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

$$\cot A,\cot B,\cot C$$    are in A.P.
$$ \Rightarrow \,\,\frac{{{b^2} + {c^2} - {a^2}}}{{2bc \cdot \frac{a}{{2R}}}},\frac{{{c^2} + {a^2} - {b^2}}}{{2ca \cdot \frac{b}{{2R}}}},\frac{{{a^2} + {b^2} - {c^2}}}{{2ab \cdot \frac{c}{{2R}}}}$$        are in A.P.
$$ \Rightarrow \,\,{b^2} + {c^2} - {a^2},{c^2} + {a^2} - {b^2},{a^2} + {b^2} - {c^2}$$        are in A.P.
$$ \Rightarrow \,\, - 2{a^2}, - 2{b^2}, - 2{c^2}$$    are in A.P. (subtracting $${{a^2} + {b^2} + {c^2}}$$   from each).

$$\eqalign{ & {\text{Since, }}A + B + C = \pi \cr & \therefore \frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi }{2} \cr & \Rightarrow \frac{A}{2} + \frac{B}{2} = \frac{\pi }{2} - \frac{C}{2} \cr & \therefore \tan \left( {\frac{A}{2} + \frac{B}{2}} \right) = \tan \left( {\frac{\pi }{2} - \frac{C}{2}} \right) = \cot \frac{C}{2} \cr & \Rightarrow \frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2}\tan \frac{B}{2}}} = \frac{1}{{\tan \frac{C}{2}}} \cr & \Rightarrow \tan \frac{A}{2}\tan \frac{C}{2} + \tan \frac{B}{2}\tan \frac{C}{2} = 1 - \tan \frac{A}{2}\tan \frac{B}{2} \cr & \Rightarrow \tan \frac{A}{2}\tan \frac{B}{2} + \tan \frac{B}{2}\tan \frac{C}{2} + \tan \frac{C}{2}\tan \frac{A}{2} = 1 \cr} $$

From the given figure we have
$$\frac{H}{{3d}} = \tan \alpha \,\,{\text{and }}\frac{H}{d} = \tan \beta $$

$$\eqalign{ & \tan \left( {\beta - \alpha } \right) = \frac{1}{2} = \frac{{\frac{H}{d} - \frac{H}{{3d}}}}{{1 + \frac{{{H^2}}}{{3{d^2}}}}} \cr & \Rightarrow {H^2} - 4dH + 3{d^2} = 0 \cr & \Rightarrow {H^2} - 80H + 3\left( {400} \right) = 0 \cr & \Rightarrow H = 20{\text{ and}}\,60\,m \cr} $$

$$\eqalign{ & A = \pi - \frac{\pi }{4} - \frac{\pi }{6} = {105^ \circ }. \cr & {\text{So, }}\frac{{\sqrt 3 + 1}}{{\sin {{105}^ \circ }}} = \frac{c}{{\sin \frac{\pi }{6}}} \cr & \Rightarrow \,\,c = \frac{{\frac{{\sqrt 3 + 1}}{2}}}{{\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}}} = \sqrt 2 \cr & \therefore \,\,{\text{area}} = \frac{1}{2}ac\sin B = \frac{1}{2} \cdot \left( {\sqrt 3 + 1} \right)\sqrt 2 \cdot \sin \frac{\pi }{4}. \cr} $$

$$\eqalign{ & 3\left( {{{\tan }^2}A + {{\tan }^2}B + {{\tan }^2}C} \right) - {\left( {\tan A + \tan B + \tan C} \right)^2} = {\left( {\tan A - \tan B} \right)^2} + {\left( {\tan B - \tan C} \right)^2} + {\left( {\tan C - \tan A} \right)^2} > 0\,\,{\text{for here }}\tan A = \tan B = \tan C = \sqrt 3 \,\,{\text{is not true}} \cr & {\text{or, }}3k - {\left( {\tan A \cdot \tan B \cdot \tan C} \right)^2} > 0\,\,{\text{because in the }}\vartriangle ABC,\tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C \cr & {\text{or, }}3k - 81 > 0\,\,\,{\text{or, }}k > 27. \cr} $$

