$$\cos B \cdot \cos C + \sin B \cdot \sin C \geqslant 1$$
because $$\sin B \cdot \sin C \cdot {\sin ^2}A$$ is positive and $${\sin ^2}A \leqslant 1$$
or, $$\cos \left( {B - C} \right) \geqslant 1.\,{\text{But}}\,\,\cos \left( {B - C} \right) > 1.\,{\text{So,}}\,\,\cos \left( {B - C} \right) = 1.$$
Therefore, $$B = C$$ and then $$\sin A = 1.$$
$$\therefore \,\,A = \frac{\pi }{2},B = C = \frac{\pi }{4}.$$
3.
In a triangle $$ABC,$$ medians $$AD$$ and $$BE$$ are drawn. If $$AD = 4,$$ $$\angle DAB = \frac{\pi }{6}$$ and $$\angle ABE = \frac{\pi }{3} ,$$ then the area of the $$\Delta \,ABC$$ is
$$\eqalign{
& AP = \frac{2}{3}AD = \frac{8}{3};\,\,PD = \frac{4}{3};\,\,{\text{Let }}PB = x \cr
& \tan {60^ \circ } = \frac{{\frac{8}{3}}}{x}\,\,\,{\text{or }}x = \frac{8}{{3\sqrt 3 }} \cr} $$
Area of $$\Delta ABD = \frac{1}{2} \times 4 \times \frac{8}{{3\sqrt 3 }} = \frac{{16}}{{3\sqrt 3 }}$$
∴ Area of $$\Delta ABC = 2 \times \frac{{16}}{{3\sqrt 3 }} = \frac{{32}}{{3\sqrt 3 }}$$
[ $$\because $$ Median of a $$\Delta $$ divides it into two $$\Delta '{\text{s}}$$ of equal area. ]
4.
If $$BD, BE$$ and $$CF$$ are the medians of a $$\vartriangle ABC$$ then $$\left( {A{D^2} + B{E^2} + C{F^2}} \right):\left( {B{C^2} + C{A^2} + A{B^2}} \right)$$ is equal to
5.
In a triangle $$ABC,$$ let $$\angle C = \frac{\pi }{2}.$$ If $$r$$ is the inradius and $$R$$ is the circumradius of the triangle $$ABC,$$ then $$2 (r+ R)$$ equals
We know by sinc rule $$\frac{c}{{\sin C}} = 2R$$
$$\eqalign{
& \Rightarrow \,\,c = 2R\sin C \cr
& \Rightarrow \,\,c = 2R\,\,\,\,\,\,\,\left( {\because \,\,\angle C = {{90}^ \circ }} \right) \cr
& {\text{Also }}\tan \frac{C}{2} = \frac{r}{{s - c}} \cr
& \Rightarrow \,\,\tan \frac{\pi }{4} = \frac{r}{{s - c}}\,\,\left( {\because \,\,\angle C = {{90}^ \circ }} \right) \cr
& \Rightarrow \,\,r = s - c = \frac{{a + b - c}}{2} \cr
& \Rightarrow \,\,2r + c = a + b \cr
& \Rightarrow \,\,2r + 2R = a + b\,\left( {{\text{using }}c = 2R} \right) \cr} $$
6.
In a $$\vartriangle ABC,A = \frac{{2\pi }}{3},b - c = 3\sqrt 3 \,cm$$ and $${\text{ar}}\left( {\vartriangle ABC} \right) = \frac{{9\sqrt 3 }}{2}\,c{m^2}.$$ Then $$a$$ is
Let us consider $$\frac{{b - c}}{a},$$ which is involved in each of the these options.
$$\eqalign{
& \frac{{b - c}}{a} = \frac{{\sin B - \sin C}}{{\sin A}} \cr
& = \frac{{2\cos \left( {\frac{{B + C}}{2}} \right)\sin \left( {\frac{{B - C}}{2}} \right)}}{{2\sin \frac{A}{2}\cos \frac{A}{2}}} \cr
& = \frac{{\sin \frac{A}{2}\sin \left( {\frac{{B - C}}{2}} \right)}}{{\sin \frac{A}{2}\cos \frac{A}{2}}} \cr
& = \frac{{\sin \left( {\frac{{B - C}}{2}} \right)}}{{\cos \frac{A}{2}}} \cr
& \therefore \,\,\left( {b - c} \right)\cos \frac{A}{2} = a\sin \left( {\frac{{B - C}}{2}} \right) \cr} $$
9.
In an equilateral triangle, 3 coins of radii 1 unit each are kept so that they touch each other and also the sides of the triangle. Area of the triangle is
The situation is as shown in the figure. For circle with centre $${C_2},BP$$ and $$BP'$$ are two tangents from $$B$$ to circle, therefore $$B{C_2}$$ must be the $$\angle$$ bisector of $$\angle B.$$ But $$\angle B = 60°$$ $$\left( {\because \,\,\Delta ABC\,\,{\text{is an equilateral}}\,\,\Delta } \right)$$
$$\tan A = \frac{4}{3}$$ and $$\tan B = \frac{12}{5}.$$ Clearly, $$\tan C$$ should be such that $$\tan A + \tan B + \tan C = \tan A\tan B\tan C$$
$$\eqalign{
& \therefore \,\,\frac{4}{3} + \frac{{12}}{5} + \tan C = \frac{4}{3} \cdot \frac{{12}}{5} \cdot \tan C\,\,\,{\text{or, }}\frac{{56}}{{15}} + \tan C = \frac{{16}}{5}\tan C \cr
& {\text{or, }}\tan C = \frac{{56}}{{33}} \cr
& \therefore \,\,\cos C = \frac{{33}}{{65}}. \cr} $$