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Question Number 8314 by tawakalitu last updated on 07/Oct/16

Commented by Tinku Tara last updated on 08/Oct/16

$$\mathrm{For}\mathrm{othercases}\mathrm{can}\mathrm{u}\mathrm{please}\mathrm{email}$$$$\mathrm{picture}\mathrm{to}\mathrm{us}\mathrm{at}\mathrm{infoattinkutara}.\mathrm{com}$$$$\mathrm{and}\mathrm{we}\mathrm{will}\mathrm{troubleshoot}\mathrm{the}\mathrm{issue}.$$$$\mathrm{Alternatively}\mathrm{try}\mathrm{to}\mathrm{post}\mathrm{image}\mathrm{as}\mathrm{a}\mathrm{new}$$$$\mathrm{post}\mathrm{to}\mathrm{see}\mathrm{if}\mathrm{that}\mathrm{also}\mathrm{fails}.$$$$\mathrm{You}\mathrm{can}\mathrm{also}\mathrm{try}\mathrm{to}\mathrm{post}\mathrm{someother}$$$$\mathrm{image}\mathrm{so}\mathrm{we}\mathrm{know}\mathrm{if}\mathrm{any}\mathrm{type}\mathrm{of}\mathrm{image}$$$$\mathrm{if}\mathrm{failing}.$$$$\mathrm{Extra}\mathrm{information}\mathrm{will}\mathrm{greatly}\mathrm{help}$$$$\mathrm{in}\mathrm{troubleshooting}.$$

Commented by Tinku Tara last updated on 08/Oct/16

$$\mathrm{Is}\mathrm{the}\mathrm{application}\mathrm{crashing}.\mathrm{Then}$$$$\mathrm{can}\mathrm{you}\mathrm{please}\mathrm{report}\mathrm{and}\mathrm{we}\mathrm{will}$$$$\mathrm{fix}\mathrm{it}\mathrm{as}\mathrm{soon}\mathrm{as}\mathrm{possible}.$$

Commented by sandy_suhendra last updated on 07/Oct/16

$$\mathrm{why}\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{upload}\mathrm{my}\mathrm{picture}\mathrm{from}\mathrm{my}{\mathrm{HP}}^{\prime }\mathrm{s}\mathrm{galery}?$$$$\mathrm{can}\mathrm{someone}\mathrm{help}\mathrm{me}?$$

Commented by tawakalitu last updated on 07/Oct/16

$$\mathrm{please}\mathrm{help}\mathrm{with}\mathrm{the}\mathrm{diagram}\mathrm{and}\mathrm{solution}$$$$\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Commented by sandy_suhendra last updated on 09/Oct/16

Commented by sandy_suhendra last updated on 09/Oct/16

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{really}\mathrm{confused}\mathrm{with}\mathrm{this}\mathrm{app}.\mathrm{I}\mathrm{can}$$$$\mathrm{upload}\mathrm{this}\mathrm{my}\mathrm{screenshoot}.\mathrm{But}\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}$$$$\mathrm{upload}\mathrm{my}\mathrm{other}\mathrm{screenshoot}.\mathrm{Why}?$$

Commented by Tinku Tara last updated on 10/Oct/16

$$\mathrm{Thanks}\mathrm{sandy}\mathrm{for}\mathrm{the}\mathrm{addl}\mathrm{inputs}.$$$$\mathrm{We}\mathrm{are}\mathrm{currently}\mathrm{doing}\mathrm{some}\mathrm{testing}$$$$\mathrm{to}\mathrm{see}\mathrm{why}\mathrm{encoding}\mathrm{of}\mathrm{particular}$$$$\mathrm{image}\mathrm{is}\mathrm{failing}.$$$$\mathrm{We}\mathrm{will}\mathrm{update}\mathrm{u}\mathrm{soon}.$$

Commented by sandy_suhendra last updated on 11/Oct/16

$$\mathrm{I}\mathrm{think},\mathrm{we}\mathrm{also}\mathrm{need}\mathrm{a}\mathrm{symbol}\mathrm{for}\mathrm{composition}$$$$\mathrm{function}\mathrm{like}\left(\mathrm{fog}\right)\left(\mathrm{x}\right)\mathrm{but}\mathrm{without}\mathrm{using}{}^{″}{\mathrm{o}}^{″}\left(\mathrm{letter}\mathrm{o}\right).$$$$\mathrm{Maybe}\mathrm{like}(\mathrm{f}\bullet \mathrm{g})\left(\mathrm{x}\right)\mathrm{but}\mathrm{not}\mathrm{a}\mathrm{black}\mathrm{spot}.$$$$\mathrm{And}\mathrm{why}\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{change}\mathrm{my}\mathrm{font}\mathrm{size}\mathrm{larger}.{\mathrm{It}}^{\prime }\mathrm{s}\mathrm{always}18\mathrm{in}\mathrm{default}\mathrm{size}.{\mathrm{Thank}}^{\prime }\mathrm{s}$$

