Question

In ΔJKL, k = 22 cm, ∠L=23° and ∠J=26°. Find the length of j, to the nearest 10th of a centimeter.

Answer 1

We have been given that in ΔJKL, k = 22 cm, ∠L=23° and ∠J=26°. We are asked to find the length of j to the nearest tenth of a centimeter.

We will use law of sines to solve for side j.

$\frac{a}{\text{sin}(A)}=\frac{b}{\text{sin}(B)}=\frac{c}{\text{sin}(C)}$, where, a, b and c are opposite sides to angles A, B and C respectively.

We need to find measure of angle K to apply law of sine to our given problem.

Using angle sum property, we will get:

$\angle J+\angle K+\angle L=180^{\circ}$

$26^{\circ}+\angle K+23^{\circ}=180^{\circ}$

$\angle K+49^{\circ}=180^{\circ}$

$\angle K+49^{\circ}-49^{\circ}=180^{\circ}-49^{\circ}$

$\angle K=131^{\circ}$

$\frac{j}{\text{sin}(J)}=\frac{k}{\text{sin}(K)}$

$\frac{j}{\text{sin}(26^{\circ})}=\frac{22}{\text{sin}(131^{\circ})}$

$\frac{j}{\text{sin}(26^{\circ})}\cdot \text{sin}(26^{\circ})=\frac{22}{\text{sin}(131^{\circ})}\cdot \text{sin}(26^{\circ})$

$j=\frac{22}{0.754709580223}\cdot 0.438371146789$

$j=\frac{9.644165229358}{0.754709580223}$

$j=12.77864423889$

Upon rounding to nearest tenth, we will get:

$j\approx 12.8$

Therefore, the length of j is approximately 12.8 cm.