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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

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

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

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

Is the application crashing. Then can you please report and we will fix it as soon as possible.

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

Commented by sandy_suhendra last updated on 07/Oct/16

why I can′t upload my picture from my HP′s galery? can someone help me?

$$\mathrm{why}\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{upload}\:\mathrm{my}\:\mathrm{picture}\:\mathrm{from}\:\mathrm{my}\:\mathrm{HP}'\mathrm{s}\:\mathrm{galery}? \\ $$$$\mathrm{can}\:\mathrm{someone}\:\mathrm{help}\:\mathrm{me}? \\ $$

Commented by tawakalitu last updated on 07/Oct/16

please help with the diagram and solution God bless you.

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

Commented by sandy_suhendra last updated on 09/Oct/16

Commented by sandy_suhendra last updated on 09/Oct/16

I′m really confused with this app. I can upload this my screenshoot. But I can′t upload my other screenshoot. Why?

$$\mathrm{I}'\mathrm{m}\:\mathrm{really}\:\mathrm{confused}\:\mathrm{with}\:\mathrm{this}\:\mathrm{app}.\:\mathrm{I}\:\mathrm{can} \\ $$$$\mathrm{upload}\:\mathrm{this}\:\mathrm{my}\:\mathrm{screenshoot}.\:\mathrm{But}\:\mathrm{I}\:\mathrm{can}'\mathrm{t} \\ $$$$\mathrm{upload}\:\mathrm{my}\:\mathrm{other}\:\mathrm{screenshoot}.\:\mathrm{Why}? \\ $$

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

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

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

Commented by sandy_suhendra last updated on 11/Oct/16

I think, we also need a symbol for composition function like (fog)(x) but without using ”o” (letter o). Maybe like (f•g)(x) but not a black spot. And why I can′t change my font size larger. It′s always 18 in default size. Thank′s

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

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

Ok. Will add a small circle. Default font size is 20sp. You can change to say 26. Tap on button with text ′default font size′. You will get a dialog box to enter new font size. We will test that functionality and fix it if needed. We will make required fixes in next 3−4 days.

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

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

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

Question for number 8300

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

Commented by tawakalitu last updated on 08/Oct/16

Thanks so much.

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

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

Your welcome

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

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

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

^• The plane containing above diagram is a horizontal plane through P and 2000m above the sea level ^• The peak A is above this plane and the peak B is below this plane. (a) Horizontal distance between the peaks A and B ∠A′PB′= 80−20=60 By cosine law (A′B′)^2 =(PA′)^2 +(PB′)^2 −2(PA′)(PB′) cos∠A′PB′ A′B′=(√((3)^2 +(1)^2 −2(3)(1) cos 60)) =(√((3)^2 +(1)^2 −2(3)(1)(1/2))) =(√(10−3))=(√7)≈2.65m (b) The peak A is above A′ , such that AA′⊥ PA′ and △PAA′ is right angled triangle ∠APA′=10 [Angle of elevation of A from P] ∠AA′P=90 PA′=3m tan(∠APA′)=((AA′)/(PA′)) tan10=((AA′)/3) ⇒AA′=3tan10 Height of A from sea level=2000+3tan10≈2000.53 −−−−−− B is below B′, such that △BB′P is right triangle with ∠BB′P=90 ∠BPB′=15 [Angle of depression of B from P] PB′=1 tan(∠BPB′)=((BB′)/(PB′)) tan15=((BB′)/1)⇒BB′=tan15 Height of B from sea level=2000−tan15≈1999.73 (c)The angle of elevation of A from B Draw a perpendicular from B to the vertical line through A meeting it A′′. △AA′′B is right triangle. ∠ABA′′ is angle of elevaion of A from B ∠AA′′B=90 AA′′=Height of A−Height of B =(2000+3tan10)−(2000−tan15) =3tan10+tan15 BA′′=Horizontal distance of A and B =(√7) tan(∠ABA′′)=((AA′′)/(BA′′)) =((3tan10+tan15)/(√7)) ∠ABA′′=tan^(−1) (((3tan10+tan15)/(√7)))≈16.76° (please confirm also this number)

