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If angle A and angle B are acute angles...

If `angle A and angle B ` are acute angles such that `cos A= cos B` then

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Let `Delta ACD and Delta BEF` be two right triangles given in such a way that cos ` A= cos B.` then ,
`cos A = cos B`
`implies(AC)/(AD)=(BE)/(BF)`
`implies (AC)/(BE)=(AD)/(BF)=K("say")`
` implies AC=K(BE)`
` and AD=K(BF).`
`therefore (CD)/(EF)=(sqrt(AD^(2)-AC^(2)))/(sqrt(BF^(2)-BE^(2)))" " ["using Pythagoras ' theorem"]`
`=(ksqrt(BF^(2)-BE^(2)))/(sqrt(BF^(2)-BE^(2)))=K["usning (i) "].`
thus , we have `:(AC)/(BE)=(AD)/(BF)=(CD)/(EF).`
`therefore Delta ACD ~Delta BEF and " hence " , angle A=angle B.`
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