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Chapter 10: Problem 1

For questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(A=36^{\circ}, B=79^{\circ}, \mathrm{AC}=11.63 \mathrm{~cm}\)

Short Answer

Expert verified
Answer: The lengths of sides AB and BC are approximately \(6.96 \: \mathrm{cm}\) and \(10.32 \: \mathrm{cm}\), respectively.

Step by step solution

01

Find angle C

To find angle C, subtract the given angles A and B from \(180^{\circ}\): \(C = 180^{\circ} - 36^{\circ} - 79^{\circ} = 65^{\circ}\)
02

Use Law of Sines to find side AB

Using the Law of Sines, we can find the length of side AB: \(\frac{\mathrm{AB}}{\sin{A}} = \frac{\mathrm{AC}}{\sin{B}}\) $\mathrm{AB} = \frac{\mathrm{AC} \times \sin{A}}{\sin{B}} \Rightarrow \mathrm{AB} = \frac{11.63 \times \sin{36^{\circ}}}{\sin{79^{\circ}}} \Rightarrow \mathrm{AB} \approx 6.96 \: \mathrm{cm}$
03

Use Law of Sines to find side BC

Using the Law of Sines, we can find the length of side BC: \(\frac{\mathrm{BC}}{\sin{C}} = \frac{\mathrm{AC}}{\sin{B}}\) $\mathrm{BC} = \frac{\mathrm{AC} \times \sin{C}}{\sin{B}} \Rightarrow \mathrm{BC} = \frac{11.63 \times \sin{65^{\circ}}}{\sin{79^{\circ}}} \Rightarrow \mathrm{BC} \approx 10.32\: \mathrm{cm}$ Now we have found all the unknown sides and angles of the triangle. The triangle ABC can be described as: - Angle A: \(36^{\circ}\) - Angle B: \(79^{\circ}\) - Angle C: \(65^{\circ}\) - Side AC: \(11.63 \: \mathrm{cm}\) - Side AB: \(\approx 6.96 \: \mathrm{cm}\) - Side BC: \(\approx 10.32 \: \mathrm{cm}\)

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Most popular questions from this chapter

\(\triangle \mathrm{ABC}\) has a right angle at \(\mathrm{A}, B=25^{\circ}\) and \(\mathrm{AC}=17.2 \mathrm{~cm} .\) Solve \(\triangle \mathrm{ABC}\).

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