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Chapter 10: Problem 7

In questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(\mathrm{BC}=17 \mathrm{~cm}, \mathrm{AC}=27 \mathrm{~cm}, C=45^{\circ}\)

Short Answer

Expert verified
Question: In triangle ABC, side BC is 17 cm, side AC is 27 cm, and angle C is 45°. Find the remaining angles and the third side of the triangle. Answer: In triangle ABC, the remaining angles are approximately A = 100.65° and B = 34.35°, and the third side AB is approximately 25.75 cm.

Step by step solution

01

Use the Law of Sines

Write down the Law of Sines formula: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ In our case, C = 45° and c = AC = 27 cm. We will find angle B using this formula.
02

Substitute the given values into the Law of Sines formula

Substitute the values into the formula: $$\frac{b}{\sin B} = \frac{27}{\sin 45^{\circ}}$$
03

Solve for angle B

To find angle B, rearrange the equation and solve for it: $$\sin B = \frac{b \cdot \sin 45^{\circ}}{27}$$
04

Calculate the value of angle B

Calculate the value of angle B using the given side length, b = BC = 17 cm: $$\sin B = \frac{17 \cdot \sin 45^{\circ}}{27}$$ $$B = \arcsin\left(\frac{17 \cdot \sin 45^{\circ}}{27}\right) \approx 34.35^{\circ}$$
05

Calculate the value of angle A

Use the property that the sum of the angles in a triangle is equal to 180°: $$A = 180^{\circ} - B - C$$ Substitute the values of B and C: $$A = 180^{\circ} - 34.35^{\circ} - 45^{\circ} \approx 100.65^{\circ}$$
06

Use the Law of Cosines to find the third side

Now, we need to find the third side, AB. We can use the Law of Cosines formula for this: $$a^2 = b^2 + c^2 - 2bc \cos A$$ In our case, a = AB, b = BC = 17 cm, c = AC = 27 cm, and A = 100.65°.
07

Substitute the values into the Law of Cosines formula

Plug the values into the formula: $$a^2 = 17^2 + 27^2 - 2\cdot 17\cdot 27\cdot \cos 100.65^{\circ}$$
08

Calculate the value of a

Calculate the value of a: $$a^2 \approx 17^2 + 27^2 - 2\cdot 17\cdot 27\cdot \cos 100.65^{\circ}$$ $$a \approx \sqrt{17^2 + 27^2 - 2\cdot 17\cdot 27\cdot \cos 100.65^{\circ}} \approx 25.75 \mathrm{~cm}$$
09

Summarize the results

In the triangle ABC, we have found the following: $$A \approx 100.65^{\circ}, B \approx 34.35^{\circ}, C = 45^{\circ}$$ $$AB \approx 25.75 \mathrm{~cm}, BC = 17 \mathrm{~cm}, AC = 27 \mathrm{~cm}$$

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Most popular questions from this chapter

In questions 16-23 solve \(\Delta \mathrm{XYZ}\) given \(Z=46^{\circ}, \mathrm{YZ}=30 \mathrm{~m}\) and \(\mathrm{XY}=25 \mathrm{~m}\).

A ship travels for \(10 \mathrm{~km}\) on a bearing of \(30^{\circ} .\) It then follows a bearing of \(60^{\circ}\) for \(20 \mathrm{~km}\). Calculate the distance of the ship from the starting position.

In questions 1-11 \(\Delta \mathrm{ABC}\) has a right angle at \(\mathrm{C}\). Calculate \(\mathrm{AC}\) given \(\mathrm{BC}=10 \mathrm{~cm}\) and \(A=40^{\circ}\).

For questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(\mathrm{AB}=15 \mathrm{~cm}, \mathrm{AC}=23 \mathrm{~cm}, B=57^{\circ}\)

In questions 1-11 \(\Delta \mathrm{ABC}\) has a right angle at \(\mathrm{C}\). Calculate \(\mathrm{AC}\) given \(\mathrm{BC}=12 \mathrm{~cm}\) and \(B=53^{\circ}\)
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