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

Here, $$BD : DC = c : b.$$
But $$BD + DC = a;$$
$$\therefore \,\,BD = \frac{c}{{b + c}} \cdot a.$$

$$\eqalign{ & {\text{In}}\,\,\vartriangle ABD,\frac{{BD}}{{\sin \frac{A}{2}}} = \frac{{AD}}{{\sin B}} \cr & \therefore \,\,AD = \frac{{ca}}{{b + c}} \cdot \frac{{\sin B}}{{\sin \frac{A}{2}}} = \frac{{2\vartriangle }}{{b + c}}{\text{cosec}}\frac{A}{2}. \cr & {\text{Also, }}\frac{{AI}}{{ID}} = \frac{{AB}}{{BD}} = \frac{c}{{\frac{{ca}}{{\left( {b + c} \right)}}}} = \frac{{b + c}}{a} \cr & \Rightarrow \,\,AI = \frac{{b + c}}{{a + b + c}} \cdot AD = \frac{\vartriangle }{s}{\text{cosec}}\frac{A}{2}. \cr} $$
Similarly for $$BI$$  and $$CI.$$

Let height of the tower be $$h\,m$$  and distance between tower and cliffbe $$x\,m.$$
$$\eqalign{ & \therefore CD = h,BD = x \cr & {\text{In }}\Delta \,ABD,\tan {45^ \circ } = \frac{{AB}}{{BD}} \cr & {\text{or }}1 = \frac{{50}}{x} \cr & x = 50\,\,\,\,.....\left( {\text{i}} \right) \cr} $$

$$\eqalign{ & {\text{In }}\Delta \,AEC \cr & \tan {30^ \circ } = \frac{{AE}}{{EC}} = \frac{{AB - EB}}{{EC}} = \frac{{AB - DC}}{{BD}}\left( {\because EB = DC,EC = BD} \right) \cr & \frac{1}{{\sqrt 3 }} = \frac{{50 - h}}{x}{\text{ or }}x = 50\sqrt 3 - h\sqrt 3 \cr & {\text{or }}50 = 50\sqrt 3 - h\sqrt 3 {\text{ or }}h\sqrt 3 = 50\sqrt 3 - 50 \cr & {\text{or }}h = \frac{{50\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 }} = 50\left( {1 - \frac{1}{{\sqrt 3 }}} \right) \cr & \therefore h = 50\left( {1 - \frac{{\sqrt 3 }}{3}} \right) \cr} $$

$$\eqalign{ & {A_0}{A_1} = 2 \times 1 \times \cos {60^ \circ } = 1 = {A_1}{A_2} \cr & \cos {120^ \circ } = \frac{{{A_0}A_1^2 + {A_1}A_2^2 - {A_0}A_2^2}}{{2 \cdot {A_0}{A_1} \cdot {A_1}{A_2}}} = \frac{{1 + 1 - {A_0}A_2^2}}{{2 \cdot 1 \cdot 1}} \cr & \therefore \,\,{A_0}{A_2} = \sqrt 3 = {A_0}{A_4} \cr & \therefore \,\,{A_0}{A_1} \times {A_0}{A_2} \times {A_0}{A_4} = 1 \times \sqrt 3 \times \sqrt 3 = 3. \cr} $$

$$\eqalign{ & 2s\left( {2s - 2c} \right) = ab\,\,\,{\text{or, }}\frac{{s\left( {s - c} \right)}}{{ab}} = \frac{1}{4} \cr & {\text{or, }}{\cos ^2}\frac{C}{2} = \frac{1}{4}\,\,\,{\text{or,}}\,\,\cos \frac{C}{2} = \frac{1}{2}\,\,\,\left( {\because \,\,\,\frac{C}{2}\,{\text{must be acute}}} \right). \cr} $$

$$\eqalign{ & \cos A + \cos B + \cos C > 1 \cr & \Rightarrow \,\,1 + 4\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} > 1 \cr & \therefore \,\,\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} > 0 \cr & \Rightarrow \,\,\frac{A}{2},\frac{B}{2},\frac{C}{2}\,\,{\text{are acute}}{\text{.}} \cr} $$

Using, $$\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}},\frac{1}{2} = \frac{{13 + {c^2} - 11}}{{2 \cdot \sqrt {13} \cdot c}}$$
So, $${c^2} = \sqrt {13} \cdot c + 2 = 0.$$     This gives two values of $$c.$$

$$\eqalign{ & \left( {150 - h} \right)\cot {30^ \circ } = 60 \cr & \Rightarrow h = 150 - 20\sqrt 3 \,m \cr} $$

$$\frac{s}{R} = \frac{{\left( {a + b + c} \right)}}{{2R}} = \frac{a}{{2R}} + \frac{b}{{2R}} + \frac{c}{{2R}} = \sin A + \sin B + \sin C.$$

Let two sides of $$\Delta $$ be $$a$$ and $$b.$$
Then $$a + b = x$$   and $$ab = y$$
Also given $${x^2} - {c^2} = y,$$   where $$c$$ is the third side of $$\Delta .$$
$$\eqalign{ & \Rightarrow \,\,{\left( {a + b} \right)^2} - {c^2} = ab \cr & \Rightarrow \,\,{a^2} + {b^2} - {c^2} = - ab \cr & \Rightarrow \,\,\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} = - \frac{1}{2} \cr & \Rightarrow \,\,\cos c = - \frac{1}{2} \cr & \Rightarrow \,\,c = {120^ \circ } \cr} $$
$$\therefore \,\,\frac{r}{R} = \frac{\Delta }{s} \times \frac{{4\Delta }}{{abc}}$$     where $$\Delta =\,\,$$ area of triangle
$$\eqalign{ & \Rightarrow \,\,\frac{r}{R} = \frac{{4{\Delta ^2}}}{{\frac{{\left( {a + b + c} \right)}}{2}abc}} \cr & = \frac{{8 \times {{\left( {\frac{1}{2}ab\sin c} \right)}^2}}}{{\left( {a + b + c} \right)abc}} \cr & \Rightarrow \,\,\frac{{2{a^2}{b^2}{{\sin }^2}{{120}^ \circ }}}{{\left( {a + b + c} \right)abc}} \cr & = \frac{{2ab \times \frac{3}{4}}}{{\left( {x + c} \right)c}} \cr & = \frac{{3y}}{{2c\left( {x + c} \right)}} \cr} $$

$$\eqalign{ & 2\cos A \cdot \sin \,C = \sin \left( {C + A} \right) = \sin \,C \cdot \cos A + \cos \,C \cdot \sin \,A \cr & \Rightarrow \,\,\sin \left( {C - A} \right) = 0 \cr & \Rightarrow \,\,C = A. \cr} $$