Question

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

Answer 1

Answer:

$\sin A=\dfrac{2}{5\sqrt5}$

$\sec B=\dfrac{5\sqrt5}{11}$

Step-by-step explanation:

Given: $a=2,b=11$

Using pythagoreous theorem:

$c^2=a^2+b^2$

$c^2=11^2+2^2$

$c=5\sqrt{5}$

Trigonometric Identities:

$\sin \theta=\dfrac{\text{Opposite}}{\text{Hypotenuse}}$

$\sin A=\dfrac{a}{c}$

$\sin A=\dfrac{2}{5\sqrt5}$

$\sec B=\dfrac{c}{b}$

$\sec B=\dfrac{5\sqrt5}{11}$

Hence, The value of trigonometric identity

$\sin A=\dfrac{2}{5\sqrt5}$

$\sec B=\dfrac{5\sqrt5}{11}$

Answer 2

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