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Question Number 218475 by lmcp1203 last updated on 10/Apr/25

Answered by Nicholas666 last updated on 10/Apr/25

•∠CBD + ∠BDC + ∠BCD =180° •x+140°+30°=180° •x+170°=180° •x=180°−170° •x=10°

$$\bullet\angle{CBD}\:+\:\angle{BDC}\:+\:\angle{BCD}\:=\mathrm{180}° \\ $$$$\bullet\boldsymbol{{x}}+\mathrm{140}°+\mathrm{30}°=\mathrm{180}° \\ $$$$\bullet\boldsymbol{{x}}+\mathrm{170}°=\mathrm{180}° \\ $$$$\bullet\boldsymbol{{x}}=\mathrm{180}°−\mathrm{170}° \\ $$$$\bullet\boldsymbol{{x}}=\mathrm{10}° \\ $$

Commented by mr W last updated on 10/Apr/25

how can you say ∠BDC=140° ? besides you have ignored the given condition AC=BD.

$${how}\:{can}\:{you}\:{say}\:\angle{BDC}=\mathrm{140}°\:? \\ $$$${besides}\:{you}\:{have}\:{ignored}\:{the}\:{given} \\ $$$${condition}\:{AC}={BD}. \\ $$

Commented by Nicholas666 last updated on 10/Apr/25

you have to look at the question carefully, it is very obvios

$${you}\:{have}\:{to}\:{look}\:{at}\:{the}\:{question}\:{carefully},\:{it}\:{is}\:{very}\:{obvios} \\ $$

Commented by mr W last updated on 10/Apr/25

i think you didn′t look carefully. when you took ∠BDC=140°, you assumed ∠DBA=40°. or how is it obvious that ∠BDC=140° ?

$${i}\:{think}\:{you}\:{didn}'{t}\:{look}\:{carefully}. \\ $$$${when}\:{you}\:{took}\:\angle{BDC}=\mathrm{140}°,\:{you} \\ $$$${assumed}\:\angle{DBA}=\mathrm{40}°.\: \\ $$$${or}\:{how}\:{is}\:{it}\:{obvious}\:{that}\:\angle{BDC}=\mathrm{140}°\:? \\ $$

Commented by Nicholas666 last updated on 10/Apr/25

I don′t really like typing at lenght,but okay, I′ll explain to you how I got 10° pay attention to the angle at A is 100° since lines AD and BD have the same sign (jagged lines), this show that triangle ABD is an isosceles triangle AD=BD in an isosceles triangle , the angles opposite the sides of the same lenght are equal. so,angle ABD=ADB calculate the angle ABD and ADB the sum of the angles in a triangle ix 180° in a triangle; ABD:∠BAD+∠ABD+∠ADB=180° 100°+∠ABD+∠ABD=180°(because∠ABD=∠ADB) 2×∠ABD=180°−100° 2×∠ABD=80° ∠ABD=40° so, ∠ADB+∠BDC=180° 40°+∠BDC=180°−40° ∠BDC=140° so,∠CBD+∠BDC+∠BCD x+140°+30°=180° x+170°=180° x=180°−170° x=10° the value of the angle is x=10°

$${I}\:{don}'{t}\:{really}\:{like}\:{typing}\:{at}\:{lenght},{but}\:{okay},\:{I}'{ll}\:{explain}\:{to}\:{you}\:{how}\:{I}\:{got}\:\mathrm{10}° \\ $$$$\:\:{pay}\:{attention}\:{to}\:{the}\:{angle}\:{at}\:{A}\:{is}\:\mathrm{100}° \\ $$$$\:\:{since}\:{lines}\:{AD}\:{and}\:{BD}\:{have}\:{the}\:{same}\:{sign} \\ $$$$\left({jagged}\:{lines}\right),\:{this}\:{show}\:{that}\:{triangle}\:{ABD} \\ $$$${is}\:{an}\:{isosceles}\:{triangle}\:{AD}={BD} \\ $$$$\:\:\:\:{in}\:{an}\:{isosceles}\:{triangle}\:, \\ $$$${the}\:{angles}\:{opposite}\:{the}\:{sides}\:\:{of}\:{the}\:{same}\:{lenght}\:{are}\:{equal}. \\ $$$$\:{so},{angle}\:{ABD}={ADB} \\ $$$$\:\:{calculate}\:{the}\:{angle}\:{ABD}\:{and}\:{ADB} \\ $$$$\:{the}\:{sum}\:{of}\:{the}\:{angles}\:{in}\:{a}\:{triangle}\:{ix}\:\mathrm{180}° \\ $$$${in}\:{a}\:{triangle};\:{ABD}:\angle{BAD}+\angle{ABD}+\angle{ADB}=\mathrm{180}° \\ $$$$\mathrm{100}°+\angle{ABD}+\angle{ABD}=\mathrm{180}°\left({because}\angle{ABD}=\angle{ADB}\right) \\ $$$$\mathrm{2}×\angle{ABD}=\mathrm{180}°−\mathrm{100}° \\ $$$$\mathrm{2}×\angle{ABD}=\mathrm{80}° \\ $$$$\angle{ABD}=\mathrm{40}° \\ $$$${so},\:\angle{ADB}+\angle{BDC}=\mathrm{180}° \\ $$$$\mathrm{40}°+\angle{BDC}=\mathrm{180}°−\mathrm{40}° \\ $$$$\angle{BDC}=\mathrm{140}° \\ $$$${so},\angle{CBD}+\angle{BDC}+\angle{BCD} \\ $$$${x}+\mathrm{140}°+\mathrm{30}°=\mathrm{180}° \\ $$$${x}+\mathrm{170}°=\mathrm{180}° \\ $$$${x}=\mathrm{180}°−\mathrm{170}° \\ $$$${x}=\mathrm{10}° \\ $$$${the}\:{value}\:{of}\:{the}\:{angle}\:{is}\:\boldsymbol{{x}}=\mathrm{10}° \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by mr W last updated on 11/Apr/25

