Properties and Solutons of Triangle MCQ Questions & Answers in Trigonometry | Maths

Clearly, $$A > B\,\,\left( {\because \,\,a > b} \right).$$
Now, $$\tan\frac{{A - B}}{2} = \frac{{a - b}}{{a + b}}\cot \frac{C}{2}$$
$$\eqalign{ & \Rightarrow \,\,\tan {30^ \circ } = \frac{1}{3}\cot \frac{C}{2} \cr & \therefore \,\,\sqrt 3 = \cot \frac{C}{2}\,\,\,{\text{or,}}\,\,\frac{C}{2} = \frac{\pi }{6}. \cr} $$

$$\cos B \cdot \cos C + \sin B \cdot \sin C \geqslant 1$$
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 $$\sin A = 1.$$
$$\therefore \,\,A = \frac{\pi }{2},B = C = \frac{\pi }{4}.$$

$$\eqalign{ & AP = \frac{2}{3}AD = \frac{8}{3};\,\,PD = \frac{4}{3};\,\,{\text{Let }}PB = x \cr & \tan {60^ \circ } = \frac{{\frac{8}{3}}}{x}\,\,\,{\text{or }}x = \frac{8}{{3\sqrt 3 }} \cr} $$
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. ]

We have, $$A{B^2} + A{C^2} = 2\left( {A{D^2} + B{D^2}} \right)$$
$$ \Rightarrow \,\,\frac{{{c^2} + {b^2}}}{2} - \frac{{{a^2}}}{4} = A{D^2},\,{\text{e}}{\text{.t}}{\text{.c}}{\text{.}}$$
Adding, $${a^2} + {b^2} + {c^2} - \frac{{{a^2} + {b^2} + {c^2}}}{4} = A{D^2} + B{E^2} + C{F^2}$$
$$ \Rightarrow \,\,\left( {A{D^2} + B{E^2} + C{F^2}} \right):\left( {{a^2} + {b^2} + {c^2}} \right) = 3:4.$$

We know by sinc rule $$\frac{c}{{\sin C}} = 2R$$
$$\eqalign{ & \Rightarrow \,\,c = 2R\sin C \cr & \Rightarrow \,\,c = 2R\,\,\,\,\,\,\,\left( {\because \,\,\angle C = {{90}^ \circ }} \right) \cr & {\text{Also }}\tan \frac{C}{2} = \frac{r}{{s - c}} \cr & \Rightarrow \,\,\tan \frac{\pi }{4} = \frac{r}{{s - c}}\,\,\left( {\because \,\,\angle C = {{90}^ \circ }} \right) \cr & \Rightarrow \,\,r = s - c = \frac{{a + b - c}}{2} \cr & \Rightarrow \,\,2r + c = a + b \cr & \Rightarrow \,\,2r + 2R = a + b\,\left( {{\text{using }}c = 2R} \right) \cr} $$

$$\eqalign{ & \frac{1}{2}bc\sin \frac{{2\pi }}{3} = \frac{{9\sqrt 3 }}{2}\,\,{\text{or, }}\frac{1}{2} \cdot \frac{{\sqrt 3 }}{2} \cdot bc = \frac{{9\sqrt 3 }}{2} \cr & \Rightarrow \,\,bc = 18. \cr & {\text{Also, }}\cos \frac{{2\pi }}{3} = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} \cr & \Rightarrow \,\, - \frac{1}{2} = \frac{{{{\left( {b - c} \right)}^2} + 2bc - {a^2}}}{{2bc}} \cr & {\text{or, }}{\left( {b - c} \right)^2} + 3bc - {a^2} = 0\,\,\,\,{\text{or, }}27 + 54 = {a^2}. \cr} $$

Clearly, $$A = {45^ \circ },B = {75^ \circ },C = {60^ \circ }.\,{\text{So, }}\frac{a}{{\sin {{45}^ \circ }}} = \frac{b}{{\sin {{75}^ \circ }}} = \frac{c}{{\sin {{60}^ \circ }}} = 2R.$$
$$\therefore \,\,a = \sqrt 2 R,b = \frac{{\sqrt 3 + 1}}{{\sqrt 2 }}R,c = \sqrt 3 R.$$
Now, $$a + b + c\sqrt 2 = \left( {\sqrt 2 + \frac{{\sqrt 3 + 1}}{{\sqrt 2 }} + \sqrt 6 } \right)R = 3b.$$

Let us consider $$\frac{{b - c}}{a},$$   which is involved in each of the these options.
$$\eqalign{ & \frac{{b - c}}{a} = \frac{{\sin B - \sin C}}{{\sin A}} \cr & = \frac{{2\cos \left( {\frac{{B + C}}{2}} \right)\sin \left( {\frac{{B - C}}{2}} \right)}}{{2\sin \frac{A}{2}\cos \frac{A}{2}}} \cr & = \frac{{\sin \frac{A}{2}\sin \left( {\frac{{B - C}}{2}} \right)}}{{\sin \frac{A}{2}\cos \frac{A}{2}}} \cr & = \frac{{\sin \left( {\frac{{B - C}}{2}} \right)}}{{\cos \frac{A}{2}}} \cr & \therefore \,\,\left( {b - c} \right)\cos \frac{A}{2} = a\sin \left( {\frac{{B - C}}{2}} \right) \cr} $$

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 $$\angle B = 60°$$   $$\left( {\because \,\,\Delta ABC\,\,{\text{is an equilateral}}\,\,\Delta } \right)$$

$$\eqalign{ & \therefore \,\,\angle {C_2}BP = {30^ \circ } \cr & \therefore \,\,\tan {30^ \circ } = \frac{1}{x} \cr & \Rightarrow \,\,x = \sqrt 3 \cr & \therefore \,\,BC = BP + PQ + QC \cr & = x + 2 + x = 2 + 2\sqrt 3 \cr & \therefore \,\,{\text{Area of }}\Delta ABC = \frac{{\sqrt 3 }}{4} \times {\left( {2 + 2\sqrt 3 } \right)^2} \cr & = 4\sqrt 3 + 6\,\,{\text{sq}}{\text{. units}}{\text{.}} \cr} $$

$$\tan A = \frac{4}{3}$$   and $$\tan B = \frac{12}{5}.$$   Clearly, $$\tan C$$  should be such that $$\tan A + \tan B + \tan C = \tan A\tan B\tan C$$
$$\eqalign{ & \therefore \,\,\frac{4}{3} + \frac{{12}}{5} + \tan C = \frac{4}{3} \cdot \frac{{12}}{5} \cdot \tan C\,\,\,{\text{or, }}\frac{{56}}{{15}} + \tan C = \frac{{16}}{5}\tan C \cr & {\text{or, }}\tan C = \frac{{56}}{{33}} \cr & \therefore \,\,\cos C = \frac{{33}}{{65}}. \cr} $$