Commented by Tinku Tara last updated on 11/Oct/16

$$\mathrm{O}k.\mathrm{Will}\mathrm{add}\mathrm{a}\mathrm{small}\mathrm{circle}.$$$$\mathrm{Default}\mathrm{font}\mathrm{size}\mathrm{is}20\mathrm{sp}.\mathrm{You}\mathrm{can}$$$$\mathrm{change}\mathrm{to}\mathrm{say}26.\mathrm{Tap}\mathrm{on}\mathrm{button}$$$$\mathrm{with}\mathrm{text}{}^{\prime }\mathrm{default}\mathrm{font}{\mathrm{size}}^{\prime }.\mathrm{You}$$$$\mathrm{will}\mathrm{get}\mathrm{a}\mathrm{dialog}\mathrm{box}\mathrm{to}\mathrm{enter}\mathrm{new}\mathrm{font}\mathrm{size}.$$$$\mathrm{We}\mathrm{will}\mathrm{test}\mathrm{that}\mathrm{functionality}$$$$\mathrm{and}\mathrm{fix}\mathrm{it}\mathrm{if}\mathrm{needed}.$$$$\mathrm{We}\mathrm{will}\mathrm{make}\mathrm{required}\mathrm{fixes}\mathrm{in}\mathrm{next}$$$$3-4\mathrm{days}.$$

