Properties and Solution of Triangles
3.0 Projection Formulae
3.0 Projection Formulae
In any triangle $$a = b\cos C + c\cos B$$
Similarly, we can write $$\begin{equation} \begin{aligned} b = c\cos A + a\cos C \\ c = a\cos B + b\cos A \\\end{aligned} \end{equation} $$
Proof: Consider the triangle $ABC$ with sides $a$, $b$ and $c$ with angles $A$, $B$ and $C$
In an acute angle triangle, draw an altitude from the vertex $A$ to the side $BC$ which intersects at a point $D$, from the figure we can write as $$a = x + y$$
From the triangle $ACD$, $$\begin{equation} \begin{aligned} \cos C = \frac{y}{b} \\ y = b\cos C...(i) \\\end{aligned} \end{equation} $$
and from triangle $ABD$, $$\begin{equation} \begin{aligned} \cos B = \frac{x}{c} \\ x = c\cos B...(ii) \\\end{aligned} \end{equation} $$
Adding $(i)$ and $(ii)$, we get $$\begin{equation} \begin{aligned} x + y = b\cos C + c\cos B \\ a = b\cos C + c\cos B \\\end{aligned} \end{equation} $$
Similarly we can find other projection rule,
In case of an obtuse angle triangle, draw an altitude fron the vertex $A$ to the side $BC$, extending $BC$ which intersects at a point $D$, from the figure we can write as
From the triangle $ADC$, $$\begin{equation} \begin{aligned} \cos C = \frac{x+a}{b} \\ x+a = b\cos C...(i) \\\end{aligned} \end{equation} $$
and from triangle $ADB$, $$\begin{equation} \begin{aligned} \cos (\pi - B) = \frac{x}{c} \\ x = - c\cos B...(ii) \\\end{aligned} \end{equation} $$
From $(i)$ and $(ii)$, we get
$$\begin{equation} \begin{aligned} - c\cos B + a = b\cos C \\ a = b\cos C + c\cos B \\\end{aligned} \end{equation} $$
Question 7. In a triangle $ABC$, prove that $$a(b\cos C - c\cos B) = {b^2} - {c^2}$$
Solution: $$\begin{equation} \begin{aligned} a(b\cos C - c\cos B) \\ = b(a\cos C) - c(a\cos B)...(i) \\\end{aligned} \end{equation} $$
From projection rule, we can write as $$\begin{equation} \begin{aligned} b = a\cos C + ccosA \\ a\cos C = b - c\cos A \\\end{aligned} \end{equation} $$
and $$\begin{equation} \begin{aligned} c = a\cos B + b\cos A \\ a\cos B = c - b\cos A \\\end{aligned} \end{equation} $$
Putting the value of $a\cos C$ and $a\cos B$ in equation $(i)$, we get $$\begin{equation} \begin{aligned} = b(b - c\cos A) - c(c - b\cos A) \\ = {b^2} - bc\cos A - {c^2} + bc\cos A \\ {b^2} - {c^2} \\\end{aligned} \end{equation} $$
Question 8. In a triangle $ABC$, prove that $$\frac{{\cos B}}{{\cos C}} = \frac{{c - b\cos A}}{{b - c\cos A}}$$
Solution: Using projection formula, $$\begin{equation} \begin{aligned} b = a\cos C + ccosA \\ a\cos C = b - c\cos A...(i) \\\end{aligned} \end{equation} $$
Similarly, we can write $$\begin{equation} \begin{aligned} c = a\cos B + b\cos A \\ a\cos B = c - b\cos A...(ii) \\\end{aligned} \end{equation} $$
Dividing $(ii)$ by $(i)$, we get $$\begin{equation} \begin{aligned} \frac{{a\cos B}}{{a\cos C}} = \frac{{c - b\cos A}}{{b - c\cos A}} \\ \frac{{\cos B}}{{\cos C}} = \frac{{c - b\cos A}}{{b - c\cos A}} \\\end{aligned} \end{equation} $$
Question 9. In a triangle $ABC$, prove that $$\frac{{\cos A}}{{c\cos B + b\cos C}} + \frac{{\cos B}}{{a\cos C + c\cos A}} + \frac{{\cos C}}{{a\cos B + b\cos A}} = \frac{{{a^2} + {b^2} + {c^2}}}{{2abc}}$$
Solution: Considering LHS,
Using Cosine formulas $$\begin{equation} \begin{aligned} \cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} \\ \cos B = \frac{{{c^2} + {a^2} - {b^2}}}{{2ac}} \\ \cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} \\\end{aligned} \end{equation} $$
and projection formulas $$\begin{equation} \begin{aligned} a = c\cos B + b\cos C \\ b = a\cos C + c\cos A \\ c = a\cos B + b\cos A \\\end{aligned} \end{equation} $$
LHS becomes $$\begin{equation} \begin{aligned} =\frac{{\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}}}{a} + \frac{{\frac{{{c^2} + {a^2} - {b^2}}}{{2ac}}}}{b} + \frac{{\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}}}{c} \\ =\frac{{{b^2} + {c^2} - {a^2}}}{{2abc}} + \frac{{{c^2} + {a^2} - {b^2}}}{{2abc}} + \frac{{{a^2} + {b^2} - {c^2}}}{{2abc}} \\ =\frac{{{a^2} + {b^2} + {c^2}}}{{2abc}} \\\end{aligned} \end{equation} $$
