In the previous section, we completed a discussion on trigonometric identities and trigonometric equations. In this section, we will see sine formula.
• Consider the triangle ABC in fig.3.39 below.
• For this triangle, we can define three angles:
♦ Angle A is the angle between sides AB and AC
♦ Angle B is the angle between sides BA and BC
♦ Angle C is the angle between sides CA and CB
• We can define three sides also:
♦ a is the length of the side BC (The side opposite angle A)
♦ b is the length of the side AC (The side opposite angle B)
♦ c is the length of the side AB (The side opposite angle C)
 |
| Fig.3.39 |
◼ Based on the above notations, we can write the sine formula as:
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
• For proving this formula, we will analyze an obtuse triangle and an acute triangle.
| • An obtuse triangle is that triangle in which one angle is obtuse and the
remaining two are acute. (Remember that, no triangle can have more than
one obtuse angles). • An acute triangle is that triangle in which all three angles are acute. |
• Both triangles can be analyzed using the same steps.
• When the formula is proved for both those types of triangles, we can confidently apply it to any triangle.
The proof can be written in 12 steps:
1. Dropping perpendiculars from the top vertex:
• For the obtuse triangle in fig.3.40(i) below, the perpendicular dropped from the top vertex B can meet the base AC only if we extend the base a bit towards the right.
 |
| Fig.3.40 |
• For the acute triangle in fig.3.40(ii), the perpendicular dropped from the top vertex B can meet the base AC with out any difficulty.
2. Name of the perpendiculars:
• For both the triangles, the perpendiculars are named as BD.
♦ D is the foot of the perpendicular.
3. Length of the perpendiculars:
• For both the triangles, the length of the perpendiculars are denoted as h.
♦ h is called the altitude of the triangle
4. Writing h in terms of A and c:
I. In fig.3.40(i), considering right triangle ABD, we have:
$\sin A = \frac{h}{c}\; \Rightarrow \; h=c \sin A$
II. In fig.3.40(ii) also, considering right triangle ABD, we have:
$\sin A = \frac{h}{c}\; \Rightarrow \; h=c \sin A$
5. Writing h in terms of a and C:
I. In fig.3.40(i), considering right triangle BCD, we have:
$\sin (180-C) = \frac{h}{a}\; \Rightarrow \; h=a \sin (180-C)$
• Using identity 9.d, we have: sin (180-C) = sin C
♦ For example,
♦ sin 30 = sin (180-30) = sin 120 = 0.5
♦ That means,
✰ taking the sine of an angle
✰ has the same effect as
✰ taking the sine of the supplement of that angle and vice versa.
• Thus we get: $h=a \sin C$
II. In fig.3.40(ii) also, considering right triangle BCD, we have:
$\sin C = \frac{h}{a}\; \Rightarrow \; h=a \sin C$
6. Relation between h, sin A, a, sin C, c:
I. For the obtuse triangle in fig(i):
• Using the results in 4.I and 5.I, we get:
h = c sin A = a sin C
$\Rightarrow \frac{\sin A}{a}=\frac{\sin C}{c}$
II. For the acute triangle in fig(ii) also:
• Using the results in 4.II and 5.II, we get:
h = c sin A = a sin C
$\Rightarrow \frac{\sin A}{a}=\frac{\sin C}{c}$
◼ We see that:
Relation between h, sin A, a, sin C and c is the same for both the triangles.
So the relation is applicable to any triangle.
7. In the above steps, we did not come across angle B and side b. The following steps from (8) to (11) will help us to bring in those items also.
8. Fig.3.41 below shows the same triangles. But this time, the perpendiculars are drawn from vertex C.
 |
| Fig.3.41 |
9. Writing h in terms of B and a:
I. In fig.3.41(i), considering right triangle BDC, we have:
$\sin B = \frac{h}{a}\; \Rightarrow \; h=a \sin B$
II. In fig.3.41(ii) also, considering right triangle BDC, we have:
$\sin B = \frac{h}{a}\; \Rightarrow \; h=a \sin B$
10. Writing h in terms of b and A:
I. In fig.3.41(i), considering right triangle ACD, we have:
$\sin A = \frac{h}{b}\; \Rightarrow \; h=b \sin A$
II. In fig.3.40(ii) also, considering right triangle ACD, we have:
$\sin A = \frac{h}{b}\; \Rightarrow \; h=b \sin A$
11. Relation between h, sin A, a, sin B, b:
I. For the obtuse triangle in fig.3.41(i):
• Using the results in 9.I and 10.I, we get:
h = a sin B = b sin A
$\Rightarrow \frac{\sin A}{a}=\frac{\sin B}{b}$
II. For the acute triangle in fig(ii) also:
• Using the results in 9.II and 10.II, we get:
h = a sin B = b sin A
$\Rightarrow \frac{\sin A}{a}=\frac{\sin B}{b}$
◼ We see that:
Relation between h, sin A, a, sin B and b is the same for both the triangles.
So the relation is applicable to any triangle.
12. Combining the results in (6) and (11), we get:
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
We see that, the sine formula has three parts. While solving simple problems, we will need only two parts at a time. Let us see some examples:
Example 1:
Fig.3.42(i) below shows triangle ABC in which one angle (C) and it's opposite side are given. One of the two remaining angles is also given.
 |
| Fig.3.42 |
• In total, we have: two angles and one side.
• We are asked to find the remaining one angle and the remaining two sides.
