Question

given $\beta = 42^{\circ}$ and $a = 6$, determine the following.

(image of a right triangle with right angle at the bottom right, angle $\beta = 42^{\circ}$ at the bottom left, side $a = 6$ (adjacent to $\beta$), side $b$ (opposite to $\beta$), hypotenuse $c$, and angle $\theta$ at the top left.)

a) what is $\theta$? round to two decimal places, if necessary. enter the degree symbol by using the toolbar or typing \deg\.
$\theta = $

b) what is $b$? round to two decimal places, if necessary.
$b = $

c) what is $c$? round to two decimal places, if necessary.
$c = $

Explanation:

Step1: Calculate $\theta$ (angle sum)

The sum of angles in a right triangle is $180^\circ$.
$\theta = 180^\circ - 90^\circ - 42^\circ = 48^\circ$

Step2: Calculate side $b$ (tangent ratio)

Use $\tan(\beta) = \frac{b}{a}$, solve for $b$.
$b = a \cdot \tan(\beta) = 6 \cdot \tan(42^\circ) \approx 6 \cdot 0.9004 = 5.40$

Step3: Calculate side $c$ (cosine ratio)

Use $\cos(\beta) = \frac{a}{c}$, solve for $c$.
$c = \frac{a}{\cos(\beta)} = \frac{6}{\cos(42^\circ)} \approx \frac{6}{0.7431} = 8.07$

Answer:

a) $48^\circ$
b) $5.40$
c) $8.07$