Question

Find sin A and sec B given a=2 b=11

Answer 1

if there is a triangle with Sid a and b then sina b/✓a2+b2and seven=✓a2+b2/a

Answer 2

`\sin A=(2)/(5\sqrt5)`

`\sec B=(5\sqrt5)/(11)`

Explanation:

Given:
`a=2,b=11`

Using pythagoreous theorem:

`c^2=a^2+b^2`

`c^2=11^2+2^2`

`c=5√(5)`

Trigonometric Identities:

`\sin \theta=\frac{\text{Opposite}}{\text{Hypotenuse}}`

`\sin A=(a)/(c)`

`\sin A=(2)/(5\sqrt5)`

`\sec B=(c)/(b)`

`\sec B=(5\sqrt5)/(11)`

Hence, The value of trigonometric identity

`\sin A=(2)/(5\sqrt5)`

`\sec B=(5\sqrt5)/(11)`