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

$$\eqalign{ & 2b = a + c \cr & \Rightarrow \,\,2\sin B = \sin A + \sin C \cr & \Rightarrow \,\,2\sin \frac{B}{2} = \cos \frac{{A - C}}{2} \cr & \Rightarrow 2\cos \frac{{A + C}}{2} = \cos \frac{{A - C}}{2} \cr & \Rightarrow \,\,\cos \frac{A}{2} \cdot \cos \frac{C}{2} = 3\sin \frac{A}{2} \cdot \sin \frac{C}{2} \cr & \Rightarrow \,\,\tan \frac{A}{2} \cdot \tan \frac{C}{2} = \frac{1}{3} \cr & \therefore \,\,\frac{{\tan \frac{A}{2} + \tan \frac{C}{2}}}{{\cot \frac{B}{2}}} = \frac{{\cos \frac{B}{2}}}{{\cos \frac{A}{2} \cdot \cos \frac{C}{2}}} \cdot \frac{{\sin \frac{B}{2}}}{{\cos \frac{B}{2}}} = \frac{{\cos \frac{{A + C}}{2}}}{{\cos \frac{A}{2} \cdot \cos \frac{C}{2}}} \cr & = 1 - \tan \frac{A}{2} \cdot \tan \frac{C}{2} = 1 - \frac{1}{3} = \frac{2}{3}. \cr} $$

$$\eqalign{ & A + B = {180^ \circ } - C = {90^ \circ } \cr & a = 2R\sin A,b = 2R\sin B,c = 2R\sin C \cr & \therefore \frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}} = \frac{{{{\sin }^2}A - {{\sin }^2}B}}{{{{\sin }^2}A + {{\sin }^2}B}} \cr & = \frac{{\sin \left( {A + B} \right)\sin \left( {A - B} \right)}}{{{{\sin }^2}A + {{\sin }^2}\left( {{{90}^ \circ } - A} \right)}}\,\,\,\left[ {\because A + B = {{90}^ \circ }} \right] \cr & = \frac{{\sin {{90}^ \circ }\sin \left( {A - B} \right)}}{{{{\sin }^2}A + {{\cos }^2}A}} = \sin \left( {A - B} \right) \cr} $$

$$\eqalign{ & \tan {15^ \circ } = \frac{{DE}}{{AE}} \cr & AE = 10\cot {15^ \circ }\,\,\,.....\left( 1 \right) \cr & \cot \left( {{{15}^ \circ }} \right) = \cot \left( {{{45}^ \circ } - {{30}^ \circ }} \right) \cr & = \frac{{\cot {{45}^ \circ }\cot {{30}^ \circ } + 1}}{{\cot {{30}^ \circ } - \cot {{45}^ \circ }}} \cr} $$

$$\eqalign{ & \cot {15^ \circ } = \frac{{1 \cdot \sqrt 3 + 1}}{{\sqrt 3 - 1}} = 2 + \sqrt 3 \cr & {\text{Putting }}\cot {15^ \circ }{\text{ in eq}}\left( 1 \right) \cr & AE = 10\cot {15^ \circ } \cr & = 10\left( {2 + \sqrt 3 } \right) \cr & = 10\left( {3.73} \right) \cr & = 37.3\,m \cr} $$

We have,
$$\eqalign{ & \frac{{2\sin P - \sin 2P}}{{2\sin P + \sin 2P}} \cr & = \frac{{2\sin P - 2\sin P\cos P}}{{2\sin P + 2\sin P\cos P}} \cr & = \frac{{1 - \cos P}}{{1 - \sin P}} \cr & = \frac{{2{{\sin }^2}\frac{P}{2}}}{{2{{\cos }^2}\frac{P}{2}}} \cr & = {\tan ^2}\frac{P}{2} \cr & = \frac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}} \cr & {\text{where }}s = \frac{{a + b + c}}{2} \cr & = \frac{{{{\left( {s - b} \right)}^2}{{\left( {s - c} \right)}^2}}}{{s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}} \cr & = \frac{{{{\left( {a + c - b} \right)}^2}{{\left( {a + b - c} \right)}^2}}}{{16.{\Delta ^2}}} \cr & = \frac{{1 \times 9}}{{16{\Delta ^2}}} \cr & = {\left( {\frac{3}{{4\Delta }}} \right)^2} \cr} $$

