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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

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

Commented by mathdanisur last updated on 05/Aug/21

$$\mathrm{Thankyou}\mathbf{S}\mathrm{er},\mathrm{cool}$$

Commented by Tawa11 last updated on 05/Aug/21

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

Commented by Tawa11 last updated on 06/Aug/21

$$\mathrm{Sir}\mathrm{mrW}\mathrm{please}\mathrm{check}\mathrm{Q}149510.\mathrm{Geometry}.\mathrm{Your}\mathrm{approach}\mathrm{too}\mathrm{is}\mathrm{needed}.$$$$\mathrm{Thanks}\mathrm{sir}\mathrm{for}\mathrm{your}\mathrm{time}.$$

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

$$thequestionisanswered.doyoumean$$$$theanswerisnotcorrect?$$

Commented by Tawa11 last updated on 06/Aug/21

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

Commented by Brahan last updated on 06/Aug/21

$$\mathit{w}\mathit{h}\mathit{y}\mathit{B}\mathit{K}=\mathit{B}\mathit{M}=1$$

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

$$sinceweonlyneedtogettheangles,$$$$we{don}^{\prime }tneedtoknowtheabsolute$$$$lengthesofthesides,weonlyneed$$$$toknowtheratiosbetweenthem.$$$$wecanalsotakeBK=BM=2orany$$$$othervalues,orjusttakeBK=BM=a.$$

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