Question

David proved the Law of Cosines with reference to ∆ABC using a different method. Arrange the steps of his proof in the correct sequence. In , (Pythagorean Theorem) In , (Pythagorean Theorem) (on simplification) (Transitive Property) Therefore, . Similarly, prove that and . (on simplification) (on simplification) Draw a perpendicular, , from B to side AC. Let and . Therefore, .

Answer 1

Answer: We can arrange the steps with help of below explanation.  

Step-by-step explanation:

Here ABC is a triangle,

Draw a perpendicular from BD to side AC ( construction)

where $D\in AC$

In the right triangle BCD, from the definition of cosine:

cos C =CD/ BC

CD= a cos C

Subtracting this from the side b, we see that

DA= b-acos C  

In the triangle BCD, from the definition of sine:

sin C =BD  / a

BD = a sin C  

In the triangle ADB, applying the Pythagorean Theorem

$c^2 =BD^2 +DA^2$  

Substituting for BD and DA from (2) and (3)

$c^2 =(asinC)^2 +(b-acosC)^2$  

⇒$c^2 =a^2sin ^2C+b^2-2ab cosC+a^2 cos ^2C$   ( On simplification)

⇒$c^2 =a^2sin ^2C+a^2 cos ^2C+b^2-2abcos C$

⇒$c^2=a^2(sin^2C+cos^2C) +b^2-2abcosC$

⇒$c^2 = a^2+b^2-2abcosC$ (because,$sin^2\theta+cos^2\theta =1$ )