Given $\sin \theta=0.31$, find $\cos \theta$ and $\tan \theta$. Where $0 \leq \theta \leq 90$ Given $\cos \theta=-0.31$, find $\sin \theta$ and $\tan \theta$ where $180 \leq \theta \leq 270$ is Find $\theta$ : - Given $\triangle A B C$ right isosceles - Where $A B=C D=5$ - $H$ is the mid point of $C E$. - angle $E=35$

Step 1 ：Given \(\sin \theta = 0.31\), we can use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\) to find \(\cos \theta\):

Step 2 ：\(\cos^2 \theta = 1 - \sin^2 \theta = 1 - (0.31)^2\)

Step 3 ：\(\cos \theta \approx 0.9507\)

Step 4 ：Next, we can find \(\tan \theta\) using the formula \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):

Step 5 ：\(\tan \theta \approx \frac{0.31}{0.9507} \approx 0.3261\)

Step 6 ：\(\boxed{\cos \theta \approx 0.9507, \tan \theta \approx 0.3261}\)

Step 7 ：Given \(\cos \theta = -0.31\), we can use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\) to find \(\sin \theta\):

Step 8 ：\(\sin^2 \theta = 1 - \cos^2 \theta = 1 - (-0.31)^2\)

Step 9 ：\(\sin \theta \approx 0.9507\)

Step 10 ：Next, we can find \(\tan \theta\) using the formula \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):

Step 11 ：\(\tan \theta \approx \frac{0.9507}{-0.31} \approx -3.0669\)

Step 12 ：\(\boxed{\sin \theta \approx 0.9507, \tan \theta \approx -3.0669}\)

Step 13 ：Given the triangle conditions, we know that \(\angle A = \angle B = 45^\circ\) and \(AB = CD = 5\). We can use the Law of Sines to find \(\theta\):

Step 14 ：\(\sin \angle C = \frac{AB}{CD} \sin \angle E\)

Step 15 ：\(\sin \angle C \approx \frac{5}{5} \sin 35^\circ \approx 0.9848\)

Step 16 ：\(\angle C \approx 100^\circ\)

Step 17 ：Since \(\angle A = \angle B = 45^\circ\), we can find \(\theta\) by subtracting the sum of \(\angle A\) and \(\angle B\) from \(\angle C\):

Step 18 ：\(\theta \approx 100^\circ - (45^\circ + 45^\circ) \approx 80^\circ\)

Step 19 ：\(\boxed{\theta \approx 80^\circ}\)