Question

If sin 37°=3/5, what is tan 16°

Answer 1

7/24

Answer 2

To find $\tan 16^\circ$ given $\sin 37^\circ = 3/5$, we follow these steps:

1. Determine $\tan 37^\circ$:
   Given $\sin 37^\circ = 3/5$.
   Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, we find $\cos 37^\circ$:
   $\cos 37^\circ = \sqrt{1 - \sin^2 37^\circ} = \sqrt{1 - (3/5)^2} = \sqrt{1 - 9/25} = \sqrt{16/25} = 4/5$.
   Now, calculate $\tan 37^\circ$:
   $\tan 37^\circ = \frac{\sin 37^\circ}{\cos 37^\circ} = \frac{3/5}{4/5} = \frac{3}{4}$.

2. Determine $\tan 53^\circ$:
   Notice that $16^\circ = 53^\circ - 37^\circ$.
   Since $37^\circ$ and $53^\circ$ are complementary angles ($37^\circ + 53^\circ = 90^\circ$), we have:
   $\sin 53^\circ = \cos 37^\circ = 4/5$
   $\cos 53^\circ = \sin 37^\circ = 3/5$
   Now, calculate $\tan 53^\circ$:
   $\tan 53^\circ = \frac{\sin 53^\circ}{\cos 53^\circ} = \frac{4/5}{3/5} = \frac{4}{3}$.

3. Apply the tangent difference formula:
   Use the identity $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.
   Let $A = 53^\circ$ and $B = 37^\circ$.
   $\tan 16^\circ = \tan(53^\circ - 37^\circ) = \frac{\tan 53^\circ - \tan 37^\circ}{1 + \tan 53^\circ \tan 37^\circ}$
   Substitute the values of $\tan 53^\circ = 4/3$ and $\tan 37^\circ = 3/4$:
   $\tan 16^\circ = \frac{\frac{4}{3} - \frac{3}{4}}{1 + \left(\frac{4}{3}\right)\left(\frac{3}{4}\right)}$
   $\tan 16^\circ = \frac{\frac{16 - 9}{12}}{1 + 1}$
   $\tan 16^\circ = \frac{\frac{7}{12}}{2}$
   $\tan 16^\circ = \frac{7}{12 \times 2}$
   $\tan 16^\circ = \frac{7}{24}$