Let area of triangle be $$\Delta ,$$ then according to question,
$$\eqalign{ & \Delta = \frac{1}{2}ax = \frac{1}{2}by = \frac{1}{2}cz \cr & \therefore \frac{{bx}}{c} + \frac{{cy}}{a} + \frac{{az}}{b} \cr & = \frac{b}{c}\left( {\frac{{2\Delta }}{a}} \right) + \frac{c}{a}\left( {\frac{{2\Delta }}{b}} \right) + \frac{a}{b}\left( {\frac{{2\Delta }}{c}} \right) \cr & = \frac{{2\Delta \left( {{b^2} + {c^2} + {a^2}} \right)}}{{abc}} \cr & = \frac{{2\left( {{a^2} + {b^2} + {c^2}} \right)}}{{abc}} \cdot \frac{{abc}}{{4R}} = \frac{{{a^2} + {b^2} + {c^2}}}{{2R}} \cr} $$

The foot of the pole is at the centroid. Because centroid is the point of intersection of medians $$AD, BE$$   and $$CF,$$  which are the lines joining a vertex with the mid point of opposite side.

$$\eqalign{ & b\,{\cos ^2}\frac{C}{2} + c\,{\cos ^2}\frac{B}{2} = \frac{b}{2}\left( {1 + \cos C} \right) + \frac{c}{2}\left( {1 + \cos B} \right) \cr & b\,{\cos ^2}\frac{C}{2} + c\,{\cos ^2}\frac{B}{2} = \frac{b}{2} + \frac{c}{2} + \frac{1}{2}\left( {b\,\cos C + c\,\cos B} \right) = \frac{{a + b + c}}{2} = \frac{k}{2}. \cr} $$

$$\eqalign{ & AB = h\left( {{\text{height of the tower}}} \right) \cr & BD = 36\,m;\,\,BC = 49\,m \cr & \angle D = {47^ \circ };\,\,\angle C = {43^ \circ } \cr} $$

$$\eqalign{ & {\text{Now, in }}\Delta \,ABD, \cr & \tan {47^ \circ } = \frac{h}{{36\,m}}\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{and in }}\Delta \,ABC,\tan {43^ \circ } = \frac{h}{{49\,m}} \cr & \tan \left( {{{90}^ \circ } - {{47}^ \circ }} \right) = \frac{h}{{49}} \cr & \therefore \cot {47^ \circ } = \frac{h}{{49}}\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Multiplying equations (i) and (ii)
$$\eqalign{ & \tan {47^ \circ } \cdot \cot {47^ \circ } = \frac{h}{{36}} \times \frac{h}{{49}} = 1 = \frac{{{h^2}}}{{36 \times 49}} \cr & h = 6 \times 7 = 42\,m \cr} $$
$$\therefore $$ Option $$\left( {B} \right)$$  is correct.

Given,
$$\eqalign{ & 1.\sin A + \sin B = \sin C \cr & a + b = c \cr & \left( {\therefore {\text{By sine law}},\frac{{\sin A}}{a} = \frac{{\sin B}}{b} = \frac{{\sin C}}{c} = k} \right) \cr} $$
Here, the sum of two sides of $$\Delta \,ABC$$   is equal to the third side, but it is not possible
(Because by triangle inequality, the sum of the length of two sides of a triangle is always greater than the length of the third side)
$$\boxed{a + b > c}$$
$$2.$$ Ratio of angles of a triangle
$$\eqalign{ & A:B:C = 1:2:3 \cr & A + B + C = {180^ \circ } \cr & \therefore A = {30^ \circ } \cr & B = {60^ \circ } \cr & C = {90^ \circ } \cr} $$
the ratio in sides according to sine rule
$$\eqalign{ & a:b:c = \sin A:\sin B:\sin C \cr & = \sin {30^ \circ }:\sin {60^ \circ }:\sin {90^ \circ } \cr & = \frac{1}{2},\frac{{\sqrt 3 }}{2},1 = \frac{1}{2}:\frac{{\sqrt 3 }}{2}:1 \cr} $$