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Question Number 179609 by Acem last updated on 31/Oct/22
Answered by mr W last updated on 31/Oct/22
Method I ((sin x)/(2 sin 15°))=((sin (x+15°))/1)=((sin x cos 15°+cos x sin 15°)/1) sin 30°+((2 sin^2 15°)/(tan x))=1 ((1−cos 30°)/(tan γ))=(1/2) tan x=2−(√3) ⇒x=15° β=360°−75°−(180°−2×15°)=135° ✓
$$\boldsymbol{{Method}}\:\boldsymbol{{I}} \\ $$$$\frac{\mathrm{sin}\:{x}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{15}°}=\frac{\mathrm{sin}\:\left({x}+\mathrm{15}°\right)}{\mathrm{1}}=\frac{\mathrm{sin}\:{x}\:\mathrm{cos}\:\mathrm{15}°+\mathrm{cos}\:{x}\:\mathrm{sin}\:\mathrm{15}°}{\mathrm{1}} \\ $$$$\mathrm{sin}\:\mathrm{30}°+\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{15}°}{\mathrm{tan}\:{x}}=\mathrm{1} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{30}°}{\mathrm{tan}\:\gamma}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:{x}=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{x}=\mathrm{15}° \\ $$$$\beta=\mathrm{360}°−\mathrm{75}°−\left(\mathrm{180}°−\mathrm{2}×\mathrm{15}°\right)=\mathrm{135}°\:\checkmark \\ $$
Commented by Acem last updated on 31/Oct/22
Great use of trigonometric formulas Thanks!
$${Great}\:{use}\:{of}\:{trigonometric}\:{formulas} \\ $$$${Thanks}! \\ $$
Commented by mr W last updated on 31/Oct/22
Commented by Acem last updated on 31/Oct/22
I′ll check it soon, thanks
$$\:{I}'{ll}\:{check}\:{it}\:{soon},\:{thanks} \\ $$
Answered by mr W last updated on 31/Oct/22
Commented by mr W last updated on 31/Oct/22
Method II set point on AB such that AE=CD, then ΔAED≡ΔCDB ⇒ED=DB, ∠ADE=∠CBD=x we have φ=x+15° φ+x=45° ⇒x+15°+x=45° ⇒x=15°, φ=30° ⇒β=135° (see Method I)
$$\boldsymbol{{Method}}\:\boldsymbol{{II}} \\ $$$${set}\:{point}\:{on}\:{AB}\:{such}\:{that}\:{AE}={CD}, \\ $$$${then}\:\Delta{AED}\equiv\Delta{CDB} \\ $$$$\Rightarrow{ED}={DB},\:\angle{ADE}=\angle{CBD}={x} \\ $$$${we}\:{have} \\ $$$$\phi={x}+\mathrm{15}° \\ $$$$\phi+{x}=\mathrm{45}° \\ $$$$\Rightarrow{x}+\mathrm{15}°+{x}=\mathrm{45}° \\ $$$$\Rightarrow{x}=\mathrm{15}°,\:\phi=\mathrm{30}° \\ $$$$\Rightarrow\beta=\mathrm{135}°\:\left({see}\:{Method}\:{I}\right) \\ $$
Commented by Acem last updated on 31/Oct/22
Hi friend, in other way i got the same angles′ size and with your this method i found three points didn′t get then. can you explain the following: 1) why DAE ∡= 15 2) why AED ∡ = CDB ∡ 3) why φ= x +15
$${Hi}\:{friend},\:{in}\:{other}\:{way}\:{i}\:{got}\:{the}\:{same}\:{angles}'\:{size} \\ $$$$\:{and}\:{with}\:{your}\:{this}\:{method}\:{i}\:{found}\:{three}\:{points} \\ $$$$\:{didn}'{t}\:{get}\:{then}.\:{can}\:{you}\:{explain}\:\:{the}\:{following}: \\ $$$$\left.\mathrm{1}\right)\:{why}\:{DAE}\:\measuredangle=\:\mathrm{15} \\ $$$$\left.\mathrm{2}\right)\:{why}\:{AED}\:\measuredangle\:=\:{CDB}\:\measuredangle \\ $$$$\left.\mathrm{3}\right)\:{why}\:\phi=\:{x}\:+\mathrm{15} \\ $$
