Properties of Triangles

Let the angles of a triangle be $x, 2 x$ and $7 x$.

Then, $x+2 x+7 x=180^{\circ} \Rightarrow x=18^{\circ}$

So, angles are $A=18^{\circ}, B=36^{\circ}$ and $C=126^{\circ}$

The least side is $a$ and the greatest side is $c$.

$ \begin{array}{rlrl}&\frac{a}{\sin A} =\frac{c}{\sin C} \\ \Rightarrow \quad & \frac{c}{a} =\frac{\sin C}{\sin A}=\frac{\sin 126^{\circ}}{\sin 18^{\circ}} \\ & =\frac{\sin \left(180^{\circ}-54^{\circ}\right)}{\sin 18^{\circ}}=\frac{\sin 54^{\circ}}{\sin 18^{\circ}} \\ & =\frac{\cos 36^{\circ}}{\sin 18^{\circ}}=\frac{\frac{(\sqrt{5}+1)}{4}}{\frac{(\sqrt{5}-1)}{4}}=\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{k+1}{k-1} \\ \therefore \quad k & =\sqrt{5} \end{array} $

In a triangle, $A+B+C=180^{\circ}$

Let $\tan \frac{A}{2}=x, \tan \frac{B}{2}=y$ and $\tan \frac{C}{2}=z$

Using the identity:

$ \begin{aligned} & \tan \frac{A}{2} \cdot \tan \frac{B}{2}+\tan \frac{B}{2} \cdot \tan \frac{C}{2}+\tan \frac{C}{2} \cdot \tan \frac{A}{2}=1 \\ \Rightarrow & \frac{2}{5} \cdot \frac{3}{7}+\frac{3}{7} \cdot z+z \cdot \frac{2}{5}=1 \\ \Rightarrow & \frac{6}{35}+\frac{15}{35} z+\frac{14}{35} z=1 \Rightarrow \frac{29}{35} z=\frac{29}{35} \\ \Rightarrow & z=1 \\ \Rightarrow & \text { Thus, } \tan \frac{C}{2}=1 \end{aligned} $

When $ A, B, $ and $ C $ are the angles of a triangle, we have $ A+B+C = \pi $. Thus, the following identity holds:

$ \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 $

Given that $ \tan \frac{A}{2} = \frac{1}{3} $ and $ \tan \frac{B}{2} = \frac{2}{3} $, let $ \tan \frac{C}{2} = x $.

Substitute these values into the equation:

$ \frac{1}{3} \times \frac{2}{3} + \frac{2}{3} \times x + x \times \frac{1}{3} = 1 $

Simplifying, we have:

$ \frac{2}{9} + \frac{2}{3}x + \frac{x}{3} = 1 $

Combine the $ x $ terms:

$ \frac{2}{9} + \frac{3x}{3} = 1 $

$ \frac{2}{9} + x = 1 $

Solving for $ x $:

$ x = 1 - \frac{2}{9} = \frac{9}{9} - \frac{2}{9} = \frac{7}{9} $

Therefore, $ \tan \frac{C}{2} = \frac{7}{9} $.

According to question,

Let $ D E=h $ and $ C D=D B=x $

In $ \triangle E B D, \quad \tan 60^{\circ}=\frac{h}{x} \Rightarrow x=\frac{h}{\sqrt{3}} $

Now, in $ \triangle A D E $

$ \tan 45^{\circ}=\frac{E D}{D A} $

$ \Rightarrow \quad D A=h $

In $ \triangle A B D $, by applying Pythagoras theorem

$ \Rightarrow\left(\frac{h}{\sqrt{3}}\right)^{2}+h^{2}=100^{2} $

$ \Rightarrow \quad \frac{4 h^{2}}{3}=10000 \Rightarrow h=50 \sqrt{3} $

$\begin{aligned} & \text { Given, } \frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)} \\ & \Rightarrow \quad \frac{\sin (B+C)}{\sin (A+C)}=\frac{\sin (A-C)}{\sin (C-B)} \\ & {[\because A+B+C=\pi \text { and } \sin (\pi-\theta)=\sin \theta]} \\ & \Rightarrow \quad \sin (B+C) \sin (C-B)=\sin (A+C) \sin (A-C) \\ & \Rightarrow \quad \sin ^2 C-\sin ^2 B=\sin ^2 A-\sin ^2 C \\ & \Rightarrow \quad 2 \sin ^2 C=\sin ^2 A+\sin ^2 B \\ & \Rightarrow \quad 2(2 R \sin C)^2=(2 R \sin A)^2+(2 R \sin B)^2 \\ & \Rightarrow \quad 2 c^2=a^2+b^2 \quad\left[\because \frac{a}{\sin A}=\frac{b}{\sin B}+\frac{c}{\sin C}=2 R\right] \\ & \Rightarrow \quad a^2, c^2, b^2 \text { are in AP. } \\ & \Rightarrow \quad b^2, c^2, a^2 \text { are in AP. } \end{aligned}$

