Question

1. solve for k and w. k = __ w = __

Explanation:

Step1: Use sine - cosine relationships in a right - triangle

In a 30 - 60 - 90 right - triangle, if the hypotenuse is $c$, the side opposite the 30 - degree angle is $a$ and the side opposite the 60 - degree angle is $b$. The relationships are $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$ and $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$. Here, the hypotenuse $c = \sqrt{5}$, the angle $\theta = 30^{\circ}$ for side $k$ and $\theta = 60^{\circ}$ for side $w$.
We know that $\sin30^{\circ}=\frac{k}{\sqrt{5}}$ and $\cos30^{\circ}=\frac{w}{\sqrt{5}}$.

Step2: Recall the values of trigonometric functions

Since $\sin30^{\circ}=\frac{1}{2}$, from $\sin30^{\circ}=\frac{k}{\sqrt{5}}$, we can solve for $k$:
$k=\sqrt{5}\times\sin30^{\circ}=\frac{\sqrt{5}}{2}$.
Since $\cos30^{\circ}=\frac{\sqrt{3}}{2}$, from $\cos30^{\circ}=\frac{w}{\sqrt{5}}$, we can solve for $w$:
$w = \sqrt{5}\times\cos30^{\circ}=\frac{\sqrt{15}}{2}$.

Answer:

$k=\frac{\sqrt{5}}{2}$
$w=\frac{\sqrt{15}}{2}$