 75 ° and ∠ PBA = 60 ° . In addition, the angle of elevation of the mountain top P from the point A is 30°. What is the height of the mountain ?

Ans. ∠APB =180° − (75° + 60°) = 45° From the law of sines AP 2 2sin 60° 3 = ∴ AP = = 2⋅ ⋅ 2 = 6 sin 60° sin 45° sin 45° 2 Since the triangle AHP is a , 6 PH = APsin30° = km 2 15