Question

Derive the sine law and the cosine law​

Answer 1

Explanation :-

Sine Law :-

Consider an oblique $\triangle ABC$

having sides a , b and c opposite of $\angle {A} , \angle {B} , \angle {C}$

Construction :-

Draw $AM \perp BC$

To prove :-

$\dfrac {a}{Sin A} = \dfrac {b}{Sin B} = \dfrac{c}{Sin C} = K$

Proof :-

In $\triangle{AMC}$

→$Sin C = \dfrac{AM}{AC}$

→$Sin C = \dfrac{AM}{b}$

→$AM = b Sin C ----1)$

In $\triangle {AMB}$

→$Sin B = \dfrac{AM}{AB}$

→$Sin B = \dfrac{AM}{c}$

→$AM = c Sin B ------2)$

• Form 1 and 2.

→$\dfrac{b}{SinB} = \dfrac{c}{SinC}= K$

Similarly when we draw perpendicular from B on AC we get,

→$\dfrac{a}{SinA}= \dfrac{c}{SinC} = K$

• Combining these equations

→$\dfrac {a}{Sin A} = \dfrac {b}{Sin B} = \dfrac{c}{Sin C} = K$

Cosine Rule :-

Construct same triangle as in Sine rule.

To prove :-

$CosB = \dfrac{a^2 + c^2 - b^2}{2ac}$

Proof :-

In $\triangle{AMC}$

→$CosC = \dfrac{MC}{AC}$

→$CosC = \dfrac{MC}{a}$

→$MC = a Cos C----1)$

In $\triangle{AMB}$

→$CosB = \dfrac{BM}{AC}$

→$CosB = \dfrac{BM}{c}$

→$BM = c CosB ---2)$

Now in $\triangle{AMC}$

→$AC^2 = AM^2 + MC^2$

• By using Pythagoras theorem

→$AC^2 = AM^2 + (BC - BM)^2$

→$AC^2 = AM^2 + BC^2 + BM^2 - 2BC .BM$

→$b^2 = c^2 + a^2 -2ac Cos B$

→$2ac CosB = c^2 + a^2 -b^2$

→$CosB = \dfrac{c^2 + a^2 -b^2}{2ac}$

Similarly you can proof others.