sorry, not AD and BD have the same sign, but AC and BD have the same sign. that means AC=BD, not AD=BD as you said. besides, even if you′re right with AD=BD, then you should not get ∠ABD=∠ADB, but get ∠ABD=∠DAB=100°. that′s definitively not right. please recheck sir.

$${sorry},\:{not}\:{AD}\:{and}\:{BD}\:{have}\:{the} \\ $$$${same}\:{sign},\:{but}\:{AC}\:{and}\:{BD}\:{have} \\ $$$${the}\:{same}\:{sign}.\:{that}\:{means}\:{AC}={BD}, \\ $$$${not}\:{AD}={BD}\:{as}\:{you}\:{said}.\:\:{besides}, \\ $$$${even}\:{if}\:{you}'{re}\:{right}\:{with}\:{AD}={BD},\: \\ $$$${then}\:{you}\:{should}\:{not}\:{get}\: \\ $$$$\angle{ABD}=\angle{ADB},\:{but}\:{get}\: \\ $$$$\angle{ABD}=\angle{DAB}=\mathrm{100}°.\:{that}'{s}\: \\ $$$${definitively}\:{not}\:{right}. \\ $$$${please}\:{recheck}\:{sir}. \\ $$

Answered by lmcp1203 last updated on 10/Apr/25

thanks

$${thanks} \\ $$

Answered by vnm last updated on 10/Apr/25

AC=BD=a ((AB)/(sin 30°))=((AC)/(sin∠ABC)) AB=(a/(2sin 50°)) E is the base of the perpendicular drawn from B to the line AC BE=ABsin 80°=((asin 80°)/(2sin 50°))=BDsin∠BDA=asin∠BDA sin∠BDA=((sin 80°)/(2sin 50°))=sin ∠BDC ∠BDC=180°−sin^(−1) (sin∠BDA) x=∠DBC=180°−(∠DCB+∠BDC)= 180°−(30°+180°−sin^(−1) (((sin 80°)/(2sin 50°))))= sin^(−1) (((sin 80°)/(2sin 50°)))−30° ((sin 80°)/(2sin 50°))=((2sin 40°cos 40°)/(2sin(90°−40°)))=sin 40° x=sin^(−1) (sin 40°)−30°=10°

$$ \\ $$$${AC}={BD}={a} \\ $$$$\frac{{AB}}{\mathrm{sin}\:\mathrm{30}°}=\frac{{AC}}{\mathrm{sin}\angle{ABC}} \\ $$$${AB}=\frac{{a}}{\mathrm{2sin}\:\mathrm{50}°} \\ $$$$\mathrm{E}\:\mathrm{is}\:\mathrm{the}\:\mathrm{base}\:\mathrm{of}\:\mathrm{the}\:\mathrm{perpendicular}\: \\ $$$$\mathrm{drawn}\:\mathrm{from}\:\mathrm{B}\:\mathrm{to}\:\mathrm{the}\:\mathrm{line}\:\mathrm{AC} \\ $$$${BE}={AB}\mathrm{sin}\:\mathrm{80}°=\frac{{a}\mathrm{sin}\:\mathrm{80}°}{\mathrm{2sin}\:\mathrm{50}°}={BD}\mathrm{sin}\angle{BDA}={a}\mathrm{sin}\angle{BDA} \\ $$$$\mathrm{sin}\angle{BDA}=\frac{\mathrm{sin}\:\mathrm{80}°}{\mathrm{2sin}\:\mathrm{50}°}=\mathrm{sin}\:\angle{BDC} \\ $$$$\angle{BDC}=\mathrm{180}°−\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\angle{BDA}\right) \\ $$$${x}=\angle{DBC}=\mathrm{180}°−\left(\angle{DCB}+\angle{BDC}\right)= \\ $$$$\mathrm{180}°−\left(\mathrm{30}°+\mathrm{180}°−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\:\mathrm{80}°}{\mathrm{2sin}\:\mathrm{50}°}\right)\right)= \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\:\mathrm{80}°}{\mathrm{2sin}\:\mathrm{50}°}\right)−\mathrm{30}° \\ $$$$\frac{\mathrm{sin}\:\mathrm{80}°}{\mathrm{2sin}\:\mathrm{50}°}=\frac{\mathrm{2sin}\:\mathrm{40}°\mathrm{cos}\:\mathrm{40}°}{\mathrm{2sin}\left(\mathrm{90}°−\mathrm{40}°\right)}=\mathrm{sin}\:\mathrm{40}° \\ $$$${x}=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{40}°\right)−\mathrm{30}°=\mathrm{10}° \\ $$

Commented by mr W last updated on 11/Apr/25

��

Answered by lmcp1203 last updated on 10/Apr/25

thank[you

$${thank}\left[{you}\:\right. \\ $$

Answered by Spillover last updated on 11/Apr/25

Commented by lmcp1203 last updated on 16/Apr/25

thank you

$${thank}\:{you} \\ $$

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