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Question Number 164709 by Tawa11 last updated on 20/Jan/22

Answered by Tawa11 last updated on 20/Jan/22

Am getting 14.28 but answer at the back is 12.7 please help.

$$\mathrm{Am}\:\mathrm{getting}\:\:\mathrm{14}.\mathrm{28}\:\:\mathrm{but}\:\mathrm{answer}\:\mathrm{at}\:\mathrm{the}\:\mathrm{back}\:\mathrm{is}\:\:\:\mathrm{12}.\mathrm{7}\:\:\:\:\mathrm{please}\:\mathrm{help}. \\ $$

Commented by mr W last updated on 21/Jan/22

how did you get 14.28? maybe you are right. this isn′t hard to you i think.

$${how}\:{did}\:{you}\:{get}\:\mathrm{14}.\mathrm{28}?\:{maybe}\:{you}\:{are} \\ $$$${right}.\:{this}\:{isn}'{t}\:{hard}\:{to}\:{you}\:{i}\:{think}. \\ $$

Commented by mr W last updated on 21/Jan/22

OP_1 =(h/(tan 40°)) OP_2 =(h/(tan 35°)) P_1 P_2 =10 m OP_1 ^2 +P_1 P_2 ^2 =OP_2 ^2 ((h/(tan 40°)))^2 +10^2 =((h/(tan 35°)))^2 h=((10)/( (√((1/(tan^2 35°))−(1/(tan^2 40°))))))≈12.71m

$${OP}_{\mathrm{1}} =\frac{{h}}{\mathrm{tan}\:\mathrm{40}°} \\ $$$${OP}_{\mathrm{2}} =\frac{{h}}{\mathrm{tan}\:\mathrm{35}°} \\ $$$${P}_{\mathrm{1}} {P}_{\mathrm{2}} =\mathrm{10}\:{m} \\ $$$${OP}_{\mathrm{1}} ^{\mathrm{2}} +{P}_{\mathrm{1}} {P}_{\mathrm{2}} ^{\mathrm{2}} ={OP}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\left(\frac{{h}}{\mathrm{tan}\:\mathrm{40}°}\right)^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} =\left(\frac{{h}}{\mathrm{tan}\:\mathrm{35}°}\right)^{\mathrm{2}} \\ $$$${h}=\frac{\mathrm{10}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\mathrm{35}°}−\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\mathrm{40}°}}}\approx\mathrm{12}.\mathrm{71}{m} \\ $$

Commented by Tawa11 last updated on 21/Jan/22

Commented by Tawa11 last updated on 21/Jan/22

Sir, please is this diagram correct? Because the angle in the triangle is more than 180. I interchange the angles and got 14.28

$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{is}\:\mathrm{this}\:\mathrm{diagram}\:\mathrm{correct}? \\ $$$$\mathrm{Because}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{in}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{more}\:\mathrm{than}\:\mathrm{180}. \\ $$$$\mathrm{I}\:\mathrm{interchange}\:\mathrm{the}\:\mathrm{angles}\:\mathrm{and}\:\mathrm{got}\:\:\mathrm{14}.\mathrm{28} \\ $$

Commented by mr W last updated on 21/Jan/22

the pole stands on the ground! it doesn′t lie on the ground! all happens in 3D world, not in 2D.

$${the}\:{pole}\:{stands}\:{on}\:{the}\:{ground}! \\ $$$${it}\:{doesn}'{t}\:{lie}\:{on}\:{the}\:{ground}! \\ $$$${all}\:{happens}\:{in}\:\mathrm{3}{D}\:{world},\:{not}\:{in}\:\mathrm{2}{D}. \\ $$

Commented by mr W last updated on 21/Jan/22

you can′t solve if you don′t think in 3D, see diagram.

$${you}\:{can}'{t}\:{solve}\:{if}\:{you}\:{don}'{t}\:{think}\:{in} \\ $$$$\mathrm{3}{D},\:{see}\:{diagram}. \\ $$

Commented by mr W last updated on 21/Jan/22

Commented by Tawa11 last updated on 21/Jan/22

Ohh. Wow. great. I just know this now. God bless you sir.

