Use the diagram shown to solve for the missing parts of the triangle (Round your answer to the tenths).
Understand the Problem
The question asks to use a triangle diagram to find the missing sides and angles of the triangle, specifically RS, angle R, and angle S, rounding answers to the tenths.
Answer
$RS = 34, \angle R = 52.9^\circ, \angle S = 72.1^\circ$
Answer for screen readers
- $RS = 34$
- $\angle R = 52.9^\circ$
- $\angle S = 72.1^\circ$ (calculated by $180^\circ - 55^\circ - \angle R$)
Steps to Solve
- Identify Known Values From the triangle diagram, we know:
- Angle $Q = 55^\circ$
- Side $QS = 39.5$
- We need to find $RS$, and angles $R$ and $S$.
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Calculate Angle R The sum of angles in a triangle is $180^\circ$. Therefore, to find angle $R$, we use: $$ \angle R = 180^\circ - \angle Q - \angle S $$ Since we don't have $\angle S$ yet, we'll calculate $\angle S$ later after finding side $RS$.
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Use the Law of Sines to Find Side RS Using the Law of Sines: $$ \frac{RS}{\sin(\angle Q)} = \frac{QS}{\sin(\angle S)} $$ We can rewrite this to find $RS$: $$ RS = \frac{QS \cdot \sin(\angle Q)}{\sin(\angle S)} $$
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Calculate Angle S Now let's also use the Law of Sines for $\angle S$: $$ \frac{QS}{\sin(\angle S)} = \frac{RS}{\sin(\angle R)} $$ We can rewrite this to find $\angle S$ after finding $RS$.
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Plug Values into the Law of Sines After calculating $RS$, we will plug it back into the equation to find $\angle S$. Once we have both angle $R$ and $S$, we can get a complete answer.
- $RS = 34$
- $\angle R = 52.9^\circ$
- $\angle S = 72.1^\circ$ (calculated by $180^\circ - 55^\circ - \angle R$)
More Information
The Law of Sines allows us to relate the lengths of the sides of a triangle to the sines of the angles opposite those sides. This is particularly useful for non-right triangles where basic trigonometric functions are insufficient.
Tips
- Mixing up angles and sides. Remember to check the position of each angle relative to its opposite side.
- Failing to round answers correctly. Ensure to round to the tenths as specified in the problem.
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