Question

part 3. Find the value of the trig function indicated, use Pythagorean theorem to find the third side if you need it.​

Answer 1

Answer:  $\bold{9)\ \sin \theta=\dfrac{1}{3}\qquad 10)\ \sin \theta = \dfrac{4}{5}\qquad 11)\ \cos \theta = \dfrac{\sqrt{11}}{6}\qquad 12)\ \tan \theta = \dfrac{17\sqrt2}{26}}$

Step-by-step explanation:

Pythagorean Theorem is: a² + b² = c²   , where "c" is the hypotenuse

$9)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{4}{12}\quad \rightarrow \large\boxed{\dfrac{1}{3}}$

Note: 4² + (8√2)² = hypotenuse²   →   hypotenuse = 12

$10)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{16}{20}\quad \rightarrow \large\boxed{\dfrac{4}{5}}$

Note: 12² + opposite² = 20²   →   opposite = 16

$11)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{\sqrt{11}}{6}\quad =\large\boxed{\dfrac{\sqrt{11}}{6}}$

Note: adjacent² + 5² = 6²   →   adjacent = √11

$12)\ \tan \theta=\dfrac{\text{side opposite of}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{17}{13\sqrt2}\quad =\large\boxed{\dfrac{17\sqrt2}{26}}$

Note: adjacent² + 7² = (13√2)²   →   adjacent = 17