$$\:^{\bullet} \mathrm{The}\:\mathrm{plane}\:\mathrm{containing}\:\mathrm{above}\:\mathrm{diagram} \\ $$$$\mathrm{is}\:\mathrm{a}\:\boldsymbol{\mathrm{horizontal}}\:\boldsymbol{\mathrm{plane}}\:\mathrm{through}\:\mathrm{P}\:\mathrm{and} \\ $$$$\mathrm{2000m}\:\mathrm{above}\:\mathrm{the}\:\mathrm{sea}\:\mathrm{level} \\ $$$$\:^{\bullet} \mathrm{The}\:\mathrm{peak}\:\mathrm{A}\:\mathrm{is}\:\mathrm{above}\:\mathrm{this}\:\mathrm{plane}\:\mathrm{and} \\ $$$$\:\:\:\:\mathrm{the}\:\mathrm{peak}\:\:\mathrm{B}\:\mathrm{is}\:\mathrm{below}\:\mathrm{this}\:\mathrm{plane}. \\ $$$$ \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Horizontal}\:\mathrm{distance}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{peaks}\:\mathrm{A}\:\mathrm{and}\:\:\mathrm{B} \\ $$$$\angle\mathrm{A}'\mathrm{PB}'=\:\mathrm{80}−\mathrm{20}=\mathrm{60} \\ $$$$\mathrm{By}\:\mathrm{cosine}\:\mathrm{law} \\ $$$$\left(\mathrm{A}'\mathrm{B}'\right)^{\mathrm{2}} =\left(\mathrm{PA}'\right)^{\mathrm{2}} +\left(\mathrm{PB}'\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{PA}'\right)\left(\mathrm{PB}'\right)\:\mathrm{cos}\angle\mathrm{A}'\mathrm{PB}'\: \\ $$$$\mathrm{A}'\mathrm{B}'=\sqrt{\left(\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{3}\right)\left(\mathrm{1}\right)\:\mathrm{cos}\:\mathrm{60}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\left(\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{3}\right)\left(\mathrm{1}\right)\left(\mathrm{1}/\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{10}−\mathrm{3}}=\sqrt{\mathrm{7}}\approx\mathrm{2}.\mathrm{65m} \\ $$$$ \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{peak}\:\mathrm{A}\:\mathrm{is}\:\mathrm{above}\:\mathrm{A}'\:,\:\mathrm{such}\:\mathrm{that}\:\mathrm{AA}'\bot\:\mathrm{PA}' \\ $$$$\mathrm{and}\:\:\bigtriangleup\mathrm{PAA}'\:\mathrm{is}\:\mathrm{right}\:\mathrm{angled}\:\mathrm{triangle} \\ $$$$\angle\mathrm{APA}'=\mathrm{10}\:\left[\mathrm{Angle}\:\mathrm{of}\:\mathrm{elevation}\:\mathrm{of}\:\mathrm{A}\:\mathrm{from}\:\mathrm{P}\right] \\ $$$$\angle\mathrm{AA}'\mathrm{P}=\mathrm{90} \\ $$$$\mathrm{PA}'=\mathrm{3m} \\ $$$$\:\mathrm{tan}\left(\angle\mathrm{APA}'\right)=\frac{\mathrm{AA}'}{\mathrm{PA}'}\: \\ $$$$\mathrm{tan10}=\frac{\mathrm{AA}'}{\mathrm{3}}\:\Rightarrow\mathrm{AA}'=\mathrm{3tan10} \\ $$$$\mathrm{Height}\:\mathrm{of}\:\mathrm{A}\:\mathrm{from}\:\mathrm{sea}\:\mathrm{level}=\mathrm{2000}+\mathrm{3tan10}\approx\mathrm{2000}.