संलग्न चित्र में , एक समकोण त्रिभुज ABC दिखाया गया है जिसमे `AM = CM = 3` मी
है | यदि `angle ACM = 15^(@)` तो AC का मान ज्ञात कीजिए |
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Explanation

यहाँ , AM=CM
`rArr angle ACM = angle CAM `
` rArr angle CAM = 15^(@)`
` therefore angle ACB = 90^(@) - 15^(@) = 75^(@)`
और ` angle BCM = anlge ACB - angle ACM = 75^(@) - 15^(@) = 60^(@)`
` Delta BCM` में
` cos (angle BCM) = (BC)/(CM) rArr cos 60^(@) = (BC)/(3)`
` rArr (1)/(2) = (BC)/(3) rArr BC = (3)/(2) ` मी
और ` sin (angle BCM) - (BM)/(CM) rArr sin 60^(@) = (BM)/(3)`
` rArr (sqrt(3))/(2) = (BM)/(3) rArr BM = (3sqrt(3))/(2) ` मी
` therefore AB = AM + BM (3 + (3sqrt(3))/(2))` मी
` Delta ABC` पाइथागोरस प्रमेय से
`AC^(2) = AB^(2) + BC^(2) = (3+ (3sqrt(3))/(2))^(2) + ((3)/(2))^(2) = 9 + (27)/(4) + 9sqrt(3) + (9)/(4)`
`= 18 + 9sqrt(3) = 9(2+ sqrt(3))= (9)/(2) (4 + 2sqrt(3))`
` rArr AC= sqrt((9(4 + 2 sqrt(3)))/(2)) = (3(sqrt(3) +1))/(sqrt(2))` मी