Math  /  Geometry

QuestionSolve the triangle. a=63,b=22,c=48a=63, b=22, c=48 AA \approx \square BB \approx \square { }^{\circ} C40C \approx 40^{\circ} (Do not round until the final answer. Then round to the nearest degree as needed.)

Studdy Solution

STEP 1

1. We are given a triangle with sides a=63 a = 63 , b=22 b = 22 , and c=48 c = 48 .
2. We need to find the angles A A , B B , and C C .
3. Angle C C is approximately 40 40^\circ .
4. Use the Law of Cosines and Law of Sines to find the unknown angles.

STEP 2

1. Use the Law of Cosines to find angle C C .
2. Use the Law of Sines to find angle A A .
3. Use the sum of angles in a triangle to find angle B B .

STEP 3

Use the Law of Cosines to find angle C C . The Law of Cosines states:
c2=a2+b22abcos(C) c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
Substitute the given values:
482=632+22226322cos(C) 48^2 = 63^2 + 22^2 - 2 \cdot 63 \cdot 22 \cdot \cos(C)

STEP 4

Calculate the left-hand side and simplify:
2304=3969+48426322cos(C) 2304 = 3969 + 484 - 2 \cdot 63 \cdot 22 \cdot \cos(C)
2304=44532772cos(C) 2304 = 4453 - 2772 \cdot \cos(C)

STEP 5

Rearrange to solve for cos(C) \cos(C) :
2772cos(C)=44532304 2772 \cdot \cos(C) = 4453 - 2304
2772cos(C)=2149 2772 \cdot \cos(C) = 2149
cos(C)=21492772 \cos(C) = \frac{2149}{2772}

STEP 6

Calculate C C using the inverse cosine function:
C=cos1(21492772) C = \cos^{-1}\left(\frac{2149}{2772}\right)
C40 C \approx 40^\circ

STEP 7

Use the Law of Sines to find angle A A . The Law of Sines states:
asin(A)=csin(C) \frac{a}{\sin(A)} = \frac{c}{\sin(C)}
Substitute the known values:
63sin(A)=48sin(40) \frac{63}{\sin(A)} = \frac{48}{\sin(40^\circ)}

STEP 8

Rearrange to solve for sin(A) \sin(A) :
sin(A)=63sin(40)48 \sin(A) = \frac{63 \cdot \sin(40^\circ)}{48}

STEP 9

Calculate A A using the inverse sine function:
A=sin1(63sin(40)48) A = \sin^{-1}\left(\frac{63 \cdot \sin(40^\circ)}{48}\right)
Calculate the value:
A82 A \approx 82^\circ

STEP 10

Use the sum of angles in a triangle to find angle B B :
A+B+C=180 A + B + C = 180^\circ
82+B+40=180 82^\circ + B + 40^\circ = 180^\circ

STEP 11

Solve for B B :
B=1808240 B = 180^\circ - 82^\circ - 40^\circ
B58 B \approx 58^\circ
The angles of the triangle are approximately:
A82 A \approx 82^\circ B58 B \approx 58^\circ C40 C \approx 40^\circ

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