# Week 17 – Precalc 11

This week in Precalculus 11, we learned about the Sine Law. It gives us a more efficient method to solving triangles without 90° angles. However, it can only be used if an angle, its opposite side length, and one other piece of information are given.

The Sine Law is commonly written in 2 forms. The first is used when solving for a side length, and the second is used when solving for an angle.

Examples

When only given one angle using Sine Law, there could be 2 different triangles. Solve for an angle, and if it is less than 90°, find its quadrant II coterminal angle. Add this to the originally given angle, and if it is less than 180°, there are 2 possible triangles.

# Week 1 – My Arithmetic Sequence

13, 26, 39, 52, 65…

$t_n$ = $t_1$ + d(n – 1)

($t_{50}$) = (13) + (13)[(50) – 1]

$t_{50}$ = 13 + 13 · 49

$t_{50}$ = 13 + 637

$t_{50}$ = 650

$S_n$ = $\frac{n}{2}$($t_1$ + $t_n$)

($S_{50}$) = $\frac{(50)}{2}$[(13) + (650)]

$S_{50}$ = 25 · 663

$S_{50}$ = 16575