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Question-226143

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Question Number 226143 by ajfour last updated on 20/Nov/25
Commented by fantastic2 last updated on 22/Nov/25
Tony: yes to all statements
$${Tony}:\:{yes}\:\:{to}\:{all}\:{statements} \\ $$
Commented by TonyCWX last updated on 22/Nov/25
Let me ask a few questions first before I attempt this. 1. Is straight line SP and RQ parallel to each other? 2. Is straight line SR and PQ parallel to each other? 3. Is straight line SP and PQ perpendicular to each other?
$${Let}\:{me}\:{ask}\:{a}\:{few}\:{questions}\:{first}\:{before}\:{I}\:{attempt}\:{this}. \\ $$$$\mathrm{1}.\:{Is}\:{straight}\:{line}\:{SP}\:{and}\:{RQ}\:{parallel}\:{to}\:{each}\:{other}? \\ $$$$\mathrm{2}.\:{Is}\:{straight}\:{line}\:{SR}\:{and}\:{PQ}\:{parallel}\:{to}\:{each}\:{other}? \\ $$$$\mathrm{3}.\:{Is}\:{straight}\:{line}\:{SP}\:{and}\:{PQ}\:{perpendicular}\:{to}\:{each}\:{other}? \\ $$
Commented by TonyCWX last updated on 22/Nov/25
Commented by TonyCWX last updated on 22/Nov/25
For a=10, b=6, k=(1/2) GE=(1/2)(6)=3 and EF=(1/2)(10)=5 As you can see, they are not necessarily parallel to each other. The value of angle may change with the position of point E.
$${For}\:{a}=\mathrm{10},\:{b}=\mathrm{6},\:{k}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${GE}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{6}\right)=\mathrm{3}\:{and}\:{EF}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{10}\right)=\mathrm{5} \\ $$$${As}\:{you}\:{can}\:{see},\:{they}\:{are}\:{not}\:{necessarily}\:{parallel}\:{to}\:{each}\:{other}. \\ $$$${The}\:{value}\:{of}\:{angle}\:{may}\:{change}\:{with}\:{the}\:{position}\:{of}\:{point}\:{E}. \\ $$
Commented by fantastic2 last updated on 22/Nov/25
see your picture and your previous statements 123
$${see}\:{your}\:{picture}\:{and}\:{your} \\ $$$${previous}\:{statements}\:\mathrm{123} \\ $$
Commented by TonyCWX last updated on 22/Nov/25
Are there any differences? If so, please point them out.
$${Are}\:{there}\:{any}\:{differences}? \\ $$$${If}\:{so},\:{please}\:{point}\:{them}\:{out}. \\ $$
Commented by fantastic2 last updated on 22/Nov/25
CG should be perpendicular to de otherwise they can't be parallel CG and EF can't be parallel
Commented by TonyCWX last updated on 22/Nov/25
But as you can see. The properties still hold.
$${But}\:{as}\:{you}\:{can}\:{see}. \\ $$$${The}\:{properties}\:{still}\:{hold}. \\ $$
Commented by fantastic2 last updated on 22/Nov/25
PQRS is a Rectangle
$${PQRS}\:{is}\:{a}\:{Rectangle} \\ $$
Commented by TonyCWX last updated on 22/Nov/25
I′ll just wait until Sir Ajfour comment.
$${I}'{ll}\:{just}\:{wait}\:{until}\:{Sir}\:{Ajfour}\:{comment}. \\ $$
Commented by fantastic2 last updated on 22/Nov/25
ok that looks like a fair option
$${ok}\:{that}\:{looks}\:{like}\:{a}\:{fair}\:{option} \\ $$
Commented by mr W last updated on 22/Nov/25
i guess ajfour sir has meant that both ABCD and SPQR are rectangle. otherwise there is no unique solution.