$$\eqalign{ & {\text{In }}\Delta CBD,\tan {60^ \circ } = \frac{{100}}{3} \cr & \Rightarrow \,\,x = \frac{{100}}{{\sqrt 3 }} \cr & {\text{In }}\Delta ACB,\tan {30^ \circ } = \frac{{100}}{{x + d}} \cr & \Rightarrow \,\,x + d = 100\sqrt 3 \cr & \Rightarrow \,\,d = 100\sqrt 3 - \frac{{100}}{{\sqrt 3 }} \cr & = \frac{{200}}{{\sqrt 3 }} \cr & = \frac{{200\sqrt 3 }}{3}m \cr} $$

Clearly, $$\tan\frac{A}{2} > \tan \frac{B}{2};$$
$$\eqalign{ & \therefore \,\,\frac{A}{2} > \frac{B}{2}\,\,\,{\text{or, }}A > B. \cr & \tan \frac{{A + B}}{2} = \frac{{\frac{5}{6} + \frac{{20}}{{37}}}}{{1 - \frac{5}{6} \cdot \frac{{20}}{{37}}}} = \frac{{305}}{{122}} \cr & {\text{or, }}\tan \frac{C}{2} = \frac{{122}}{{305}}. \cr & {\text{As, }}\frac{{20}}{{37}} > \frac{{122}}{{305}},\,{\text{we get,}} \cr & \tan \frac{B}{2} > \tan \frac{C}{2} \cr & \Rightarrow \,\,\frac{B}{2} > \frac{C}{2}\,{\text{i}}{\text{.e}}{\text{., }}B > C. \cr & \therefore \,\,A > B > C\,\,\,\,\,\,\,\,\therefore \,\,a > b > c. \cr & \therefore \,\,2a > b + c\,\,{\text{and }}a + b > 2c. \cr} $$

We know that $$A + B + C = 180°$$
⇒ $$A + C- B = 180° - 2B$$
Now $$2ac\sin \left[ {\frac{1}{2}\left( {A - B + C} \right)} \right]$$
$$\eqalign{ & = 2ac\sin \left( {{{90}^ \circ } - B} \right) \cr & = 2ac\cos B \cr & = \frac{{2ac\left( {{a^2} + {c^2} - {b^2}} \right)}}{{2ac}} \cr & = {a^2} + {c^2} - {b^2} \cr} $$

Here, $$CD = p\cot C\,\,{\text{and }}AD = p\cot A = p\tan C$$
Adding, $$4p = p\cot C + p\tan C$$
$$\eqalign{ & \Rightarrow \,\,\tan C + \cot C = 4 \cr & {\text{or, }}\frac{1}{{\sin C \cdot \cos C}} = 4\,\,{\text{or, }}\sin 2C = \frac{1}{2} \cr & \Rightarrow \,\,C = {15^ \circ }. \cr} $$

From the figure
$$\eqalign{ & \tan {60^ \circ } = \frac{y}{x} \cr & \Rightarrow \,\,y = \sqrt 3 x\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & \tan {30^ \circ } = \frac{y}{{x + 40}} \cr & \Rightarrow \,\,y = \frac{{x + 40}}{{\sqrt 3 }}\,\,\,\,\,.....\left( 2 \right) \cr} $$
From (1) and (2), $$\sqrt 3 x = \frac{{x + 40}}{{\sqrt 3 }}$$
$$ \Rightarrow \,\,x = 20m$$

$$\frac{{{b^2} - {c^2}}}{{2aR}} = \frac{{4{R^2}\left( {{{\sin }^2}B - {{\sin }^2}C} \right)}}{{4{R^2}\sin A}} = \frac{{\sin \left( {B + C} \right)\sin \left( {B - C} \right)}}{{\sin A}} = \sin \left( {B - C} \right).$$