Question

: If tan A=- 9/13 and 0 < A < 180, find, without using tables or calculators, the value of

Answer 1

Answer:  $\bold{\sin A=\dfrac{9\sqrt{10}}{50}\qquad \cos A=-\dfrac{13\sqrt{10}}{50}}$

Step-by-step explanation:

0° < A < 180° so it is in Quadrant I or II

since tan A is negative, it is in Quadrant II ---> (-x, +y)

$\text{Given}:\tan A=\dfrac{y}{x}\ =\dfrac{9}{-13}$

Next, find the hypotenuse given that x = -13 and y = 9

(-13)² + (9)² = r²

         250 = r²

         5√10 = r

$\sin A = \dfrac{y}{r}\quad =\dfrac{9}{5\sqrt{10}}\bigg(\dfrac{\sqrt{10}}{\sqrt{10}}\bigg)\quad =\large\boxed{\dfrac{9\sqrt{10}}{50}}\\\\\\\\\cos A = \dfrac{x}{r}\quad =\dfrac{-13}{5\sqrt{10}}\bigg(\dfrac{\sqrt{10}}{\sqrt{10}}\bigg)\quad =\large\boxed{-\dfrac{13\sqrt{10}}{50}}$