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If the base angles of triangle are ((22)...

If the base angles of triangle are `((22)/(12))^@` and `(112 1/2)^@` , then prove that the altitude of the triangle is equal to `1/2` of its base.

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`/_BAC = 180-112.5-22.5 = 45^@`
`/_ACL = 180-112.5 = 67.5^@`
Now, in `Delta ABC`,
From sine law,
`(BC)/(sin45^@) =( AC)/(sin22.5^@)`
`=>AC = (BC sin22.5^@)/sin45^@`
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