$$\mathrm{Ohh}.\:\mathrm{Wow}.\:\mathrm{great}.\:\mathrm{I}\:\mathrm{just}\:\mathrm{know}\:\mathrm{this}\:\mathrm{now}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 21/Jan/22

The diagram now strange to me to find h.

$$\mathrm{The}\:\mathrm{diagram}\:\mathrm{now}\:\mathrm{strange}\:\mathrm{to}\:\mathrm{me}\:\mathrm{to}\:\mathrm{find}\:\:\mathrm{h}. \\ $$

Commented by Tawa11 last updated on 21/Jan/22

Sir, please find h. I still have more questions on this aspect. I will solve them all after your wirkings. I just know this now from you. God bless you sir. Even if it is short workings to get h. I will do the rest.

$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{find}\:\:\mathrm{h}. \\ $$$$\mathrm{I}\:\mathrm{still}\:\mathrm{have}\:\mathrm{more}\:\mathrm{questions}\:\mathrm{on}\:\mathrm{this}\:\mathrm{aspect}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{solve}\:\mathrm{them}\:\mathrm{all}\:\mathrm{after}\:\mathrm{your} \\ $$$$\mathrm{wirkings}.\:\mathrm{I}\:\mathrm{just}\:\mathrm{know}\:\mathrm{this}\:\mathrm{now}\:\mathrm{from}\:\mathrm{you}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{Even}\:\mathrm{if}\:\mathrm{it}\:\mathrm{is}\:\mathrm{short}\:\mathrm{workings}\:\mathrm{to}\:\mathrm{get}\:\mathrm{h}. \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{do}\:\mathrm{the}\:\mathrm{rest}. \\ $$

Commented by Tawa11 last updated on 21/Jan/22

God bless you sir. I understand better now. Am happy. I can solve the rest. That 3D you told me open my eye to drawings now. God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{understand}\:\mathrm{better}\:\mathrm{now}. \\ $$$$\mathrm{Am}\:\mathrm{happy}.\:\mathrm{I}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{rest}. \\ $$$$\mathrm{That}\:\:\mathrm{3D}\:\:\mathrm{you}\:\mathrm{told}\:\mathrm{me}\:\mathrm{open}\:\mathrm{my}\:\mathrm{eye}\:\mathrm{to}\:\mathrm{drawings}\:\mathrm{now}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by mr W last updated on 21/Jan/22

i thought you were even much further than the book with your answer 14.28m. as you can see above, the answer 12.7 m assumed the the student observes the top of the pole from the level of the ground, i.e. his eyes must directly lie on the ground. this is naturally not possible. in fact he can only observe when he also stands. his eyes have a height over the ground. if this height is 1.57 m, which is possible for a student with normal body size, then the height of the pole is actually 1.57+12.71=14.28m. i thought you had considered this. thus i didn′t say that your answer 14.28m is wrong.

$${i}\:{thought}\:{you}\:{were}\:{even}\:{much}\:{further}\: \\ $$$${than}\:{the}\:{book}\:{with}\:{your}\:{answer} \\ $$$$\mathrm{14}.\mathrm{28}{m}. \\ $$$${as}\:{you}\:{can}\:{see}\:{above},\:{the}\:{answer}\:\mathrm{12}.\mathrm{7} \\ $$$${m}\:{assumed}\:{the}\:{the}\:{student}\:{observes} \\ $$$${the}\:{top}\:{of}\:{the}\:{pole}\:{from}\:{the}\:{level} \\ $$$${of}\:{the}\:{ground},\:{i}.{e}.\:{his}\:{eyes}\:{must}\: \\ $$$${directly}\:{lie}\:{on}\:{the}\:{ground}.\:{this}\:{is} \\ $$$${naturally}\:{not}\:{possible}.\:{in}\:{fact}\:{he} \\ $$$${can}\:{only}\:{observe}\:{when}\:{he}\:{also}\:{stands}. \\ $$$${his}\:{eyes}\:{have}\:{a}\:{height}\:{over}\:{the}\:{ground}. \\ $$$${if}\:{this}\:{height}\:{is}\:\mathrm{1}.\mathrm{57}\:{m},\:{which}\:{is} \\ $$$${possible}\:{for}\:{a}\:{student}\:{with}\:{normal} \\ $$$${body}\:{size},\:{then}\:{the}\:{height}\:{of}\:{the}\:{pole} \\ $$$${is}\:{actually}\:\mathrm{1}.\mathrm{57}+\mathrm{12}.\mathrm{71}=\mathrm{14}.\mathrm{28}{m}.\: \\ $$$${i}\:{thought}\:{you}\:{had}\:{considered}\:{this}.\:{thus} \\ $$$${i}\:{didn}'{t}\:{say}\:{that}\:{your}\:{answer}\:\mathrm{14}.\mathrm{28}{m} \\ $$$${is}\:{wrong}. \\ $$

