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Chapter 10: Problem 5

For questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(B=18^{\circ}, C=110^{\circ}, \mathrm{BC}=12.3 \mathrm{~cm}\)

Short Answer

Expert verified
Answer: Angle A is 52 degrees, side AC is approximately 9.61 cm, and side AB is approximately 3.84 cm.

Step by step solution

01

Find Angle A

To find angle A, we will use the fact that the sum of angles in a triangle is 180 degrees. Since we have angle B and angle C, we can find angle A by subtracting the sum of angle B and angle C from 180 degrees: \[A = 180 - B - C\] \[A = 180 - 18 - 110\] \[A = 52^{\circ}\] So, angle A is 52 degrees.
02

Use the Law of Sines to find side AC

Now that we have all three angles, we can use the Law of Sines to find side AC. The Law of Sines states that \(\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}}\), where a, b, and c are the side lengths opposite to angles A, B, and C respectively. We are given side BC (c) and its corresponding angle C. We also know angle A and want to find side AC (a). We can set up the following equation using the Law of Sines: \[\frac{a}{\sin{52^{\circ}}} = \frac{12.3}{\sin{110^{\circ}}}\] Now, we can solve for side AC (a): \[a = \frac{12.3 \times \sin{52^{\circ}}}{\sin{110^{\circ}}}\] \[a \approx 9.61 \mathrm{~cm}\] So, side AC is approximately 9.61 cm.
03

Use the Law of Sines to find side AB

Similarly, we can use the Law of Sines to find side AB (b). We know angle B and want to find the side length opposite to it: \[\frac{b}{\sin{18^{\circ}}} = \frac{12.3}{\sin{110^{\circ}}}\] Now, we can solve for side AB (b): \[b = \frac{12.3 \times \sin{18^{\circ}}}{\sin{110^{\circ}}}\] \[b \approx 3.84 \mathrm{~cm}\] So, side AB is approximately 3.84 cm. In conclusion, we have found all the angles and side lengths of the triangle: Angle A: \(52^{\circ}\) Angle B: \(18^{\circ}\) Angle C: \(110^{\circ}\) Side AC: \(\approx 9.61\) cm Side AB: \(\approx 3.84\) cm Side BC: \(12.3\) cm

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Most popular questions from this chapter

For questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(\mathrm{BC}=14 \mathrm{~cm}, \mathrm{AB}=20 \mathrm{~cm}, C=50^{\circ}\)

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A point \(\mathrm{A}\) has a bearing of \(45^{\circ}\) and is \(10 \mathrm{~km}\) distant from O. A point B has a bearing of \(160^{\circ}\) and is \(20 \mathrm{~km}\) distant from \(\mathrm{O}\). A point \(\mathrm{C}\) is mid-way between A and B. Calculate (a) the bearing of \(\mathrm{C}\) from \(\mathrm{O}\) (b) the distance of \(\mathrm{C}\) from \(\mathrm{O}\).

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