Derivation of Sine Law

For any triangles with vertex angles and corresponding opposite sides are A, B, C and a, b, c, respectively, the sine law is given by the formula...
 

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

 

Derivation
To derive the formula, erect an altitude through B and label it hB as shown below. Expressing hB in terms of the side and the sine of the angle will lead to the formula of the sine law.
 

First figure for the derivation of sine law

 

$\sin A = \dfrac{h_B}{c}$

$h_B = c \sin A$
 

$\sin C = \dfrac{h_B}{a}$

$h_B = a \sin C$
 

Equate the two hB's above:
$h_B = h_B$

$c \sin A = a \sin C$

$\dfrac{c}{\sin C} = \dfrac{a}{\sin A}$
 

To include angle B and side b in the above relationship, construct an altitude through C and label it hC as shown below.
 

derivation-of-sine-law-02.jpg

 

$\sin A = \dfrac{h_C}{b}$

$h_C = b \sin A$
 

$\sin B = \dfrac{h_C}{a}$

$h_C = a \sin B$
 

$h_C = h_C$

$b \sin A = a \sin B$

$\dfrac{b}{\sin B} = \dfrac{a}{\sin A}$
 

Thus,

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

Therefore, the ratio of one side to the sine of its opposite angle is constant.
 

Note:
The constant ratio above is the diameter of the circumscribing circle about the triangle. See the proof (not available for now) for this note.
 

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