Subjects geometry

Angle C D40C8E

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Question: Find the measure of angle $C$. $A$ $b = 8$ $c = 5$ $C = ?^\circ$ $a = 9$ $B$ Law of Cosines: $$c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$$ Round your answer to the nearest hundredth. graph: triangle with vertices $C$ at bottom-left, $A$ at top, and $B$ at bottom-right; side labels $b = 8$ on the left slanted side from $C$ to $A$, $c = 5$ on the right slanted side from $A$ to $B$, and $a = 9$ on the bottom side from $C$ to $B$; the unknown angle $C$ is at the bottom-left, position_hint = center
1. **State the problem:** We need to find the measure of angle $C$ in a triangle with sides $a=9$, $b=8$, and $c=5$. 2. **Recall the Law of Cosines formula:** $$c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$$ This formula relates the lengths of the sides of a triangle to the cosine of one of its angles. 3. **Substitute the known values:** $$5^2 = 9^2 + 8^2 - 2 \times 9 \times 8 \cdot \cos(C)$$ 4. **Calculate the squares:** $$25 = 81 + 64 - 144 \cdot \cos(C)$$ 5. **Simplify the right side:** $$25 = 145 - 144 \cdot \cos(C)$$ 6. **Isolate the cosine term:** $$25 - 145 = -144 \cdot \cos(C)$$ $$-120 = -144 \cdot \cos(C)$$ 7. **Divide both sides by $-144$:** $$\frac{-120}{-144} = \cancel{\frac{-144}{-144}} \cdot \cos(C)$$ $$\frac{120}{144} = \cos(C)$$ 8. **Simplify the fraction:** $$\cos(C) = \frac{5}{6} \approx 0.8333$$ 9. **Find angle $C$ by taking the inverse cosine:** $$C = \cos^{-1}(0.8333)$$ 10. **Calculate the angle:** $$C \approx 33.56^\circ$$ **Final answer:** The measure of angle $C$ is approximately $33.56^\circ$.
CABb=8c=5a=9C