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

$$\eqalign{ & \Delta = \frac{1}{2}{p_1}a = \frac{1}{2}{p_2}b = \frac{1}{2}{p_3}b \cr & {p_1},{p_2},{p_3}\,\,{\text{are in H}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\,\frac{{2\Delta }}{a},\frac{{2\Delta }}{b},\frac{{2\Delta }}{c}\,\,{\text{are in H}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\,\frac{1}{a},\frac{1}{b},\frac{1}{c}\,\,{\text{are in H}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\,a,b,c\,\,{\text{are in A}}{\text{.P}}{\text{.}} \cr} $$
$$ \Rightarrow \,\,K\sin A,K\sin B,K\sin C$$     are in A.P.
$$ \Rightarrow \,\,\sin A,\sin B,\sin C\,\,{\text{are in A}}{\text{.P}}{\text{.}}$$

$$\eqalign{ & \frac{{c + b}}{{c - b}}\tan \frac{A}{2} = \frac{{\sin C + \sin B}}{{\sin C - \sin B}}\tan \frac{A}{2} = \frac{{2\sin \frac{{B + C}}{2} \cdot \cos \frac{{C - B}}{2}}}{{2\cos \frac{{B + C}}{2} \cdot \sin \frac{{C - B}}{2}}} \cdot \tan \frac{A}{2} \cr & \frac{{c + b}}{{c - b}}\tan \frac{A}{2} = \tan \frac{{B + C}}{2} \cdot \cot \frac{{C - B}}{2} \cdot \tan \frac{A}{2} = \cot \frac{{C - B}}{2} = \cot \frac{{\pi - A - B - B}}{2} \cr & \frac{{c + b}}{{c - b}}\tan \frac{A}{2} = \cot\left( {\frac{\pi }{2} - \overline {\frac{A}{2} + B} } \right) = \tan \left( {\frac{A}{2} + B} \right). \cr} $$

$$\left( {H + a} \right)\cot b = \left( {H - a} \right)\cot a$$
using componendo and dividendo
$$ \Rightarrow H = \frac{{a\sin \left( {\alpha + \beta } \right)}}{{\sin \left( {\beta - \alpha } \right)}}$$

$$\eqalign{ & 1 + b + c = 6.\frac{{\sin A + \sin B + \sin C}}{3} = 2\left( {\frac{1}{{2R}} + \frac{b}{{2R}} + \frac{c}{{2R}}} \right) = \frac{1}{R}\left( {1 + b + c} \right) \cr & \therefore \,\,R = 1\,\,\,\,\,\left( {\because \,1 + b + c \ne 0} \right). \cr & {\text{So}},\frac{1}{{\sin A}} = 2R = 2 \cr & \Rightarrow \,\,A = \frac{\pi }{6}. \cr} $$

$$\eqalign{ & {a^2} + {b^2} + {c^2} = 8{R^2} \cr & \Rightarrow \,\,{\sin ^2}A + {\sin ^2}B + {\sin ^2}C = 2 \cr & \Rightarrow \cos 2A + \cos 2B + \cos 2C + 1 = 0 \cr & \therefore - 4\cos A \cdot \cos B \cdot \cos C = 0 \cr & \Rightarrow \,\,A\,\,{\text{or }}B\,\,{\text{or }}C = \frac{\pi }{2}. \cr} $$

$$c\cos \theta - a\sin \theta = b.$$    Therefore, $$c\cos\alpha - a\sin \alpha = c\cos \beta - a\sin \beta ,$$       where $$\alpha ,\beta $$  are the other two angles of the triangle.
$$\eqalign{ & \therefore \,\,c\left( {\cos \alpha - \cos \beta } \right) = a\left( {\sin \alpha - \sin \beta } \right) \cr & {\text{or, }}\frac{c}{a} = \frac{{2\cos \frac{{\alpha + \beta }}{2}\sin \frac{{\alpha - \beta }}{2}}}{{2\sin \frac{{\alpha + \beta }}{2}\sin \frac{{\beta - \alpha }}{2}}} = - \cot \frac{{\alpha + \beta }}{2} \cr & \therefore \,\,\tan \frac{{\alpha + \beta }}{2} = \frac{{ - a}}{c} \cr & \therefore \,\,\tan \left( {\alpha + \beta } \right) = \frac{{2 \times \left( {\frac{{ - a}}{c}} \right)}}{{1 - \frac{{{a^2}}}{{{c^2}}}}} = \frac{{2ac}}{{{a^2} - {c^2}}} \cr & \therefore \,\,\tan \left( {\pi - \frac{{3\pi }}{4}} \right) = \frac{{2ac}}{{{a^2} - {c^2}}}\,\,\,\,{\text{or, }}{a^2} - {c^2} = 2ac. \cr} $$

$$\eqalign{ & {\text{If}}\,A + B + C = \pi , \cr & {\text{then}}\,\,\cos \,mA + \cos \,mB + \cos \,mC \cr & = 1 - 4\sin \frac{{mA}}{2}\sin \frac{{mB}}{2}\sin \frac{{mC}}{2} \cr & \therefore {\text{For }}\,m = 2:\cos 2A + \cos 2B + \cos 2C \cr & = 1 - 4\sin A\sin B\sin C \cr & \Rightarrow \cos 2A + \cos 2B + \cos 2C + 4\sin A\sin B\sin C = 1 \cr} $$

$$\eqalign{ & {\sin ^{ - 1}}\frac{1}{3} = {\cot ^{ - 1}}2\sqrt 2 \cr & {\text{and }}\,{\sin ^{ - 1}}\frac{1}{{\sqrt 5 }} = {\cot ^{ - 1}}2 \cr} $$
If $$C$$ is the foot of the tower and $$h$$ is the height, then
$$AC = h \cdot 2\sqrt 2 ,CB = h \cdot 2,h\left( {2\sqrt 2 + 2} \right) = 50$$

$$ \Rightarrow h = 25\left( {\sqrt 2 - 1} \right)$$

$$\eqalign{ & r = \frac{\vartriangle }{s} = \frac{{\left( {\frac{1}{2}} \right) \cdot ac}}{{\left( {\frac{1}{2}} \right) \cdot \left( {a + b + c} \right)}} = \frac{{ac}}{{a + b + c}} = \frac{{ac\left( {c + a - b} \right)}}{{{{\left( {c + a} \right)}^2} - {b^2}}} = \frac{{ac\left( {c + a - b} \right)}}{{{c^2} + 2ca + {a^2} - {b^2}}} \cr & r = \frac{{ac\left( {c + a - b} \right)}}{{2ca + {b^2} - {b^2}}} = \frac{{c + a - b}}{2}\,\,\,\left( {\because \,\,{a^2} + {c^2} = {b^2}} \right). \cr} $$

$$\eqalign{ & {a^2}\sin 2B + {b^2}\sin 2A = 2{a^2}\sin B \cdot \cos B + 2{b^2}\sin A\cos A \cr & {a^2}\sin 2B + {b^2}\sin 2A = \frac{{{a^2}b}}{R}\cos B + \frac{{{b^2}a}}{R}\cos A \cr & {a^2}\sin 2B + {b^2}\sin 2A = \frac{{ab}}{R}\left( {a\cos B + b\cos A} \right) = \frac{{abc}}{R} = 2bc\sin A \cr & {a^2}\sin 2B + {b^2}\sin 2A = 4\left( {\frac{1}{2}bc\sin A} \right) = 4\lambda . \cr} $$