Commented by mr W last updated on 21/Jan/22

Commented by mr W last updated on 21/Jan/22

Commented by Tawa11 last updated on 21/Jan/22

Sir, I have used your 3D thinking to interprete all the complex diagrams in my book. I got everything. Am happy. God bless you sir.

$$\mathrm{Sir},\:\mathrm{I}\:\mathrm{have}\:\mathrm{used}\:\mathrm{your}\:\:\mathrm{3D}\:\:\mathrm{thinking}\:\mathrm{to}\:\mathrm{interprete}\:\mathrm{all}\:\mathrm{the}\:\mathrm{complex} \\ $$$$\mathrm{diagrams}\:\mathrm{in}\:\mathrm{my}\:\mathrm{book}.\:\mathrm{I}\:\mathrm{got}\:\mathrm{everything}.\:\mathrm{Am}\:\mathrm{happy}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 21/Jan/22

Wow. true sir. This is what I thought.

$$\mathrm{Wow}.\:\mathrm{true}\:\mathrm{sir}.\:\:\mathrm{This}\:\mathrm{is}\:\mathrm{what}\:\mathrm{I}\:\mathrm{thought}. \\ $$

Commented by Tawa11 last updated on 21/Jan/22

I understand everything better now sir. God bless you more. I really appreciate your time.

$$\mathrm{I}\:\mathrm{understand}\:\mathrm{everything}\:\mathrm{better}\:\mathrm{now}\:\mathrm{sir}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{more}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}. \\ $$

Commented by mr W last updated on 22/Jan/22

Q1. what is the answer if the student moves 10 m not due south, but due west?

$${Q}\mathrm{1}.\:{what}\:{is}\:{the}\:{answer}\:{if}\:{the}\:{student} \\ $$$${moves}\:\mathrm{10}\:{m}\:{not}\:{due}\:{south},\:{but} \\ $$$${due}\:{west}? \\ $$

Commented by mr W last updated on 22/Jan/22

Q2. what is the answer if the student moves 10 m not due south, but due south−west?

$${Q}\mathrm{2}.\:{what}\:{is}\:{the}\:{answer}\:{if}\:{the}\:{student} \\ $$$${moves}\:\mathrm{10}\:{m}\:{not}\:{due}\:{south},\:{but} \\ $$$${due}\:{south}−{west}? \\ $$

Commented by Tawa11 last updated on 22/Jan/22

Commented by Tawa11 last updated on 22/Jan/22

tan 40 = (h/x) , ∴ x = (h/(tan 40)) tan 35 = (h/(10 + x)) 10 + x = (h/(tan 35)) ∴ x = (h/(tan 35)) − 10 ∴ (h/(tan 40)) = (h/(tan 35)) − 10 ∴ h tan 35 = h tan 40 − 10 tan 40 tan 35 ∴ h(tan 40 − tan 35) = ((10 tan 40 tan 35)/1) ∴ h = ((10 tan 40 tan 35)/(tan 40 − tan 35)) = 42.30m

