$$\eqalign{ & H = \left( {10\tan {{60}^ \circ } + 1.5} \right) \cr & = \left( {10\sqrt 3 + 1.5} \right)m \cr} $$

$$\eqalign{ & {\text{Since }}AP = 2AB \cr & \Rightarrow \,\,\frac{{AB}}{{AP}} = \frac{1}{2}\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{Let }}\angle APC = \alpha \cr & \tan \alpha = \frac{{AC}}{{AP}} \cr & = \frac{1}{2}\frac{{AB}}{{AP}} \cr & = \frac{1}{4} \cr & \left( {\because \,C\,\,{\text{is the mid point }}\therefore \,AC = \frac{1}{2}AB} \right) \cr & \Rightarrow \,\,\tan \alpha = \frac{1}{4} \cr} $$

$$\eqalign{ & {\text{As }}\tan \left( {\alpha + \beta } \right) = \frac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }} \cr & \Rightarrow \,\,\frac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }} \cr} $$
\[ = \frac{1}{2}\left[ \begin{gathered} \because \,\tan \left( {\alpha + \beta } \right) = \frac{{AB}}{{AP}} \hfill \\ \tan \left( {\alpha + \beta } \right) = \frac{1}{2}\left[ {{\text{From}}\left( 1 \right)} \right] \hfill \\ \end{gathered} \right]\]
$$\eqalign{ & \Rightarrow \,\,\frac{{\frac{1}{4} + \tan \beta }}{{1 - \frac{1}{4}\tan \beta }} = \frac{1}{2} \cr & \therefore \,\,\tan \beta = \frac{2}{9} \cr} $$

$$\eqalign{ & \frac{{2\sin P - 2\sin P\cos P}}{{2\sin P + 2\sin P\cos P}} \cr & = \frac{{1 - \cos P}}{{1 + \cos P}} = \frac{{2{{\sin }^2}\frac{P}{2}}}{{2{{\cos }^2}\frac{P}{2}}} \cr & = {\tan ^2}\frac{P}{2} = \frac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}} = \frac{{{{\left( {\left( {s - b} \right)\left( {s - c} \right)} \right)}^2}}}{{{\Delta ^2}}} \cr & = \frac{{{{\left( {\left( {\frac{1}{2}} \right)\left( {\frac{3}{2}} \right)} \right)}^2}}}{{{\Delta ^2}}} = {\left( {\frac{3}{{4\Delta }}} \right)^2} \cr} $$

We know by Sine law in $$\Delta ABC\,\,{\text{as}}$$
$$\eqalign{ & \frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin \left( {\pi - A - B} \right)}} = 2R \cr & \Rightarrow \,\,\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin \left( {A + B} \right)}} = 2R \cr} $$
(a) If we know $$a, \sin A, \sin B$$    we can find $$b, c ;$$  values of $$\angle 's\,A,B$$   and $$C$$ all.
(b) Using $$a, b, c$$  we can find $$\angle A,\angle B,\angle C$$   using cosine law.
(c) $$a, \sin B, R$$   are given then $$\sin A, b$$   and hence sin$$(A + B)$$  and then $$C$$ can be found.
(d) If we know $$a, \sin A , R$$   then we know only the ratio $$\frac{b}{{\sin B}} = \frac{c}{{\sin \left( {A + B} \right)}};$$     we can not determine the values of $$b, c, \sin B, \sin C$$    separately.
$$∴ \Delta $$ can not be determined in this case.

$$\sum {\frac{1}{{{r_1}}} = \sum {\frac{{s - a}}{\vartriangle } = \frac{{3s - \left( {a + b + c} \right)}}{\vartriangle }} = \frac{s}{\vartriangle }.} $$
∴ HM of exradii $$ = \frac{1}{{\frac{{\frac{{\sum 1 }}{{{r_1}}}}}{3}}} = \frac{3}{{\frac{s}{\vartriangle }}} = \frac{{3\vartriangle }}{s} = 3r.$$

