6.0 Area of triangle

  Properties and Solution of Triangles

Area of a triange $ABC$ is given as

$$\Delta = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B = \sqrt {s(s - a)(s - b)(s - c)} $$

Question 12. In a triangle $ABC$, if $b\sin C(b\cos C + c\cos B) = 42$, then find the area of triangle.

Solution: $$b\sin C(b\cos C + c\cos B) = 42...(i)$$

Using projection rule $$a = b\cos C + c\cos B$$ so put this in equation $(i)$, we get $$ab\sin C = 42$$

We know that area of triangle is given as $$\frac{1}{2}ab\sin C$$

So area of triangle is $21$square units.

Question 13. In a triangle $ABC$, prove that $$(a + b + c)(\tan \frac{A}{2} + \tan \frac{B}{2}) = 2c\cot \frac{C}{2}$$

Solution: Considering LHS $$\begin{equation} \begin{aligned} (a + b + c)(\tan \frac{A}{2} + \tan \frac{B}{2}) = 2c\cot \frac{C}{2} \\ \tan \frac{A}{2} = \sqrt {\frac{{(s - b)(s - c)}}{{s(s - a)}}} ,\tan \frac{B}{2} = \sqrt {\frac{{(s - c)(s - a)}}{{s(s - b)}}} \\ (a + b + c)(\tan \frac{A}{2} + \tan \frac{B}{2}) = (a + b + c)[\{ \sqrt {\frac{{(s - b)(s - c)}}{{s(s - a)}}} + \sqrt {\frac{{(s - c)(s - a)}}{{s(s - b)}}} \} ] \\ 2s\sqrt {\frac{{s - c}}{s}} [\sqrt {\frac{{(s - b)}}{{(s - a)}}} + \sqrt {\frac{{(s - a)}}{{(s - b)}}} ] \\\end{aligned} \end{equation} $$ using $2s = a + b + c$

$$2\sqrt {s(s - c)} [\frac{{s - b + s - a}}{{\sqrt {(s - a)(s - b)} }}]$$

Using $$\begin{equation} \begin{aligned} 2s = a + b + c \\ 2s - b - a = c \\\end{aligned} \end{equation} $$

so we get $$2c\sqrt {\frac{{s(s - c)}}{{(s - b)(s - a)}}} $$

And we know that $$\cot \frac{C}{2} = \sqrt {\frac{{s(s - c)}}{{(s - b)(s - a)}}} $$

therefore,

$$ = 2c\cot \frac{C}{2}$$

which is equal to RHS.