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Leonardodvin

Sine Rule & Practice Problems

Author: Leonardodvin

Friends, in this lesson I am going discuss about the "Sine Rule" and the "Extended Rule of Sine" which are quite simple rules to prove but the questions based on these rules, sometimes, may be a bit confusing. So let us take a look at the proof of this rule and after that we will practice some questions as well.

Sine Rule: In any triangle the sides are proportional to the sines of the opposite angles. i.e.

$\frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC} \Leftrightarrow a:b:c = SinA:SinB:SinC$

Let's look at a simple proof of this before we move to the practice questions.

Let ABC be any plane triangle. Draw the perpendicular h from one of the vertices C of the triangle to the opposite side AB.

From the right triangle ACD, we have

$SinA = \frac{h}{b} \Leftrightarrow h = bSinA ......(1)$

Similarly, from the right triangle BCD, we have

$SinB = \frac{h}{a} \Leftrightarrow h = aSinB ......(2)$

From equations (1) and (2), we have

$bSinA = aSinB \Leftrightarrow \frac{a}{SinA} = \frac{b}{SinB}$

Similarly, by drawing a perpendicular from B to the opposite side, we obtain

$\frac{a}{SinA} = \frac{c}{SinC}.$

$\therefore \frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC} \Leftrightarrow a:b:c = SinA:SinB:SinC$

Note: In the above proof I considered an acute angled triangle to prove the Sine Rule. We can also consider an obtuse angled triangle and a similar proof can be worked out. You can look at the adjacent figure for an obtuse angled triangle

Extended Rule of Sine: The ratio of any side of a triangle to the sine of the angle opposite is numerically equal to the diameter (D) of the circumscribed circle.. i.e.

$\frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC} = D$

Circumscribe a circle about the triangle ABC and denote by D the diameter BA', drawn through one of the vertices, as B. Join A' and C. A'BC is a right triangle, and therefore

$SinA' = \frac{a}{D} \Leftrightarrow D = \frac{a}{SinA'}$

But angle A' = angle A (angles inscribed in the same arc are equal), hence

$D = \frac{a}{SinA'} = \frac{a}{SinA}$

Similarly,

$D = \frac{b}{SinB} \quad and \quad D = \frac{c}{SinC}$

$\therefore \frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}=D$

Based on the "Sine Rule", we have the following formula to calculate the Area of a Triangle

$\triangle = \frac{abSinC}{2} = \frac{bcSinA}{2} = \frac{caSinB}{2}$

Now let's get into some practice problems.

Practice Problem:1
$In \quad \triangle ABC, \quad \angle ABC = 45^\circ . \quad Point \quad D \quad is \quad on \quad segment \quad BC \quad such \quad that \quad 2|BD|=|CD| \quad and \quad \angle DAB = 15^\circ. \quad Find \quad \angle ACB.$

Solution: We construct this triangle in the following way:

Fix segment BC, choose point D on segment BC such that 2|BD|=|CD| (FIGURE-1), and construct ray BP such that PBC = 45â-¦

Let A be a point on ray BP that moves from B in the direction of the ray. It is not difï¬cult to see that DAB decreases as A moves away fromB. Hence, there is a unique position for A such that DAB = 15â-¦. This completes our construction of triangle ABC.
$Set \quad \angle CAD = \alpha$
$Note \quad that \quad \angle CDA = \angle CBA + \angle DAB = 60^\circ$
This ï¬gure brings to mind the proof of the angle-bisector theorem.We apply the law of sines to triangles ACD and ABC. We have

FIGURE - 1

$\frac{|CD|}{Sin\alpha} = \frac{|CA|}{Sin60^\circ} \quad and \quad \frac{|BC|}{Sin(\alpha + 15^\circ)} = \frac{|CA|}{Sin45^\circ}$

$Dividing \quad the \quad first \quad equation \quad by \quad the \quad second \quad equation \quad gives$

$\frac{|CD|Sin(\alpha + 15^\circ)}{|BC|Sin\alpha} = \frac{Sin45^\circ}{Sin60^\circ} -------(A)$

$It \quad is \quad given \quad that \quad \frac{|CD|}{|BC|} = \frac{2}{3} \quad we \quad know \quad that \quad \frac{Sin45^\circ}{Sin60^\circ} = \sqrt{\frac{2}{3}}$ $Hence \quad replacing \quad \frac{|CD|}{|BC|} \quad with \quad [\frac{Sin45^\circ}{Sin60^\circ}]^2 \quad in \quad in \quad the \quad equation \quad (A) \quad we \quad get$ $\frac{Sin(\alpha + 15^\circ)}{Sin\alpha} = \frac{Sin45^\circ}{Sin60^\circ}$ $It \quad is \quad clear \quad that \quad \alpha = 45^\circ \quad is \quad a \quad solution \quad for \quad the \quad above \quad equation$ $Hence \quad in \quad the \quad triangle ABC, \quad \angle CAB = 60^\circ \quad and \quad \angle CBA = 45^\circ$ $\therefore \angle ACB = 75^\circ$

Practice Problem:2
$In \quad \triangle ABC, \quad |AB| = 8, |AC| = 5 \quad and \quad the \quad area \quad of \quad the \quad triangle \quad is \quad 10 \sqrt{3}.$

$Find \quad all \quad the \quad possible \quad values \quad of \quad \angle A.$

Solution

$We \quad know \quad that \quad area \quad of \quad the \quad triangle \quad ABC = \frac{|AB||AC|SinA}{2}$ $Given \quad that \quad \frac{8.5.SinA}{2} = 10 \sqrt{3}.$
$\Rightarrow SinA = \frac{\sqrt{3}}{2}$
$\therefore \angle A = 60^\circ \quad or \quad 120^\circ$

For more practice problems and a quick review go through the presentation
Sine Rule and Practice Problems

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