Answered by ridwan balatif last updated on 08/Oct/16

Commented by ridwan balatif last updated on 08/Oct/16

$$\mathrm{Question}\mathrm{for}\mathrm{number}8300$$

Commented by tawakalitu last updated on 08/Oct/16

$$\mathrm{Thanks}\mathrm{so}\mathrm{much}.$$

Commented by ridwan balatif last updated on 08/Oct/16

$$\mathrm{Your}\mathrm{welcome}$$

Answered by Rasheed Soomro last updated on 09/Oct/16

Commented by Rasheed Soomro last updated on 09/Oct/16

$${}^{\bullet }\mathrm{The}\mathrm{plane}\mathrm{containing}\mathrm{above}\mathrm{diagram}$$$$\mathrm{is}\mathrm{a}\mathbf{horizontal}\mathbf{plane}\mathrm{through}\mathrm{P}\mathrm{and}$$$$2000\mathrm{m}\mathrm{above}\mathrm{the}\mathrm{sea}\mathrm{level}$$$${}^{\bullet }\mathrm{The}\mathrm{peak}\mathrm{A}\mathrm{is}\mathrm{above}\mathrm{this}\mathrm{plane}\mathrm{and}$$$$\mathrm{the}\mathrm{peak}\mathrm{B}\mathrm{is}\mathrm{below}\mathrm{this}\mathrm{plane}.$$$$$$$$\left(\mathrm{a}\right)\mathrm{Horizontal}\mathrm{distance}\mathrm{between}$$$$\mathrm{the}\mathrm{peaks}\mathrm{A}\mathrm{and}\mathrm{B}$$$$\mathrm{\angle }{\mathrm{A}}^{\prime }{\mathrm{PB}}^{\prime }=80-20=60$$$$\mathrm{By}\mathrm{cosine}\mathrm{law}$$$${\left({\mathrm{A}}^{\prime }{\mathrm{B}}^{\prime }\right)}^2={\left({\mathrm{PA}}^{\prime }\right)}^2+{\left({\mathrm{PB}}^{\prime }\right)}^2-2\left({\mathrm{PA}}^{\prime }\right)\left({\mathrm{PB}}^{\prime }\right)\cos \mathrm{\angle }{\mathrm{A}}^{\prime }{\mathrm{PB}}^{\prime }$$$${\mathrm{A}}^{\prime }{\mathrm{B}}^{\prime }=\sqrt{{\left(3\right)}^2+{\left(1\right)}^2-2\left(3\right)\left(1\right)\cos 60}$$$$=\sqrt{{\left(3\right)}^2+{\left(1\right)}^2-2\left(3\right)\left(1\right)\left(1/2\right)}$$$$=\sqrt{10-3}=\sqrt{7}\approx 2.65\mathrm{m}$$$$$$$$\left(\mathrm{b}\right)\mathrm{The}\mathrm{peak}\mathrm{A}\mathrm{is}\mathrm{above}{\mathrm{A}}^{\prime },\mathrm{such}\mathrm{that}{\mathrm{AA}}^{\prime }\mathrm{\perp }{\mathrm{PA}}^{\prime }$$$$\mathrm{and}△{\mathrm{PAA}}^{\prime }\mathrm{is}\mathrm{right}\mathrm{angled}\mathrm{triangle}$$$$\mathrm{\angle }{\mathrm{APA}}^{\prime }=10\left[\mathrm{Angle}\mathrm{of}\mathrm{elevation}\mathrm{of}\mathrm{A}\mathrm{from}\mathrm{P}\right]$$$$\mathrm{\angle }{\mathrm{AA}}^{\prime }\mathrm{P}=90$$$${\mathrm{PA}}^{\prime }=3\mathrm{m}$$$$\tan \left(\mathrm{\angle }{\mathrm{APA}}^{\prime }\right)=\frac{{\mathrm{AA}}^{\prime }}{{\mathrm{PA}}^{\prime }}$$$$\tan 10=\frac{{\mathrm{AA}}^{\prime }}{3}\Rightarrow {\mathrm{AA}}^{\prime }=3\tan 10$$$$\mathrm{Height}\mathrm{of}\mathrm{A}\mathrm{from}\mathrm{sea}\mathrm{level}=2000+3\tan 10\approx 2000.53$$$$------$$$$\mathrm{B}\mathrm{is}\mathrm{below}{\mathrm{B}}^{\prime },\mathrm{such}\mathrm{that}△{\mathrm{BB}}^{\prime }\mathrm{P}\mathrm{is}\mathrm{right}\mathrm{triangle}$$$$\mathrm{with}\mathrm{\angle }{\mathrm{BB}}^{\prime }\mathrm{P}=90$$$$\mathrm{\angle }{\mathrm{BPB}}^{\prime }=15\left[\mathrm{Angle}\mathrm{of}\mathrm{depression}\mathrm{of}\mathrm{B}\mathrm{from}\mathrm{P}\right]$$$${\mathrm{PB}}^{\prime }=1$$$$\tan \left(\mathrm{\angle }{\mathrm{BPB}}^{\prime }\right)=\frac{{\mathrm{BB}}^{\prime }}{{\mathrm{PB}}^{\prime }}$$$$\tan 15=\frac{{\mathrm{BB}}^{\prime }}{1}\Rightarrow {\mathrm{BB}}^{\prime }=\tan 15$$$$\mathrm{Height}\mathrm{of}\mathrm{B}\mathrm{from}\mathrm{sea}\mathrm{level}=2000-\tan 15\approx 1999.73$$$$$$$$\left(\mathrm{c}\right)\mathrm{The}\mathrm{angle}\mathrm{of}\mathrm{elevation}\mathrm{of}\mathrm{A}\mathrm{from}\mathrm{B}$$$$\mathrm{Draw}\mathrm{a}\mathrm{perpendicular}\mathrm{from}\mathrm{B}\mathrm{to}\mathbf{the}\mathbf{vertical}$$$$\mathbf{line}\mathbf{through}\mathbf{A}\mathrm{meeting}\mathrm{it}{\mathrm{A}}^{″}.$$$$△{\mathrm{AA}}^{″}\mathrm{B}\mathrm{is}\mathrm{right}\mathrm{triangle}.$$$$\mathrm{\angle }{\mathrm{ABA}}^{″}\mathrm{is}\mathrm{angle}\mathrm{of}\mathrm{elevaion}\mathrm{of}\mathrm{A}\mathrm{from}\mathrm{B}$$$$\mathrm{\angle }{\mathrm{AA}}^{″}\mathrm{B}=90$$$${\mathrm{AA}}^{″}=\mathrm{Height}\mathrm{of}\mathrm{A}-\mathrm{Height}\mathrm{of}\mathrm{B}$$$$=(2000+3\tan 10)-(2000-\tan 15)$$$$=3\tan 10+\tan 15$$$${\mathrm{BA}}^{″}=\mathrm{Horizontal}\mathrm{distance}\mathrm{of}\mathrm{A}\mathrm{and}\mathrm{B}$$$$=\sqrt{7}$$$$\tan \left(\mathrm{\angle }{\mathrm{ABA}}^{″}\right)=\frac{{\mathrm{AA}}^{″}}{{\mathrm{BA}}^{″}}$$$$=\frac{3\tan 10+\tan 15}{\sqrt{7}}$$$$\mathrm{\angle }{\mathrm{ABA}}^{″}={\tan }^{-1}\left(\frac{3\tan 10+\tan 15}{\sqrt{7}}\right)\approx 16.76°(\mathrm{please}\mathrm{confirm}$$$$\mathrm{also}\mathrm{this}\mathrm{number})$$

Commented by tawakalitu last updated on 09/Oct/16

$$\mathrm{Thanks}\mathrm{so}\mathrm{much}.\mathrm{it}\mathrm{is}\mathrm{correct}.$$

Commented by tawakalitu last updated on 10/Oct/16

$$\mathrm{i}\mathrm{really}\mathrm{appreciate}.\mathrm{thanks}\mathrm{sir}.$$

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