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

Expression $$ = \frac{{{b^2} + {c^2} - {a^2}}}{{2abc}} + \frac{{{c^2} + {a^2} - {b^2}}}{{2abc}} + \frac{{{a^2} + {b^2} - {c^2}}}{{2abc}} = \frac{{{a^2} + {b^2} + {c^2}}}{{2abc}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left( {a + b + c} \right)}^2} - 2\left( {ab + bc + ca} \right)}}{{2abc}} = \frac{{{{11}^2} - 2 \cdot 38}}{{2 \cdot 40}} = \frac{9}{{16}}.$$

Let height of tower $$MN = h$$
In $$\Delta QMN$$   we have
$$\eqalign{ & \tan {30^ \circ } = \frac{{MN}}{{QM}} \cr & \therefore \,\,QM = \sqrt 3 \,h = MR\,\,\,.....\left( 1 \right) \cr & {\text{Now in }}\Delta MNP \cr & MN = PM\,\,\,\,\,\,\,.....\left( 2 \right) \cr & {\text{In }}\Delta PMQ\,\,{\text{we have :}} \cr & MP = \sqrt {{{\left( {200} \right)}^2} - {{\left( {\sqrt 3 \,h} \right)}^2}} \cr & \therefore \,\,{\text{From }}\left( 2 \right),{\text{ we get :}} \cr & \sqrt {{{\left( {200} \right)}^2} - {{\left( {\sqrt 3 \,h} \right)}^2}} = h \cr & \Rightarrow \,\,h = 100\,m \cr} $$

Here, $$a = 4k, b = 5k, c = 6k.$$
$$\eqalign{ & \therefore \,\,s = \frac{{15k}}{2}. \cr & \therefore \,\,\vartriangle = \sqrt {\frac{{15k}}{2}\left( {\frac{{15k}}{2} - 4k} \right)\left( {\frac{{15k}}{2} - 5k} \right)\left( {\frac{{15k}}{2} - 6k} \right)} = \frac{{15\sqrt 7 }}{4}{k^2}. \cr & {\text{But }}\,R = \frac{{abc}}{{4\vartriangle }} = \frac{{4k \cdot 5k \cdot 6k}}{{15\sqrt 7 {k^2}}} = \frac{8}{{\sqrt 7 }}k \cr & {\text{and }}\,r = \frac{\vartriangle }{s} = \frac{{15\sqrt 7 }}{4}{k^2} \cdot \frac{2}{{15k}} = \frac{{\sqrt 7 }}{2}k. \cr & \therefore \,\,\frac{R}{r} = \frac{{\frac{{8k}}{{\sqrt 7 }}}}{{\frac{{\sqrt 7 k}}{2}}} = \frac{{16}}{7}. \cr} $$

$$\eqalign{ & {\text{Given that}}\,{\text{ }}A > B \cr & {\text{and }}3\sin x - 4{\sin ^3}x - k = 0,\,\,\,\,\,\,0 < k < 1 \cr & \Rightarrow \,\,\sin \,3x = k \cr} $$
As $$A$$ and $$B$$ satisfy above eq. (given)
$$\eqalign{ & \therefore \,\,\sin \,3A\, = k,\,\,\sin \,3B = k \cr & \Rightarrow \,\sin \,3A - \sin \,3B = 0 \cr & \Rightarrow \,\,2\,\cos \,\frac{{3A + 3B}}{2}\sin \,\frac{{3A - 3B}}{2} = 0 \cr & \Rightarrow \,\,\cos \left( {\frac{{3A + 3B}}{2}} \right) = 0\,\,{\text{or }}\sin \left( {\frac{{3A - 3B}}{2}} \right) = 0 \cr & \Rightarrow \,\,\frac{{3A + 3B}}{2} = {90^ \circ }\,\,{\text{or }}\frac{{3A - 3B}}{2} = 0 \cr & \Rightarrow \,\,A + B = {60^ \circ }\,\,{\text{or }}A = B \cr} $$
But given that $$A > B,\,\,\therefore \,\,A \ne B$$
Thus, $$A + B = 60°$$
But $$A + B + C = 180°$$
⇒ $$C = 180° - 60° = 120°$$
∴ $$C = \frac{{2\pi }}{3}$$