$${i}\:{guess}\:{ajfour}\:{sir}\:{has}\:{meant}\:{that} \\ $$$${both}\:{ABCD}\:{and}\:{SPQR}\:{are} \\ $$$${rectangle}.\:{otherwise}\:{there}\:{is}\:{no} \\ $$$${unique}\:{solution}. \\ $$
Commented by fantastic2 last updated on 22/Nov/25
i also think that.But lets see what ajfour sir says
$${i}\:{also}\:{think}\:{that}.{But}\:{lets}\:{see} \\ $$$${what}\:{ajfour}\:{sir}\:{says} \\ $$
Answered by fantastic2 last updated on 20/Nov/25
i get θ=cos^(−1) ((b/a))
$${i}\:{get}\:\theta=\mathrm{cos}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right) \\ $$
Answered by fantastic2 last updated on 20/Nov/25
Commented by fantastic2 last updated on 20/Nov/25
btan θ+kacos θ=a ⇒k=((a−btan θ)/(acos θ)) ka+kbtan θ=acos θ ⇒k=((acos θ)/(a+btan θ)) so ((a−btan θ)/(acos θ))=((acos θ)/(a+btan θ)) a^2 cos^2 θ=a^2 −b^2 tan^2 θ ⇒a^2 sin^2 θ=b^2 tan^2 θ ⇒cos θ=(b/a)
$${b}\mathrm{tan}\:\theta+{ka}\mathrm{cos}\:\theta={a} \\ $$$$\Rightarrow{k}=\frac{{a}−{b}\mathrm{tan}\:\theta}{{a}\mathrm{cos}\:\theta} \\ $$$${ka}+{kb}\mathrm{tan}\:\theta={a}\mathrm{cos}\:\theta \\ $$$$\Rightarrow{k}=\frac{{a}\mathrm{cos}\:\theta}{{a}+{b}\mathrm{tan}\:\theta} \\ $$$${so} \\ $$$$\frac{{a}−{b}\mathrm{tan}\:\theta}{{a}\mathrm{cos}\:\theta}=\frac{{a}\mathrm{cos}\:\theta}{{a}+{b}\mathrm{tan}\:\theta} \\ $$$${a}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \theta \\ $$$$\Rightarrow{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta={b}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \theta \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{{b}}{{a}} \\ $$
Commented by fantastic2 last updated on 22/Nov/25
Commented by fantastic2 last updated on 22/Nov/25
what i get ((kb)/(cos θ))+kasin θ=b ⇒k=((bcos θ)/(b+acos θsin θ)) btan θ+kacos θ=a ⇒k=((a−btan θ)/(acos θ)) kb+asin θ=(b/(sin θ)) ⇒k=((b−asin^2 θ)/(bsin θ)) kbtan θ+ka=acos θ ⇒k=((acos θ)/(a+btan θ))
$${what}\:{i}\:{get} \\ $$$$\frac{{kb}}{\mathrm{cos}\:\theta}+{ka}\mathrm{sin}\:\theta={b} \\ $$$$\Rightarrow{k}=\frac{{b}\mathrm{cos}\:\theta}{{b}+{a}\mathrm{cos}\:\theta\mathrm{sin}\:\theta} \\ $$$${b}\mathrm{tan}\:\theta+{ka}\mathrm{cos}\:\theta={a} \\ $$$$\Rightarrow{k}=\frac{{a}−{b}\mathrm{tan}\:\theta}{{a}\mathrm{cos}\:\theta} \\ $$$${kb}+{a}\mathrm{sin}\:\theta=\frac{{b}}{\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{k}=\frac{{b}−{a}\mathrm{sin}\:^{\mathrm{2}} \theta}{{b}\mathrm{sin}\:\theta} \\ $$$${kb}\mathrm{tan}\:\theta+{ka}={a}\mathrm{cos}\:\theta \\ $$$$\Rightarrow{k}=\frac{{a}\mathrm{cos}\:\theta}{{a}+{b}\mathrm{tan}\:\theta} \\ $$
Commented by fantastic2 last updated on 22/Nov/25
i was right
$${i}\:{was}\:{right} \\ $$

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