1. **Stating the problem:** We have a right triangle ABC with a right angle at C. Given that $\sin A = \frac{4}{5}$, we need to determine which of the given statements are true. 2. **Recall important trigonometric rules for right triangles:** - $\sin A = \frac{\text{opposite side to } A}{\text{hypotenuse}}$ - $\cos A = \frac{\text{adjacent side to } A}{\text{hypotenuse}}$ - $\tan A = \frac{\text{opposite side}}{\text{adjacent side}}$ - Complementary angles satisfy $\sin(90^\circ - A) = \cos A$ and $\cos(90^\circ - A) = \sin A$ - Since $\triangle ABC$ is right angled at C, angles A and B are complementary: $A + B = 90^\circ$ 3. **Find $\cos A$ and $\tan A$ using $\sin A = \frac{4}{5}$:** Using Pythagoras theorem for the sides opposite and adjacent to angle A: $$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$$ $$\tan A = \frac{\sin A}{\cos A} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3}$$ 4. **Evaluate each statement:** - **cos A = 3/5**: True, as calculated above. - **sin (90° - A) = 4/5**: Since $\sin(90^\circ - A) = \cos A = \frac{3}{5}$, this statement is False. - **tan B = 3/4**: Since $B = 90^\circ - A$, $\tan B = \tan(90^\circ - A) = \cot A = \frac{1}{\tan A} = \frac{1}{\frac{4}{3}} = \frac{3}{4}$, so True. - **cot A = 3/4**: $\cot A = \frac{1}{\tan A} = \frac{1}{\frac{4}{3}} = \frac{3}{4}$, True. - **sin A > cos B**: Since $B = 90^\circ - A$, $\cos B = \cos(90^\circ - A) = \sin A = \frac{4}{5}$. So $\sin A = \cos B$, not greater. False. 5. **Final answers:** - cos A = 3/5: True - sin (90° - A) = 4/5: False - tan B = 3/4: True - cot A = 3/4: True - sin A > cos B: False