Question

An acute angle 0 is in a right triangle with cos 0=9/10. What is the value of sec 0

Answer 1

The value of sec O is
`(10)/(9)`

Explanation:

Given as for right angle triangle

cos O =
`(9)/(10)`

Now from Pythagoras theorem, for right angled triangle

H² = P² + B²

∵ cos O =
`(9)/(10)` =
`(Base)/(hypotenuse)`

So Perpendicular² = Hypotenuse² - Base²

Or, Perpendicular² = 10² - 9²

Or, Perpendicular² = 100 - 81 = 19

∴ Perpendicular ( P ) =
`√(19)`

∵ , Sec Ф =
`(hypotenuse)/(base)`

So , Sec O =
`(10)/(9)`

Hence The value of sec O is
`(10)/(9)` Answer