In the adjoining frgure ,a right - angled triangle ABC is shown un which AM=CM=3m . If `angleACM=15^(@)` , then find AC.

Explanation

Here , AM=CM
`rArrltACM=ltCAM` " " (opposite angles of equal sides)

`rArrltCAM=15^(@)`
`:. ltACB=90^(@)-15^(@)=75^(@)`
and `ltBCM=ltACB- ltACM=75^(@)-15^(@)=60^(@)`
In `DeltaBCM`,
`cos(ltBCM)=(BC)/(CM)rArrcos60^(@)=(BC)/(3)`
`rArr(1)/(2)=(BC)/(3) rArr BC=(3)/(2)m`
and `sin (ltBCM)=(BM)/(CM)rArr sin 60^(@)=(BM)/(3)`
`rArr(sqrt(3))/(2)=(BM)/(3)rArrBM=(3sqrt(3))/(2)m`
`:. AB=AM+BM=(3+(3sqrt(3))/(2))m`
Now , in `DeltaABC` , from Pythagoras theorem
`AC^(2)=AB^(2)+BC^(2)=(3+(3sqrt(3))/(2))^(2)+((3)/(2))^(3)=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))`m