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

$$\eqalign{ & {\text{Given that, }}{r_1} = 2{r_2} = 3{r_3} \cr & \therefore \frac{\Delta }{{s - a}} = \frac{{2\Delta }}{{s - b}} = \frac{{3\Delta }}{{s - c}} = \frac{\Delta }{k} \cr & {\text{Then, }}s - a = k,s - b = 2k,s - c = 3k \cr & \Rightarrow 3s - \left( {a + b + c} \right) = 6k \cr & \Rightarrow s = 6k \cr & \therefore \frac{a}{5} = \frac{b}{4} = \frac{c}{3} = k \cr & {\text{Now, }}\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = \frac{5}{4} + \frac{4}{3} + \frac{3}{5} \cr & = \frac{{75 + 80 + 36}}{{60}} = \frac{{191}}{{60}} \cr} $$

From the question, $$\sin 3A = k,\sin 3B = k$$
$$\eqalign{ & \Rightarrow \,\,\sin 3A = \sin 3B \cr & \Rightarrow \,\,3A + 3B = \pi \cr & \Rightarrow \,\,3\left( {\pi - C} \right) = \pi . \cr} $$

\[\left| {\begin{array}{*{20}{c}} a&b&c \\ b&c&a \\ c&a&b \end{array}} \right| = \left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}} 1&1&1 \\ b&c&a \\ c&a&b \end{array}} \right|\]
\[\left| {\begin{array}{*{20}{c}} a&b&c \\ b&c&a \\ c&a&b \end{array}} \right| = \left( {a + b + c} \right)\left( {bc + ca + ab - {a^2} - {b^2} - {c^2}} \right)\]
\[\left| {\begin{array}{*{20}{c}} a&b&c \\ b&c&a \\ c&a&b \end{array}} \right| = - \left( {{a^3} + {b^3} + {c^3} - 3abc} \right)\]
\[\left| {\begin{array}{*{20}{c}} a&b&c \\ b&c&a \\ c&a&b \end{array}} \right| = - 8{R^3}\left( {{{\sin }^3}A + {{\sin }^3}B + {{\sin }^3}C - 3\sin A\sin B\sin C} \right)\]
\[\left| {\begin{array}{*{20}{c}} a&b&c \\ b&c&a \\ c&a&b \end{array}} \right| = 0,\,{\text{from the question}}{\text{.}}\]

According to sine rule,
$$\eqalign{ & \frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} \cr & \therefore \frac{a}{{\sin A}} = \frac{c}{{\sin C}} \cr & \Rightarrow \sin C = \frac{{c \cdot \sin A}}{a} = \frac{{2 \cdot \sin {{45}^ \circ }}}{{2\sqrt 2 }} \cr & = \frac{1}{{\sqrt 2 }} \cdot \frac{1}{{\sqrt 2 }} = \frac{1}{2} = \sin {30^ \circ } \cr & \therefore C = {30^ \circ } \cr} $$

$$\eqalign{ & \frac{{\sin A}}{{\sin C}} = \frac{{\sin \left( {A - B} \right)}}{{\sin \left( {B - C} \right)}} \cr & \Rightarrow \frac{{\sin \left( {B + C} \right)}}{{\sin \left( {A + B} \right)}} = \frac{{\sin \left( {A - B} \right)}}{{\sin \left( {B - C} \right)}} \cr & \Rightarrow {\sin ^2}B - {\sin ^2}C = {\sin ^2}A - {\sin ^2}B \cr} $$
$$ \Rightarrow {\sin ^2}A,{\sin ^2}B,{\sin ^2}C$$     and hence $${a^2},{b^2},{c^2}$$  are in A.P.

$$\eqalign{ & {\tan ^2}\frac{C}{2} = \frac{{\left( {s - a} \right)\left( {s - b} \right)}}{{s\left( {s - c} \right)}} = \frac{{\left( {9 + c - 10} \right)\left( {9 + c - 8} \right)}}{{\left( {9 + c} \right)\left( {9 - c} \right)}} = \frac{{{c^2} - 1}}{{81 - {c^2}}} \cr & \therefore \,\,\frac{7}{9} = \frac{{{c^2} - 1}}{{81 - {c^2}}} \cr & \Rightarrow \,\,c = 6. \cr} $$

Clearly, the roots of $${x^2} + \sqrt 2 x + 1 = 0$$    are nonreal complex. So, one root common implies both roots are common.
So, $$\frac{a}{1} = \frac{b}{{\sqrt 2 }} = \frac{c}{1} = k.$$
$$\therefore \,\,\cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} = \frac{{{k^2} + 2{k^2} - {k^2}}}{{2 \cdot k \cdot \sqrt 2 k}} = \frac{1}{{\sqrt 2 }}.$$

From figure, we can deduce
$$H = l\tan \alpha \cot \beta .$$

Given that $$A : B : C = 4 : 1 : 1$$
Let $$A = 4x, B = x, C = x$$
But $$A + B + C = 180°$$
⇒ $$4x + x + x = 180°$$
⇒ $$x = 30°$$
∴ $$A = 120°, B = 30°, C = 30°$$
By sine law, $$\frac{a}{{\sin {{120}^ \circ }}} = \frac{b}{{\sin {{30}^ \circ }}} = \frac{c}{{\sin {{30}^ \circ }}}$$
$$\eqalign{ & \Rightarrow \,\,\frac{a}{{\sqrt {\frac{3}{2}} }} = \frac{b}{{\frac{1}{2}}} = \frac{c}{{\frac{1}{2}}} \cr & \Rightarrow \,\,a:b:c = \sqrt 3 :1:1 \cr} $$
∴ Ratio of longest side to the perimeter
$$\eqalign{ & = \sqrt 3 :1 + 1 + \sqrt 3 \cr & = \sqrt 3 :2 + \sqrt 3 \cr} $$

By Sine law in $$\Delta ABC$$
$$\eqalign{ & \frac{x}{{\sin {{30}^ \circ }}} = \frac{a}{{\sin {{120}^ \circ }}} \cr & \Rightarrow \,\,a = x\sqrt 3 \cr} $$

$$\eqalign{ & \therefore \,\,\Delta = \frac{1}{2} \times x \times x\sin {120^ \circ } \cr & = \frac{{\sqrt 3 }}{4}{x^2} \cr & {\text{Also, }}\sqrt 3 = \frac{\Delta }{s} \cr & \Rightarrow \,\,\frac{{\left( {2x + a} \right)}}{2}\sqrt 3 \cr & = \frac{{\sqrt 3 }}{4}{x^2} \cr & \Rightarrow \,\,x = 2\left( {2 + \sqrt 3 } \right) \cr & \Rightarrow \,\,\Delta = \frac{{\sqrt 3 }}{4} \times 4\left( {4 + 3 + 4\sqrt 3 } \right) \cr & \Rightarrow \,\,\Delta = 7\sqrt 3 + 12\,\,{\text{sq}}{\text{. units}}{\text{.}} \cr} $$