Commented by mr W last updated on 31/Oct/22
1) AC=CB ⇒∠A=∠B=45° ∠DAE=45°−α=45°−30°=15°
$$\left.\mathrm{1}\right) \\ $$$${AC}={CB}\:\Rightarrow\angle{A}=\angle{B}=\mathrm{45}° \\ $$$$\angle{DAE}=\mathrm{45}°−\alpha=\mathrm{45}°−\mathrm{30}°=\mathrm{15}° \\ $$
Commented by mr W last updated on 31/Oct/22
2) AD=CB ∠EAD=∠DCB=15° AE=CD ⇒ΔEAD≡ΔDCB ⇒∠AED=∠CDB
$$\left.\mathrm{2}\right) \\ $$$${AD}={CB} \\ $$$$\angle{EAD}=\angle{DCB}=\mathrm{15}° \\ $$$${AE}={CD} \\ $$$$\Rightarrow\Delta{EAD}\equiv\Delta{DCB} \\ $$$$\Rightarrow\angle{AED}=\angle{CDB} \\ $$
Commented by mr W last updated on 31/Oct/22
3) ∠BED=∠EDA+∠EAD ⇒φ=x+15°
$$\left.\mathrm{3}\right) \\ $$$$\angle{BED}=\angle{EDA}+\angle{EAD} \\ $$$$\Rightarrow\phi={x}+\mathrm{15}° \\ $$
Commented by Acem last updated on 31/Oct/22
1st, I shouldn′t have asked about ∡DAE^( 15°) “someone forgot the story of AC, CB ” ≈^(x x) and that made me deem that the 2 triangles are differents 2nd, φ= 15+ x , this idea reflects an intense attention! through crowded lines. 3rd, Thanks for you! Also, you can check the 3rd method, all of them are good to have knowledge.
$$\mathrm{1}{st},\:{I}\:{shouldn}'{t}\:{have}\:{asked}\:{about}\:\measuredangle{DAE}^{\:\mathrm{15}°} \: \\ $$$$\:“{someone}\:{forgot}\:{the}\:{story}\:{of}\:{AC},\:{CB}\:''\:\:\overset{{x}\:{x}} {\approx}\: \\ $$$$\:{and}\:{that}\:{made}\:{me}\:{deem}\:{that}\:{the}\:\mathrm{2}\:{triangles}\: \\ $$$$\:{are}\:{differents} \\ $$$$\: \\ $$$$\mathrm{2}{nd},\:\phi=\:\mathrm{15}+\:{x}\:,\:{this}\:{idea}\:{reflects} \\ $$$$\:\:{an}\:{intense}\:{attention}!\:{through}\:{crowded}\:{lines}. \\ $$$$\: \\ $$$$\mathrm{3}{rd},\:{Thanks}\:{for}\:{you}!\: \\ $$$$ \\ $$$${Also},\:{you}\:{can}\:{check}\:{the}\:\mathrm{3}{rd}\:{method},\:{all}\:{of}\:{them} \\ $$$$\:{are}\:{good}\:{to}\:{have}\:{knowledge}.\: \\ $$$$ \\ $$
Answered by Acem last updated on 31/Oct/22
Method III:
$$\boldsymbol{{Method}}\:\boldsymbol{{III}}: \\ $$
Commented by Acem last updated on 31/Oct/22
Commented by Acem last updated on 31/Oct/22
ϕ= 60 ⇒ θ= β_1 = 60, θ_1 = 30 ⇒ ACD is equilateral triangle i.e. CD= DB ⇒ β_2 = 75 then 𝛃= 135°
$$\:\varphi=\:\mathrm{60}\:\Rightarrow\:\theta=\:\beta_{\mathrm{1}} =\:\mathrm{60},\:\theta_{\mathrm{1}} =\:\mathrm{30} \\ $$$$\Rightarrow\:{ACD}\:{is}\:{equilateral}\:{triangle}\:{i}.{e}.\:{CD}=\:{DB} \\ $$$$\Rightarrow\:\beta_{\mathrm{2}} =\:\mathrm{75}\:{then}\:\boldsymbol{\beta}=\:\mathrm{135}° \\ $$

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