Question

1. find the length of a. (2 points)

triangle abc with angle at a being 110°, ab = 12, ac = 15, and bc = a

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines for side $a$ (opposite angle $A$) is:
$$a^2 = b^2 + c^2 - 2bc\cos(A)$$
Here, $b=15$, $c=12$, $A=110^\circ$.

Step2: Substitute the known values

$$a^2 = 15^2 + 12^2 - 2(15)(12)\cos(110^\circ)$$
Calculate each term:
$15^2=225$, $12^2=144$, $2(15)(12)=360$, $\cos(110^\circ)\approx-0.3420$
$$a^2 = 225 + 144 - 360(-0.3420)$$

Step3: Compute the right-hand side

$$a^2 = 369 + 123.12 = 492.12$$

Step4: Take square root to find $a$

$$a = \sqrt{492.12}$$

Answer:

$\approx22.18$