In a right angle triangle if sine a =9/10

Step 1: In a right-angle triangle, the sine of an angle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. Given $\sin A = \frac{9}{10}$. This means: Opposite side $= 9$ units Hypotenuse $= 10$ units

Step 2: Use the Pythagorean theorem to find the length of the adjacent side. Let the adjacent side be $x$. $(Opposite)^2 + (Adjacent)^2 = (Hypotenuse)^2$ $9^2 + x^2 = 10^2$ $81 + x^2 = 100$

Step 3: Solve for $x$. $x^2 = 100 - 81$ $x^2 = 19$ $x = \sqrt{19}$ So, the adjacent side is $\sqrt{19}$ units.

Step 4: Now we can find the other trigonometric ratios for angle $A$. The cosine of angle $A$ is $\frac{Adjacent}{Hypotenuse}$. $\cos A = \frac{\sqrt{19}}{10}$ The tangent of angle $A$ is $\frac{Opposite}{Adjacent}$. $\tan A = \frac{9}{\sqrt{19}}$ To rationalize the denominator for $\tan A$: $\tan A = \frac{9}{\sqrt{19}} \times \frac{\sqrt{19}}{\sqrt{19}} = \frac{9\sqrt{19}}{19}$

The adjacent side is $\sqrt{19}$, $\cos A = \frac{\sqrt{19}}{10}$, and $\tan A = \frac{9\sqrt{19}}{19}$.

\text{Adjacent side = \sqrt{19}, \cos A = \frac{\sqrt{19}}{10}, \tan A = \frac{9\sqrt{19}}{19}}