Trigonometry Final Exam Review – Step-by-Step Guidance

Q19. Solve the right triangle given: A = 41.3°, b = 2.5 m. Find the missing sides and angles. (Round side lengths to one decimal place.)

Background

Topic: Solving Right Triangles

This question tests your ability to use trigonometric ratios and the properties of right triangles to find unknown sides and angles when given one angle and one side.

Key Terms and Formulas:

• Right triangle: A triangle with one 90° angle.

• Trigonometric ratios: $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$, $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$

• Pythagorean Theorem: $a^2 + b^2 = c^2$

Step-by-Step Guidance

1. Identify the given values: $A = 41.3^\circ$, $b = 2.5$ m, and $C = 90^\circ$ (since it's a right triangle).

2. Find angle $B$ using the triangle angle sum: $A + B + C = 180^\circ$.

3. Use the definition of sine to find side $a$: $\sin A = \frac{a}{c}$, so $a = c \sin A$.

4. Use the definition of cosine to find side $b$: $\cos A = \frac{b}{c}$, so $c = \frac{b}{\cos A}$.

5. Once you have $c$, use it to find $a$ from the earlier step.

Try solving on your own before revealing the answer!

Final Answer:

$B = 48.7^\circ$, $a = 2.2$ m, $c = 3.3$ m

We used the triangle sum to find $B$, then used trigonometric ratios and the given side to solve for the other sides.