$$\mathrm{tan}\:\mathrm{40}\:\:\:\:=\:\:\:\frac{\mathrm{h}}{\mathrm{x}}\:,\:\:\:\:\:\: \\ $$$$\therefore\:\:\:\:\mathrm{x}\:\:\:=\:\:\:\frac{\mathrm{h}}{\mathrm{tan}\:\mathrm{40}} \\ $$$$\mathrm{tan}\:\mathrm{35}\:\:\:=\:\:\:\frac{\mathrm{h}}{\mathrm{10}\:\:\:+\:\:\:\mathrm{x}} \\ $$$$\mathrm{10}\:\:+\:\:\mathrm{x}\:\:\:\:=\:\:\:\frac{\mathrm{h}}{\mathrm{tan}\:\mathrm{35}} \\ $$$$\therefore\:\:\:\:\:\:\mathrm{x}\:\:\:\:=\:\:\:\frac{\mathrm{h}}{\mathrm{tan}\:\mathrm{35}}\:\:\:−\:\:\:\mathrm{10} \\ $$$$\therefore\:\:\:\:\:\:\:\frac{\mathrm{h}}{\mathrm{tan}\:\mathrm{40}}\:\:\:\:=\:\:\:\frac{\mathrm{h}}{\mathrm{tan}\:\mathrm{35}}\:\:\:−\:\:\:\mathrm{10} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\mathrm{h}\:\mathrm{tan}\:\mathrm{35}\:\:\:\:=\:\:\:\:\mathrm{h}\:\mathrm{tan}\:\mathrm{40}\:\:\:\:−\:\:\:\mathrm{10}\:\mathrm{tan}\:\mathrm{40}\:\mathrm{tan}\:\mathrm{35} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{h}\left(\mathrm{tan}\:\mathrm{40}\:\:−\:\:\mathrm{tan}\:\mathrm{35}\right)\:\:\:=\:\:\:\frac{\mathrm{10}\:\mathrm{tan}\:\mathrm{40}\:\mathrm{tan}\:\mathrm{35}}{\mathrm{1}} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{h}\:\:\:=\:\:\:\frac{\mathrm{10}\:\mathrm{tan}\:\mathrm{40}\:\mathrm{tan}\:\mathrm{35}}{\mathrm{tan}\:\mathrm{40}\:\:\:−\:\:\:\mathrm{tan}\:\mathrm{35}}\:\:\:\:=\:\:\:\mathrm{42}.\mathrm{30m} \\ $$

Commented by Tawa11 last updated on 22/Jan/22

Please check sir.

$$\mathrm{Please}\:\mathrm{check}\:\mathrm{sir}. \\ $$

Commented by mr W last updated on 22/Jan/22

h≈42.30 m

$${h}\approx\mathrm{42}.\mathrm{30}\:{m} \\ $$

Commented by Tawa11 last updated on 22/Jan/22

Wow, am correct sir. I have changed it. I did not press calculator directly before. God bless you sir.

$$\mathrm{Wow},\:\mathrm{am}\:\mathrm{correct}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{have}\:\mathrm{changed}\:\mathrm{it}.\:\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{press} \\ $$$$\mathrm{calculator}\:\mathrm{directly}\:\mathrm{before}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 22/Jan/22

Sir, also in Bearing and Distance. I should be working in 3D too??? Becsuse I will start soon.

$$\mathrm{Sir},\:\mathrm{also}\:\mathrm{in}\:\mathrm{Bearing}\:\mathrm{and}\:\mathrm{Distance}.\:\:\mathrm{I}\:\mathrm{should}\:\mathrm{be}\:\mathrm{working}\:\mathrm{in}\:\:\mathrm{3D}\:\:\mathrm{too}??? \\ $$$$\mathrm{Becsuse}\:\mathrm{I}\:\mathrm{will}\:\mathrm{start}\:\mathrm{soon}. \\ $$

Commented by mr W last updated on 22/Jan/22

what about Q2 from above?

$${what}\:{about}\:{Q}\mathrm{2}\:{from}\:{above}? \\ $$

Commented by Tawa11 last updated on 22/Jan/22

Commented by Tawa11 last updated on 22/Jan/22

Sir, please check my diagram first. Hahahaha.

$$\mathrm{Sir},\:\:\mathrm{please}\:\mathrm{check}\:\mathrm{my}\:\mathrm{diagram}\:\mathrm{first}.\:\:\mathrm{Hahahaha}. \\ $$

Commented by mr W last updated on 22/Jan/22

maybe correct. your diagram is not clear. you only need to draw a ground view.

$${maybe}\:{correct}.\:{your}\:{diagram}\:{is}\:{not} \\ $$$${clear}.\:{you}\:{only}\:{need}\:{to}\:{draw}\:{a}\:{ground} \\ $$$${view}. \\ $$