Consider a triangle $$ABC.$$
Given, angles of a triangle are in the ratio $$4 : 1 : 1.$$
Angles are $$4x, x$$  and $$x.$$
$${\text{i}}{\text{.e}}{\text{., }}\angle A = 4x,\angle B = x,\angle C = x$$
Now, by angle sum property of $$\Delta ,$$ we have
$$\eqalign{ & \angle A + \angle B + \angle C = {180^ \circ } \cr & \Rightarrow 4x + x + x = {180^ \circ } \cr & \Rightarrow x = \frac{{{{180}^ \circ }}}{6} = {30^ \circ } \cr & \therefore \angle A = {120^ \circ },\angle B = {30^ \circ },\angle C = {30^ \circ } \cr} $$
We know, ratio of sides of $$\Delta \,ABC$$  is given by
$$\eqalign{ & \sin A:\sin B:\sin C = \sin {120^ \circ }:\sin {30^ \circ }:\sin {30^ \circ } \cr & = \frac{{\sqrt 3 }}{2}:\frac{1}{2}:\frac{1}{2} = \sqrt 3 :1:1 \cr} $$
Required ration $$ = \frac{{\sqrt 3 }}{{1 + 1 + \sqrt 3 }} = \frac{{\sqrt 3 }}{{2 + \sqrt 3 }}.$$

In the $$\Delta \,AOB,\angle \,AOB = {60^ \circ },$$     $$\angle \,OBA = \angle \,OAB$$    (since $$OA = OB = AB$$     radius of same circle).
∴ $$\Delta \,AOB$$   is a equilateral triangle. Let the height of tower is $$h$$

$$m,$$ Given distance between two points $$A$$ & $$B$$  lie on boundary of circular park, subtends an angle of 60° at the foot of the tower is $$AB$$  i.e. $$AB = a. A$$   tower $$OC$$  stands at the centre of a circular park. Angle of elevation of the top of the tower from $$A$$ and $$B$$ is 30°.
In $$\Delta \,OAC$$
$$\eqalign{ & \tan {30^ \circ } = \frac{h}{a} \cr & \Rightarrow \,\,\frac{1}{{\sqrt 3 }} = \frac{h}{a} \cr & \Rightarrow \,\,h = \frac{a}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & \cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \frac{{{a^2}{{\cos }^2}A - {a^2}}}{{2bc}} = \frac{{{a^2}\left( {{{\cos }^2}A - 1} \right)}}{{2bc}} < 0. \cr & {\text{So, }}A > \frac{\pi }{2}. \cr} $$

$$r \cdot {r_1} \cdot {r_2} \cdot {r_3} = \frac{\vartriangle }{s} \cdot \frac{\vartriangle }{{s - a}} \cdot \frac{\vartriangle }{{s - b}} \cdot \frac{\vartriangle }{{s - c}} = \frac{{{\vartriangle ^4}}}{{{\vartriangle ^2}}} = {\vartriangle ^2}.$$

In $$\Delta ABD,$$   applying Sine law, we get
$$\eqalign{ & \frac{{AD}}{{\sin \frac{\pi }{3}}} = \frac{x}{{\sin \alpha }} \cr & \Rightarrow \,\,AD = \frac{{\sqrt 3 x}}{{2\sin \alpha }}\,\,\,\,\,.....\left( 1 \right) \cr} $$
In $$\Delta ACD,$$   applying Sine law, we get
$$\eqalign{ & \frac{{AD}}{{\sin \frac{\pi }{4}}} = \frac{{3x}}{{\sin \beta }} \cr & \Rightarrow \,\,AD = \frac{{3x}}{{\sqrt 2 \sin \beta }}\,\,\,\,\,.....\left( 2 \right) \cr} $$
From (i) and (ii) $$\frac{{\sqrt 3 x}}{{2\sin \alpha }} = \frac{{3x}}{{\sqrt 2 \sin \beta }}$$
$$ \Rightarrow \,\,\frac{{\sin \alpha }}{{\sin \beta }} = \frac{1}{{\sqrt 6 }}$$

$$\eqalign{ & \frac{1}{2}pa = {\text{area}} = \frac{1}{2}bc \cdot \sin \frac{\pi }{4}; \cr & \therefore \,\,p = \frac{{bc}}{{a\sqrt 2 }}. \cr & {\text{Also, }}\frac{a}{{\sin \frac{{ - \pi }}{4}}} = \frac{b}{{\sin \frac{\pi }{8}}} = \frac{c}{{\sin \frac{{5\pi }}{8}}} \cr & \Rightarrow \,\,b = \sqrt 2 a\sin \frac{\pi }{8},c = \sqrt 2 a\sin \frac{{5\pi }}{8} \cr & \therefore \,\,p = \frac{{2{a^2} \cdot \sin \frac{\pi }{8} \cdot \sin \frac{{5\pi }}{8}}}{{a\sqrt 2 }} = \frac{a}{{\sqrt 2 }}\left( {\cos \frac{{4\pi }}{8} - \cos \frac{{6\pi }}{8}} \right). \cr} $$