\mathrm{53} \\ $$$$−−−−−− \\ $$$$\mathrm{B}\:\mathrm{is}\:\mathrm{below}\:\mathrm{B}',\:\mathrm{such}\:\mathrm{that}\:\bigtriangleup\mathrm{BB}'\mathrm{P}\:\mathrm{is}\:\mathrm{right}\:\mathrm{triangle} \\ $$$$\mathrm{with}\:\angle\mathrm{BB}'\mathrm{P}=\mathrm{90} \\ $$$$\angle\mathrm{BPB}'=\mathrm{15}\:\:\:\left[\mathrm{Angle}\:\mathrm{of}\:\mathrm{depression}\:\mathrm{of}\:\mathrm{B}\:\mathrm{from}\:\mathrm{P}\right] \\ $$$$\:\mathrm{PB}'=\mathrm{1}\: \\ $$$$\mathrm{tan}\left(\angle\mathrm{BPB}'\right)=\frac{\mathrm{BB}'}{\mathrm{PB}'}\: \\ $$$$\mathrm{tan15}=\frac{\mathrm{BB}'}{\mathrm{1}}\Rightarrow\mathrm{BB}'=\mathrm{tan15}\: \\ $$$$\mathrm{Height}\:\mathrm{of}\:\mathrm{B}\:\mathrm{from}\:\mathrm{sea}\:\mathrm{level}=\mathrm{2000}−\mathrm{tan15}\approx\mathrm{1999}.\mathrm{73} \\ $$$$ \\ $$$$\left(\mathrm{c}\right)\mathrm{The}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{elevation}\:\mathrm{of}\:\mathrm{A}\:\mathrm{from}\:\mathrm{B} \\ $$$$\:\:\:\mathrm{Draw}\:\mathrm{a}\:\mathrm{perpendicular}\:\mathrm{from}\:\mathrm{B}\:\mathrm{to}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{vertical}} \\ $$$$\boldsymbol{\mathrm{line}}\:\boldsymbol{\mathrm{through}}\:\boldsymbol{\mathrm{A}}\:\mathrm{meeting}\:\mathrm{it}\:\mathrm{A}''. \\ $$$$\bigtriangleup\mathrm{AA}''\mathrm{B}\:\mathrm{is}\:\mathrm{right}\:\mathrm{triangle}. \\ $$$$\angle\mathrm{ABA}''\:\mathrm{is}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{elevaion}\:\mathrm{of}\:\mathrm{A}\:\mathrm{from}\:\mathrm{B} \\ $$$$\angle\mathrm{AA}''\mathrm{B}=\mathrm{90} \\ $$$$\mathrm{AA}''=\mathrm{Height}\:\mathrm{of}\:\mathrm{A}−\mathrm{Height}\:\mathrm{of}\:\mathrm{B} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{2000}+\mathrm{3tan10}\right)−\left(\mathrm{2000}−\mathrm{tan15}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3tan10}+\mathrm{tan15} \\ $$$$\mathrm{BA}''=\mathrm{Horizontal}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{7}} \\ $$$$\mathrm{tan}\left(\angle\mathrm{ABA}''\right)=\frac{\mathrm{AA}''}{\mathrm{BA}''} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3tan10}+\mathrm{tan15}}{\sqrt{\mathrm{7}}} \\ $$$$\angle\mathrm{ABA}''=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3tan10}+\mathrm{tan15}}{\sqrt{\mathrm{7}}}\right)\approx\mathrm{16}.\mathrm{76}°\:\left(\mathrm{please}\:\mathrm{confirm}\right. \\ $$$$\left.\mathrm{also}\:\mathrm{this}\:\mathrm{number}\right) \\ $$

Commented by tawakalitu last updated on 09/Oct/16

Thanks so much. it is correct.

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

Commented by tawakalitu last updated on 10/Oct/16

i really appreciate. thanks sir.

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

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