Question

1. find the value of the unknown length (x ) in these triangles. round your answer to 2 decimal places where necessary.

Explanation:

Step1: Use sine - rule for right - triangles

For a right - triangle with hypotenuse $c$ and an angle $\theta$, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$.

a)

In the first right - triangle, the hypotenuse $c = 2$ and the angle $\theta=30^{\circ}$, and we want to find the side opposite the $30^{\circ}$ angle.
Using the formula $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, we have $\sin30^{\circ}=\frac{x}{2}$.
Since $\sin30^{\circ}=\frac{1}{2}$, then $x = 2\times\sin30^{\circ}=2\times\frac{1}{2}=1$.

b)

In the second right - triangle, the hypotenuse $c = 5$ and the angle $\theta = 25^{\circ}$, and we want to find the side opposite the $25^{\circ}$ angle.
Using the formula $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, we have $x = 5\times\sin25^{\circ}$.
We know that $\sin25^{\circ}\approx0.4226$, so $x=5\times0.4226 = 2.113\approx2.11$.

c)

In the third right - triangle, the side adjacent to the $60^{\circ}$ angle is $a = 3$, and we want to find the hypotenuse $x$.
We know that $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, so $\cos60^{\circ}=\frac{3}{x}$.
Since $\cos60^{\circ}=\frac{1}{2}$, we can rewrite the equation as $\frac{1}{2}=\frac{3}{x}$.
Cross - multiplying gives us $x=6$.

Answer:

a. $x = 1$
b. $x\approx2.11$
c. $x = 6$