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Question Number 149481 by mathdanisur last updated on 05/Aug/21

Answered by mr W last updated on 05/Aug/21

say AB=CK=x, BM=BK=1 AC=x−1+x=2x−1 (2x−1)^2 =x^2 +(x+1)^2 x^2 =3x ⇒x=3 ⇒AB=3,BC=4,CA=5 AK=(√(3^2 +1^2 ))=(√(10)) AK×AH=AM^2 (√(10))×AH=2^2 AH=((2(√(10)))/5) ⇒HK=(√(10))−((2(√(10)))/5)=((3(√(10)))/5) HC^2 =3^2 +(((3(√(10)))/5))^2 +2×3×((3(√(10)))/5)×(1/( (√(10)))) HC^2 =((81)/5) HC=(9/( (√5))) ((sin θ)/3)=(3/( (√(10))))×((√5)/9) sin θ=(1/( (√2))) ⇒θ=45°

$${say}\:{AB}={CK}={x},\:{BM}={BK}=\mathrm{1} \\ $$$${AC}={x}−\mathrm{1}+{x}=\mathrm{2}{x}−\mathrm{1} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} =\mathrm{3}{x} \\ $$$$\Rightarrow{x}=\mathrm{3} \\ $$$$\Rightarrow{AB}=\mathrm{3},{BC}=\mathrm{4},{CA}=\mathrm{5} \\ $$$${AK}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }=\sqrt{\mathrm{10}} \\ $$$${AK}×{AH}={AM}^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{10}}×{AH}=\mathrm{2}^{\mathrm{2}} \\ $$$${AH}=\frac{\mathrm{2}\sqrt{\mathrm{10}}}{\mathrm{5}} \\ $$$$\Rightarrow{HK}=\sqrt{\mathrm{10}}−\frac{\mathrm{2}\sqrt{\mathrm{10}}}{\mathrm{5}}=\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{5}} \\ $$$${HC}^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} +\left(\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{2}×\mathrm{3}×\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{5}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}} \\ $$$${HC}^{\mathrm{2}} =\frac{\mathrm{81}}{\mathrm{5}} \\ $$$${HC}=\frac{\mathrm{9}}{\:\sqrt{\mathrm{5}}} \\ $$$$\frac{\mathrm{sin}\:\theta}{\mathrm{3}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}×\frac{\sqrt{\mathrm{5}}}{\mathrm{9}} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\theta=\mathrm{45}° \\ $$

Commented by mathdanisur last updated on 05/Aug/21

Thankyou Ser, cool

$$\mathrm{Thankyou}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{cool} \\ $$

Commented by Tawa11 last updated on 05/Aug/21

great sir

$$\mathrm{great}\:\mathrm{sir} \\ $$

Commented by Tawa11 last updated on 06/Aug/21

Sir mrW please check Q149510. Geometry. Your approach too is needed. Thanks sir for your time.

$$\mathrm{Sir}\:\:\mathrm{mrW}\:\:\:\mathrm{please}\:\mathrm{check}\:\mathrm{Q149510}.\:\mathrm{Geometry}.\:\:\mathrm{Your}\:\mathrm{approach}\:\mathrm{too}\:\mathrm{is}\:\mathrm{needed}. \\ $$$$\mathrm{Thanks}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}. \\ $$

Commented by mr W last updated on 06/Aug/21

the question is answered. do you mean the answer is not correct?

$${the}\:{question}\:{is}\:{answered}.\:{do}\:{you}\:{mean} \\ $$$${the}\:{answer}\:{is}\:{not}\:{correct}? \\ $$

Commented by Tawa11 last updated on 06/Aug/21

No sir. Just to see another approach.

$$\mathrm{No}\:\mathrm{sir}.\:\mathrm{Just}\:\mathrm{to}\:\mathrm{see}\:\mathrm{another}\:\mathrm{approach}. \\ $$

Commented by Brahan last updated on 06/Aug/21

why BK=BM=1

$$\boldsymbol{{why}}\:\boldsymbol{{BK}}=\boldsymbol{{BM}}=\mathrm{1} \\ $$

Commented by mr W last updated on 07/Aug/21

since we only need to get the angles, we don′t need to know the absolute lengthes of the sides, we only need to know the ratios between them. we can also take BK=BM=2 or any other values, or just take BK=BM=a.

$${since}\:{we}\:{only}\:{need}\:{to}\:{get}\:{the}\:{angles}, \\ $$$${we}\:{don}'{t}\:{need}\:{to}\:{know}\:{the}\:{absolute} \\ $$$${lengthes}\:{of}\:{the}\:{sides},\:{we}\:{only}\:{need} \\ $$$${to}\:{know}\:{the}\:{ratios}\:{between}\:{them}. \\ $$$${we}\:{can}\:{also}\:{take}\:{BK}={BM}=\mathrm{2}\:{or}\:{any} \\ $$$${other}\:{values},\:{or}\:{just}\:{take}\:{BK}={BM}={a}. \\ $$

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