Commented by Tawa11 last updated on 22/Jan/22

Alright sir. Let me draw it.

$$\mathrm{Alright}\:\mathrm{sir}.\:\mathrm{Let}\:\mathrm{me}\:\mathrm{draw}\:\mathrm{it}. \\ $$

Commented by Tawa11 last updated on 22/Jan/22

Commented by mr W last updated on 22/Jan/22

you seem not to realise what is meant with “he moves 10 m due south−west”.

$${you}\:{seem}\:{not}\:{to}\:{realise}\:{what}\:{is} \\ $$$${meant}\:{with}\:``{he}\:{moves}\:\mathrm{10}\:{m}\:{due} \\ $$$${south}−{west}''. \\ $$

Commented by Tawa11 last updated on 22/Jan/22

tan 40 = (h/x) ∴ x = (h/(tan 40))

$$\mathrm{tan}\:\mathrm{40}\:\:\:=\:\:\:\frac{\mathrm{h}}{\mathrm{x}} \\ $$$$\therefore\:\:\:\:\mathrm{x}\:\:\:\:=\:\:\:\frac{\mathrm{h}}{\mathrm{tan}\:\mathrm{40}} \\ $$

Commented by mr W last updated on 22/Jan/22

wrong!

$${wrong}! \\ $$

Commented by Tawa11 last updated on 22/Jan/22

Ok sir. I will try again

$$\mathrm{Ok}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{try}\:\mathrm{again} \\ $$

Commented by Tawa11 last updated on 22/Jan/22

The trigonometry is hard for me sir. But is my diagram correct first?

$$\mathrm{The}\:\mathrm{trigonometry}\:\mathrm{is}\:\mathrm{hard}\:\mathrm{for}\:\mathrm{me}\:\mathrm{sir}. \\ $$$$\mathrm{But}\:\mathrm{is}\:\mathrm{my}\:\mathrm{diagram}\:\mathrm{correct}\:\mathrm{first}? \\ $$

Commented by Tawa11 last updated on 22/Jan/22

Sir, in your own diagram. How is TOP_2 = 90

$$\mathrm{Sir},\:\mathrm{in}\:\mathrm{your}\:\mathrm{own}\:\mathrm{diagram}.\:\mathrm{How}\:\mathrm{is}\:\:\mathrm{TOP}_{\mathrm{2}} \:\:\:=\:\:\:\mathrm{90} \\ $$

Commented by mr W last updated on 22/Jan/22

when the pole stands on the ground, that means it stands vertically, i.e. with a 90° angle to the horozontal. ∠TOP_1 =90°, ∠TOP_2 =90°, .... these are very basic things, i think.

$${when}\:{the}\:{pole}\:{stands}\:{on}\:{the}\:{ground}, \\ $$$${that}\:{means}\:{it}\:{stands}\:{vertically},\:{i}.{e}. \\ $$$${with}\:{a}\:\mathrm{90}°\:{angle}\:{to}\:{the}\:{horozontal}. \\ $$$$\angle{TOP}_{\mathrm{1}} =\mathrm{90}°,\:\angle{TOP}_{\mathrm{2}} =\mathrm{90}°,\:.... \\ $$$${these}\:{are}\:{very}\:{basic}\:{things},\:{i}\:{think}. \\ $$

Commented by Tawa11 last updated on 22/Jan/22

Commented by Tawa11 last updated on 22/Jan/22

Ohhh. I just see it now sir. God bless you.

$$\mathrm{Ohhh}.\:\mathrm{I}\:\mathrm{just}\:\mathrm{see}\:\mathrm{it}\:\mathrm{now}\:\mathrm{sir}.\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

Commented by Tawa11 last updated on 22/Jan/22

Is my last diagram interprete what you mean by 10 m due − south west sir?

$$\mathrm{Is}\:\mathrm{my}\:\mathrm{last}\:\mathrm{diagram}\:\mathrm{interprete}\:\mathrm{what}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{by}\:\:\mathrm{10}\:\mathrm{m}\:\:\mathrm{due}\:−\:\mathrm{south}\:\mathrm{west} \\ $$$$\mathrm{sir}? \\ $$

