Law of Sines

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proportionalities

$\blacktriangleright $ Acute Triangles

An acute triangle is one where all three interior angles measure less than $90^{\circ }$.

In the diagram below, we drop an altitude from the top corner (angle $C$). This altitude (length $h$) is perpendicular to the opposite side, forming two right triangles inside the acute triangle.

From these two right triangles we can deduce the following.

\[ \frac {h}{b} = \sin (A) \, \text { and } \, \frac {h}{a} = \sin (B) \]

\[ h = b \sin (A) \, \text { and } \, h = a \sin (B) \]

\[ b \sin (A) = a \sin (B) \]

\[ \frac {\sin (A)}{a} = \frac {\sin (B)}{b} \]

The same argument with an altitude from angle $B$ to side $b$ shows a similar relationship with angle $C$, giving us

\[ \frac {\sin (A)}{a} = \frac {\sin (B)}{b} = \frac {\sin (C)}{c} \]

$\blacktriangleright $ Obtuse Triangles

An obtuse triangle is one with one angle greater than $90^{\circ }$.

We can establish the same relationship with an altitude on the outside.

Note: Straight angles measure $180^{\circ }$. $\angle B$ is part of a straight angle, which makes the measurement of the angle on the other side $180^{\circ } - B$.

Additional Note: The $y$-coordinates are the same as you move down the unit circle on either side from $90^{\circ }$. Therefore, $\sin (180^{\circ } - B) = \sin (B)$

\[ \frac {h}{b} = \sin (A) \, \text { and } \, \frac {h}{a} = \sin (180^{\circ } - B) = \sin (B) \]

\[ h = b \sin (A) \, \text { and } \, h = a \sin (B) \]

\[ b \sin (A) = a \sin (B) \]

\[ \frac {\sin (A)}{a} = \frac {\sin (B)}{b} \]

The same argument could be made with angle $C$, giving us

\[ \frac {\sin (A)}{a} = \frac {\sin (B)}{b} = \frac {\sin (C)}{c} \]

Law of Sines

For any triangle with angles $A$, $B$, and $C$, and opposite sides $a$, $b$, and $c$, respectively,

\[ \frac {\sin (A)}{a} = \frac {\sin (B)}{b} = \frac {\sin (C)}{c} \]

General Triangles

Suppose we have a triangle with $A=46^{\circ }$, $B=65^{\circ }$, and $a=14$.

Figure out the other three measurements. (3 decimals)

$A + B + C = 46^{\circ } + 65^{\circ } + C = \answer {180}^{\circ }$

$C = \answer {69}^{\circ }$

$\frac {\sin (46^{\circ })}{14} = \frac {\sin (65^{\circ })}{b}$

$b = \frac {14 \sin (65^{\circ })}{\sin (46^{\circ })} \approx \answer [tolerance=0.001]{17.63882523}$

$\frac {\sin (46^{\circ })}{14} = \frac {\sin (69^{\circ })}{c}$

$c = \frac {14 \sin (69^{\circ })}{\sin (46^{\circ })} \approx \answer [tolerance=0.001]{18.16961325}$

No Triangle

Suppose we have a triangle with $C=55^{\circ }$, $a=4$, and $c=2$.

Figure out the other three measurements.

$\frac {\sin (55^{\circ })}{2} = \frac {\sin (A)}{4}$

$\sin (A) = \frac {4 \sin (55^{\circ })}{2} \approx 1.638304086$

This is greater than $1$. $\sin (A)$ cannot be greater than $1$.

Therefore, there is no triangle with $C=55^{\circ }$, $a=4$, and $c=2$.

Two Triangles

Suppose we have a triangle with $A=37^{\circ }$, $a=12$, and $b=16$.

Figure out the other three measurements.

$\frac {\sin (37^{\circ })}{12} = \frac {\sin (B)}{16}$

$\sin (B) = \frac {16 \sin (37^{\circ }}{12}) \approx \answer [tolerance=0.001]{0.8024200309}$

There are two angles whose sine is $0.8024200309$. One in the first quadrant and one in the second quadrant.

Therefore, there are two triangles with $A=37^{\circ }$, $a=12$, and $b=16$.

Using the calculator, the two angles are

$B \approx SIN^{-1}(0.8024200309) \approx \answer [tolerance=0.01]{53.36}^{\circ }$
$B \approx 180^{\circ } - 53.36^{\circ } = 126.64^{\circ }$

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more examples can be found by following this link
More Examples of Triangles

2025-07-02 00:56:51