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प्रश्न
In the given figure, AD is perpendicular to BC. Find:
`(3)/("sin" x) + (4)/("cos" y) - 4 "tan" y`
बेरीज
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उत्तर

ΔADB is a right-angled triangle.
∴ AB2
= AB2 + BD2
= 122 + 162
= 144 + 256
= 400
⇒ AB = 20cm
ΔADC is a right-angled triangle.
∴ AC2
= AD2 + DC2
= 122 + 92
= 144 + 81
= 225
⇒ AC = 15cm
`(3)/("sin" x) + (4)/("cos" y) - 4 "tan" y`
= `(3)/("AD"/"AB") + (4)/("AD"/"AC") - 4 xx "CD"/"AD"`
= `(3)/(12/20) + (4)/(12/15) - 4 xx (9)/(12)`
= `(60)/(12) + (60)/(12) - 3`
= 5 + 5 - 3
= 7.
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