Commented by mr W last updated on 22/Jan/22

Commented by Tawa11 last updated on 22/Jan/22

I see sir. It look like my first diagram

$$\mathrm{I}\:\mathrm{see}\:\mathrm{sir}.\:\mathrm{It}\:\mathrm{look}\:\mathrm{like}\:\mathrm{my}\:\mathrm{first}\:\mathrm{diagram} \\ $$

Commented by mr W last updated on 22/Jan/22

what′s the result for h in Q2?

$${what}'{s}\:{the}\:{result}\:{for}\:{h}\:{in}\:{Q}\mathrm{2}? \\ $$

Commented by Tawa11 last updated on 22/Jan/22

C to the foot of the pole ((sin 10)/(10)) = ((sin 135)/y) ∴ y = ((10 sin 135)/(sin 10)) = 40.52 ∴ tan 35 = (h/(40.52)) ∴ h = 40.52 tan 35 ∴ h = 28.37m This will be wrong because I did not use 40 degrees in the question.

$$\mathrm{C}\:\:\mathrm{to}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pole} \\ $$$$\:\:\:\:\:\:\frac{\mathrm{sin}\:\mathrm{10}}{\mathrm{10}}\:\:\:\:=\:\:\:\:\frac{\mathrm{sin}\:\mathrm{135}}{\mathrm{y}} \\ $$$$\therefore\:\:\:\:\:\:\:\mathrm{y}\:\:\:=\:\:\:\frac{\mathrm{10}\:\mathrm{sin}\:\mathrm{135}}{\mathrm{sin}\:\mathrm{10}}\:\:\:\:=\:\:\:\:\mathrm{40}.\mathrm{52} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\mathrm{tan}\:\mathrm{35}\:\:\:\:=\:\:\:\:\frac{\mathrm{h}}{\mathrm{40}.\mathrm{52}} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{h}\:\:\:\:=\:\:\:\mathrm{40}.\mathrm{52}\:\mathrm{tan}\:\mathrm{35} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{h}\:\:\:\:=\:\:\:\mathrm{28}.\mathrm{37m} \\ $$$$ \\ $$$$\mathrm{This}\:\mathrm{will}\:\mathrm{be}\:\mathrm{wrong}\:\mathrm{because}\:\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{use}\:\:\mathrm{40}\:\mathrm{degrees}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}. \\ $$$$ \\ $$

Commented by mr W last updated on 23/Jan/22

i can′t help you to think and to understand in 3D space. one needs some ability for spatia immagination for such questions. ((h/(tan 40°))+10 cos 45°)^2 +(10 sin 45°)^2 =((h/(tan 35°)))^2 h≈32.2239 m

$${i}\:{can}'{t}\:{help}\:{you}\:{to}\:{think}\:{and}\:{to} \\ $$$${understand}\:{in}\:\mathrm{3}{D}\:{space}. \\ $$$${one}\:{needs}\:{some}\:{ability}\:{for}\:{spatia}\: \\ $$$${immagination}\:{for}\:{such}\:{questions}. \\ $$$$ \\ $$$$\left(\frac{{h}}{\mathrm{tan}\:\mathrm{40}°}+\mathrm{10}\:\mathrm{cos}\:\mathrm{45}°\right)^{\mathrm{2}} +\left(\mathrm{10}\:\mathrm{sin}\:\mathrm{45}°\right)^{\mathrm{2}} =\left(\frac{{h}}{\mathrm{tan}\:\mathrm{35}°}\right)^{\mathrm{2}} \\ $$$${h}\approx\mathrm{32}.\mathrm{2239}\:{m} \\ $$

Commented by Tawa11 last updated on 23/Jan/22

Thanks sir. I really appreciate. I will study more on 3D space. God bless you sir.

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{study}\:\mathrm{more}\:\mathrm{on}\:\:\mathrm{3D}\:\mathrm{space}.\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by ajfour last updated on 23/Jan/22

good pursual at least...

$${good}\:{pursual}\:{at}\:{least}... \\ $$

Commented by Tawa11 last updated on 23/Jan/22

Thanks sir. I learnt new thing.

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{learnt}\:\mathrm{new}\:\mathrm{thing}. \\ $$

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