Solution:
1. We have the initial table:
$\begin{eqnarray}
A = -- \,&{|}& B = 32&{|}& C = 95 \\
a = -- \,&{|}& b = -- &{|}& c = 21 \
\end{eqnarray}$
2. Using the sine formula, we have:
$\begin{eqnarray}
&{}& \frac{\sin B}{b}=\frac{\sin C}{c} \\
&\Rightarrow& \frac{\sin 32}{b}=\frac{\sin 95}{21} \\
&\Rightarrow& b=\frac{\sin 32}{1} \times \frac{21}{\sin 95} \\
&\Rightarrow& b=11.17 \; \text{cm} \
\end{eqnarray}$
• Now the table becomes:
$\begin{eqnarray}
A = -- \,&{|}& B = 32&{|}& C = 95 \\
a = -- \,&{|}& b = 11.17 &{|}& c = 21 \
\end{eqnarray}$
3. Using the sine formula again, we have:
$\frac{\sin A}{a}=\frac{\sin C}{c}$
• We have: A = [180 - (95+32)=53]
• Now the table becomes:
$\begin{eqnarray}
A = 53 \,&{|}& B = 32&{|}& C = 95 \\
a = -- \,&{|}& b = 11.17 &{|}& c = 21 \
\end{eqnarray}$
• Thus we get:
$\begin{eqnarray}
&{}& \frac{\sin 53}{a}=\frac{\sin 95}{21} \\
&\Rightarrow& a=\frac{\sin 53}{1} \times \frac{21}{\sin 95} \\
&\Rightarrow& a=16.84 \; \text{cm} \
\end{eqnarray}$
4. We can check the result in (3) using the relation:
$\frac{\sin A}{a}=\frac{\sin B}{b}$
• Thus we get:
$\begin{eqnarray}
&{}& \frac{\sin 53}{a}=\frac{\sin 32}{11.17} \\
&\Rightarrow& a=\frac{\sin 53}{1} \times \frac{11.17}{\sin 32} \\
&\Rightarrow& a=16.84 \; \text{cm} \
\end{eqnarray}$
• This is same as the result in (3)
• So the final table is:
$\begin{eqnarray}
A = 53 \,&{|}& B = 32&{|}& C = 95 \\
a = 16.84 \,&{|}& b = 11.17 &{|}& c = 21 \
\end{eqnarray}$
Example 2:
Fig.3.42(ii) above shows triangle ABC in which one
angle (C) and it's opposite side are given. One of the two remaining
sides is also given.
• So in total, we have: two sides and one angle.
• We are asked to find the remaining one side and the remaining two angles.
Solution:
1. We have the initial table:
$\begin{eqnarray}
A = -- \,&{|}& B = --&{|}& C = 100 \\
a = 3.6 \,&{|}& b = -- &{|}& c = 5.1 \
\end{eqnarray}$
2. Using the sine formula, we have:
$\begin{eqnarray}
&{}& \frac{\sin A}{a}=\frac{\sin C}{c} \\
&\Rightarrow& \frac{\sin A}{3.6}=\frac{\sin 100}{5.1} \\
&\Rightarrow& \sin A=3.6 \times \frac{\sin 100}{5.1}=0.6952 \\
&\Rightarrow& A=44.04^{\text{o}} \
\end{eqnarray}$
• Consider the identity: sin x = sin (180-x). We get:
♦ sin 44.04 = sin (180-44.04) = sin 135.96 = 0.6952
♦ That means, sin A = 0.6952 will give two values for A: 44.04 and 135.96
• In our present case, angle at C is already an obtuse angle. No triangle can have two obtuse angles. So the appropriate value for A is 44.04
• Now the table becomes:
$\begin{eqnarray}
A = 44.04 \,&{|}& B = --&{|}& C = 100 \\
a = 3.6 \,&{|}& b = -- &{|}& c = 5.1 \
\end{eqnarray}$
3. Using the sine formula again, we have:
$\frac{\sin A}{a}=\frac{\sin B}{b}$
• We have: B = [180 - (100+44.04)=35.96]
• Now the table becomes:
$\begin{eqnarray}
A = 44.04 \,&{|}& B = 35.96&{|}& C = 100 \\
a = 3.6 \,&{|}& b = -- &{|}& c = 5.1 \
\end{eqnarray}$
• Thus we get:
$\begin{eqnarray}
&{}& \frac{\sin 44.04}{3.6}=\frac{\sin 35.96}{b} \\
&\Rightarrow& b=\frac{\sin 35.96}{1} \times \frac{3.6}{\sin 44.04} \\
&\Rightarrow& b=3.04 \; \text{cm} \
\end{eqnarray}$
4. We can check the result in (3) using the relation:
$\frac{\sin B}{b}=\frac{\sin C}{c}$
• Thus we get:
$\begin{eqnarray}
&{}& \frac{\sin 35.96}{b}=\frac{\sin 100}{5.1} \\
&\Rightarrow& b=\sin 35.96 \times \frac{5.1}{\sin 100} \\
&\Rightarrow& b=3.04 \; \text{cm} \
\end{eqnarray}$
• This is same as the result in (3)
• So the final table is:
$\begin{eqnarray}
A = 44.04 \,&{|}& B = 35.96&{|}& C = 100 \\
a = 3.6 \,&{|}& b = 3.04 &{|}& c = 5.1 \
\end{eqnarray}$
• The sine formula can be used when:
(i) We are given any two angles and any one side.
(ii) We are given any two sides and a non- included angle.
• If we are given two sides and the angle included between those two given sides, we will not be able to apply the sine formula.
• Also, if we are given all the tree sides but no angles, we will not be able to apply the sine formula.
• In such cases, we can use the cosine formula. We will see it in
the next
section.
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