Projection Formula & Overview

Projection is used to find the analyses of one side as per the other side. This concept is particularly useful in solving problems involving triangles, vectors, and their components. It connect the sides and angles of the triangle. The law of tangents or tangent rule is a statement about the relationship between the tangents of two angles of a triangle and the lengths of the opposing sides. The law of tangents can be used in any case where two sides and the included angle, or two angles and a side, are known. In real life, we use the Law of Tangents to calculate the angle of celestial objects such as stars, moon, etc.

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Projection Formula

In a geometric sense, projecting a vector onto another vector involves finding the shadow or footprint of one vector along the direction of another. Given two vectors a and $\mathbf{b}$, the projection of $\mathbf{a}$ onto $\mathbf{b}$ is a vector that lies along $\mathbf{b}$ and represents the component of $\mathbf{a}$ in the direction of b\mathbf\{b\}b.

In the $\mathrm{△}ABC$,
Projection of $AB$ on $BC$ is $BD$ and Projection of $AC$ on $BC$ is $DC$

Now,

$\begin{array}{rl}& BD=c\cos B\text{and}DC=b\cos C\\ & \text{and,}BC=a=BD+DC\\ & =c\cos B+b\cos C\\ & a=c\cos B+b\cos C\end{array}$

In a similar way, other projection formula can be derived
1. $a=c\cos B+b\cos C$
2. $b=c\cos A+a\cos C$
3. $c=b\cos A+a\cos B$

The projection formula is significant because it provides a clear and mathematical method to decompose vectors into components along different directions. This decomposition is essential in simplifying complex problems in physics, engineering, and other sciences. It allows for a better understanding of the influence of vector components in various applications, leading to more precise calculations and analyses.

Summary: The projection tool is powerful/ important in trigonometry. The projection formula can be used in any case where two sides and the included angle, or two angles and a side, are known. The law of tangent is not only used to derive various formulas in trigonometry but also helps in the fields of physics and astrology. Knowledge of tangent rules is useful in solving complex problems.

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Solved Examples Based on Projection Formula:

Example 1: In Triangle ABC , point D is on BC is such that $AD\perp BC$ and E is the middle point of BC and $b^2+2a^2=c^2$ distance between D and E is?
1) b
2) $\frac{a+b+c}{2}$
3) $\frac{b+c}{2}$
4) a

Solution:

$\begin{array}{rl}DE& =EC-DC\\ & =\frac{a}{2}-b\cos C\\ & =\frac{a}{2}-b\frac{a^2+b^2-c^2}{2ab}\\ & =\frac{a}{2}-b\frac{a^2-(c^2-b^2)}{2ab}\\ & =\frac{a}{2}-\frac{a^2-\left(2a^2\right)}{2a}\\ & =a\end{array}$

Example 2: In triangle $ABC$ if
$a+c-\frac{3b}{2}=a{\sin }^2\left(\frac{C}{2}\right)+c{\sin }^2\left(\frac{A}{2}\right)$ then a,b,c are in ?

1) A.P.

2) G.P.

3) H.P.

4)

None of these

Solution

Diagram-

$\begin{array}{rl}& a+c-\frac{3b}{2}=a{\sin }^2\left(\frac{C}{2}\right)+c{\sin }^2\left(\frac{A}{2}\right)\\ & a+c-\frac{3b}{2}=a\left[\frac{1-\cos C}{2}\right]+c\left[\frac{1-\cos A}{2}\right]\\ & c\cos A+a\cos C=3b-a-c\\ & b=3b-a-c\end{array}$

by above triangle $c\cos A+a\cos C=b$
a+c=2 b
$$

$a,b,c$ in $A.P$.
Example 3: The value of $b\cos C+c\cos B$ is:
1) $-a$
2) a
3) b
4) c

Solution
Given that,

$b\cos C+c\cos B$

Using the law of cosine,

$\begin{array}{rl}& b\cos C+c\cos B=b\left[\frac{(a^2+b^2-c^2)}{2ab}\right]+c\left[\frac{(c^2+a^2-b^2)}{2ca}\right]\\ & b\cos C+c\cos B=\left[\frac{(a^2+b^2-c^2+c^2+a^2+b^2)}{2a}\right]\end{array}$

$b\cos C+c\cos B=\frac{2a^2}{2a}$

$b\cos C+c\cos B=a$
Hence, the answer is the option (2).

Example 4: If the sides of the $\mathrm{△}ABC$ are $a=3,b=2$ and $c=4$, then the value of $3\cos B+2\cos C$ is:
1) $\frac{4}{5}$
2) $-\frac{4}{5}$
3) $\frac{2}{5}$
4) $-\frac{2}{5}$

Solution
Given that,
$\begin{array}{rl}& 3\cos B+2\cos C\\ & \cos B=\frac{a^2+c^2-b^2}{2ac}\\ & \cos B=\frac{9+16-4}{24}\\ & \cos B=\frac{21}{24}\end{array}$

$\begin{array}{rl}& \cos C=\frac{a^2+b^2-c^2}{2ab}\\ & \cos C=\frac{-1}{4}\end{array}$
Therefore,

$\begin{array}{rl}& 3\cos B+2\cos C=2\times \frac{21}{24}+2\times \frac{-1}{4}\\ & 3\cos B+2\cos C=\frac{21}{12}-\frac{1}{2}\\ & 3\cos B+2\cos C=\frac{4}{5}\end{array}$
Hence, the answer is the option (1).

Example 5: If the area of the $\mathrm{△}ABC$ be $\mathrm{\Delta }$, then the value of $b^2\sin 2C+c^2\sin 2B$ is:
1) $3\mathrm{\Delta }$
2) $2\mathrm{\Delta }$
3) $4\mathrm{\Delta }$
4) $5\mathrm{\Delta }$

Solution
Given that,

$\begin{array}{rl}& b^2\sin 2C+c^2\sin 2B\\ & b^2\sin 2C+c^2\sin 2B=b^{2}2\sin C\cos C+c^{2}2\sin B\cos B\\ & b^2\sin 2C+c^2\sin 2B=2b\cos C\cdot b\sin C+2c\cos B\cdot c\sin B\end{array}$

We know that,

$\frac{a}{\sin B}=\frac{c}{\sin C}\Rightarrow b\sin C=c\sin B$

Thus,

$\begin{array}{rl}& b^2\sin 2C+c^2\sin 2B=2b\cos C\cdot c\sin B+2c\cos B\cdot c\sin B\\ & b^2\sin 2C+c^2\sin 2B=2c\sin B(b\cos C+c\cos B)\end{array}$

We know that,

$a=b\cos C+c\cos B$
Therefore,

$b^2\sin 2C+c^2\sin 2B=2c\sin B\cdot a^{b^2\sin 2C+c^2\sin 2B=4\cdot \frac{1}{2}ac\sin B}{b}^2\sin 2C+c^2\sin 2B=4\mathrm{\Delta }$

Hence, the answer is the option (3).