We have,

${16^{{{\sin }^2}\theta }} + {16^{{{\cos }^2}\theta }} = 10$

$ \Rightarrow {16^{{{\sin }^2}\theta }} + {16^{1 - {{\sin }^2}\theta }} = 10$

$ \Rightarrow x + {{16} \over x} = 10$, where $x = {16^{{{\sin }^2}\theta }}$

$ \Rightarrow {x^2} - 10x + 16 = 0 \Rightarrow x = 2,8$

$\therefore$ ${16^{{{\sin }^2}\theta }} = 2,8 \Rightarrow {2^{4{{\sin }^2}\theta }} = 2,{2^3}$

$ \Rightarrow 4{\sin ^2}\theta = 1,3 \Rightarrow {\sin ^2}\theta = {1 \over 4},{\left( {{{\sqrt 3 } \over 2}} \right)^2}$

$ \Rightarrow \sin \theta = {1 \over 2},{{\sqrt 3 } \over 2} \Rightarrow \theta = {\pi \over 6},{\pi \over 3}$

Let the altitude of the sun be $\theta$. Then,

$\tan \theta = {h \over {h/\sqrt 3 }} \Rightarrow \tan \theta = \sqrt 3 $

$ \Rightarrow \theta = \pi /3 \Rightarrow \theta = 2\alpha $

Let PQ be the house subtending a right angle at the window B of opposite house AB

In $\Delta$PAB, we have

tan60$^\circ$ = $\frac{AB}{6}$ = AB = 6$\sqrt3$ m

In $\Delta$CBQ, we have

$\tan 30^\circ = {{h - CP} \over {BC}}$

$ \Rightarrow {1 \over {\sqrt 3 }} = {{h - 6\sqrt 3 } \over 6}$ [$\because$ $AB = CP,BC = AP$]

$ \Rightarrow h = 6\left( {\sqrt 3 + {1 \over {\sqrt 3 }}} \right)$

$ \Rightarrow h = 8\sqrt 3 $ m

${{\sin 3B} \over {\sin B}} = {{3\sin B - 4{{\sin }^3}B} \over {\sin B}}$

$ = 3 - 4{\sin ^2}B$

$ = 3 - 4(1 - {\cos ^2}B)$

$ = - 1 + {{4{{({a^2} + {c^2} - {b^2})}^2}} \over {4{{(ac)}^2}}}$

$ = - 1 + {{{{\left( {{{{a^2} + {c^2}} \over 2}} \right)}^2}} \over {{{(ac)}^2}}}$

$ = - 1 + {{{{({a^2} + {c^2})}^2}} \over {4{{(ac)}^2}}}$

$ = {\left( {{{{c^2} - {a^2}} \over {2ac}}} \right)^2}$

Let PQ be the chimney whose height is h meters.

In $\Delta$BPQ, $\tan 60^\circ = {{PQ} \over {BP}} \Rightarrow \sqrt 3 = {h \over x} \Rightarrow h = \sqrt 3 x$ ..... (i)

and in $\Delta$APQ, $\tan 30^\circ = {{PQ} \over {AP}}$

$ \Rightarrow {1 \over {\sqrt 3 }} = {h \over {50 + x}} \Rightarrow 50 + x = h\sqrt 3 $

$ \Rightarrow 50 + x = 3x \Rightarrow x = 25$ [using Eq. (i)]

$\therefore$ Height of the chimney = $25\sqrt 3 $ m

Let AB = h = height of vertical pole

OA = x = Horizontal distance from O to pole

In $\Delta$OAB

$\tan 45^\circ = {{AB} \over {OA}} = 1 = {{20} \over x} \Rightarrow x = 20$ m

In $\Delta$OPQ

$\tan 30^\circ = {{PQ} \over {OP}} \Rightarrow {1 \over {\sqrt 3 }} = {{20} \over {x + y}}$

$ \Rightarrow x + y = 20\sqrt 3 $

$ \Rightarrow y = 20\sqrt 3 - x = 20\sqrt 3 - 20 = 20(\sqrt 3 - 1)$

$\therefore$ Speed of bird is $20(\sqrt 3 - 1)$ m/s.