If the circumradius of triangle $$ABC$$  be $$R,$$ then
$$R = \frac{a}{{2\sin A}} = \frac{b}{{2\sin B}} = \frac{c}{{2\sin C}}$$
where $$a, b, c$$  has their usual meanings.
Given $$\Delta \,ABC$$   is isoceles such that

$$AB = AC$$
Let circumradius be $$R,$$ then
$$\eqalign{ & R = \frac{{AC}}{{2\sin B}} = AB = AC \cr & \Rightarrow \frac{{AC}}{{2\sin B}} = AC\sin B = \frac{1}{2} \cr & \Rightarrow \sin B = \sin \frac{\pi }{6} \cr & \Rightarrow \angle B = \frac{\pi }{6} = \angle C \cr & {\text{we know that}}\,\angle A + \angle B + \angle C = {180^ \circ } = \pi \cr & \angle A + \frac{\pi }{6} + \frac{\pi }{6} = \pi \cr & \Rightarrow \angle A + \frac{\pi }{3} = \pi \cr & \Rightarrow \angle A = \pi - \frac{\pi }{3} = \frac{{2\pi }}{3} = \frac{{2 \times 180}}{3} \cr & \Rightarrow \angle A = {120^ \circ } \cr} $$

Let the speed be $$y\,\, m/sec.$$
Let $$AC$$  be the vertical pole of height $$20\, m.$$
Let $$O$$ be the point on the ground such that $$\angle \,AOC = {45^ \circ }$$
Let $$OC = x$$

$$\eqalign{ & {\text{Time }}t = 1\,{\text{s}} \cr & {\text{From }}\Delta \,AOC,\tan {45^ \circ } = \frac{{20}}{x}\,\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{and from }}\Delta \,BOD,\tan {30^ \circ } = \frac{{20}}{{x + y}}\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
From (i) and (ii), we have $$x = 20\,\,{\text{and }}\frac{1}{{\sqrt 3 }} = \frac{{20}}{{x + y}}$$
$$\eqalign{ & \Rightarrow \,\,\frac{1}{{\sqrt 3 }} = \frac{{20}}{{20 + y}} \cr & \Rightarrow \,\,20 + y = 20\sqrt 3 \cr} $$
$${\text{So, }}y = 20\left( {\sqrt 3 - 1} \right)$$     i.e., speed $$ = 20\left( {\sqrt 3 - 1} \right){\text{m}}\,{\text{/s}}$$

$$\eqalign{ & {\text{In }}\Delta \,ABC \cr & \frac{h}{x} = \tan {60^ \circ } = \sqrt 3 \cr & \Rightarrow \,\,x = \frac{h}{{\sqrt 3 }} \cr & {\text{In }}\Delta \,ABD\frac{h}{{x + 7}} \cr & = \tan {45^ \circ } = 1 \cr & \Rightarrow \,\,h = x + 7 \cr & \Rightarrow \,\,h - \frac{h}{{\sqrt 3 }} = 7 \cr & \Rightarrow \,\,h = \frac{{7\sqrt 3 }}{{\sqrt 3 - 1}} \times \frac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}} \cr & \Rightarrow \,\,h = \frac{{7\sqrt 3 }}{2}\left( {\sqrt 3 + 1} \right)m \cr} $$

$$\eqalign{ & \frac{{a\cos A + b\cos B + c\cos C}}{{a + b + c}} = \frac{{2R\sin A\cos A + 2R\sin B\cos B + 2r\sin C\cos C}}{{2s}} \cr & \frac{{a\cos A + b\cos B + c\cos C}}{{a + b + c}} = \frac{R}{{2s}} \cdot \left( {\sin 2A + \sin 2B + \sin 2C} \right) \cr & \frac{{a\cos A + b\cos B + c\cos C}}{{a + b + c}} = \frac{R}{{2s}} \cdot 4\sin A\sin B\sin C \cr & \frac{{a\cos A + b\cos B + c\cos C}}{{a + b + c}} = \frac{{2R}}{s} \cdot \frac{{abc}}{{8{R^3}}} = \frac{{abc}}{{4s{R^2}}}. \cr} $$
But, $$R = \frac{{abc}}{{4\vartriangle }},r = \frac{\vartriangle }{s}.$$    So, the value $$ = \frac{{4\vartriangle R}}{{4 \cdot \frac{\vartriangle }{r} \cdot {R^2}}} = \frac{r}{R}.$$

$$\eqalign{ & {\sin ^2}A + {\sin ^2}B + {\sin ^2}C \cr & = 1 - {\cos ^2}A + 1 - {\cos ^2}B + {\sin ^2}C \cr & = 2 - {\cos ^2}A - \cos \left( {B + C} \right)\cos \left( {B - C} \right) \cr & = 2 - \cos A\left[ {\cos A - \cos \left( {B - C} \right)} \right] \cr & = 2 - \cos A\left[ { - \cos \left( {B + C} \right) - \cos \left( {B - C} \right)} \right] \cr & = 2 + \cos A \cdot 2\cos B\cos C \cr & \therefore {\sin ^2}A + {\sin ^2}B + {\sin ^2}C - 2\cos A\cos B\cos C \cr & = 2 \cr} $$