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Chapter 10: Problem 5

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

Short Answer

Expert verified
Answer: In the right triangle ABC, the length of side AB is approximately 7.31 cm, the length of side BC is approximately 15.44 cm, and the measure of angle C is 65°.

Step by step solution

01

Find side AB using sine function

To find side AB, we use the sine function with the given angle B and the given side AC. The formula is as follows: AB = AC * sin(B) Plugging in our given values, we get: AB = 17.2 * sin(25°) Approximately, AB = 7.31 cm
02

Find side BC using cosine function

To find side BC, we use the cosine function with the same angle B and side AC. The formula reads: BC = AC * cos(B) Now, we plug in our values: BC = 17.2 * cos(25°) Approximately, BC = 15.44 cm
03

Find angle C

Since the sum of angles in a triangle adds up to 180 degrees and we know that angle A is a right angle (90°), and angle B is 25°, we can find angle C by subtracting these angles from 180°. C = 180° - A - B C = 180° - 90° - 25° C = 65° We have now solved the triangle. The right triangle ABC has sides AB ≈ 7.31 cm, BC ≈ 15.44 cm and angle C is 65°.

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Most popular questions from this chapter

A tower has a bearing of \(37^{\circ} 00^{\prime}\) when measured from a point \(\mathrm{O}\), and is \(973 \mathrm{~m}\) distant from \(\mathrm{O}\). A chimney has a bearing of \(100^{\circ} 30^{\prime}\) when measured from \(\mathrm{O}\) and is \(1042 \mathrm{~m}\) distant from O. Calculate the distance from the tower to the chimney.

For questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(B=42^{\circ}, C=93^{\circ}, \mathrm{BC}=13 \mathrm{~cm}\)

In questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(\mathrm{AC}=105 \mathrm{~cm}, \mathrm{AB}=76 \mathrm{~cm}, A=29^{\circ}\)

In questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(\mathrm{AB}=69 \mathrm{~cm}, \mathrm{BC}=52 \mathrm{~cm}, \mathrm{AC}=49 \mathrm{~cm}\)

In questions 16-23 solve \(\Delta \mathrm{XYZ}\) given \(\mathrm{XY}=17 \mathrm{~cm}, \mathrm{YZ}=23 \mathrm{~cm}\) and \(\mathrm{XZ}=31 \mathrm{~cm}\).
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