Solve triangle ABC. (If an answer does not exist, enter DNE. Round your answers to one decimal place.)
a = 63, c = 50, ∠C = 53°
Well, we have side a, side c, and angle C. Time to get solving!
Given that angle C is 53°, we can use the Law of Sines to find the length of side b:
sin(A) / a = sin(C) / c
sin(A) / 63 = sin(53°) / 50
sin(A) = (63 * sin(53°)) / 50
sin(A) = 0.8
A = sin^(-1)(0.8)
A ≈ 53.1°
Now, we can use the fact that the sum of angles in a triangle is 180° to find angle B:
B = 180° - A - C
B ≈ 180° - 53.1° - 53°
B ≈ 73.9°
Therefore, the measures of angles A, B, and C are approximately 53.1°, 73.9°, and 53°, respectively.
To solve the triangle ABC, we can use the Law of Sines and the Law of Cosines.
First, let's find angle A using the Law of Sines.
Step 1: Use the Law of Sines to find sin(A):
sin(A) / a = sin(C) / c
sin(A) / 63 = sin(53) / 50
Step 2: Rearrange the equation and solve for sin(A):
sin(A) = (63 * sin(53)) / 50
sin(A) ≈ 0.832
Step 3: Find angle A by taking the inverse sine of sin(A):
A ≈ arcsin(0.832)
A ≈ 56.9°
Next, let's find angle B using the fact that the sum of the angles in a triangle is 180°:
Step 4: Find angle B:
B = 180 - A - C
B = 180 - 56.9 - 53
B ≈ 70.1°
Now, let's find side b using the Law of Cosines:
Step 5: Use the Law of Cosines to find b:
b^2 = a^2 + c^2 - 2ac * cos(B)
b^2 = 63^2 + 50^2 - 2 * 63 * 50 * cos(70.1)
Step 6: Solve for b by taking the square root of both sides:
b ≈ sqrt(63^2 + 50^2 - 2 * 63 * 50 * cos(70.1))
b ≈ sqrt(3969 + 2500 - 6300 * cos(70.1))
Finally, we can calculate the value of side b:
Step 7: Calculate b:
b ≈ sqrt(3969 + 2500 - 6300 * cos(70.1))
b ≈ sqrt(6469 - 6300 * cos(70.1))
b ≈ sqrt(6469 - 6300 * 0.340)
b ≈ sqrt(6469 - 2142)
b ≈ sqrt(4327)
b ≈ 65.8
So, the triangle ABC has angles A ≈ 56.9°, B ≈ 70.1°, and C = 53°, and sides a = 63, b ≈ 65.8, and c = 50.
To solve triangle ABC, we can use the Law of Sines and the Law of Cosines.
First, let's identify the given information:
Side a = 63
Side c = 50
Angle C = 53°
Step 1: Find angle A using the Law of Sines.
The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant.
So, we have:
sin A / a = sin C / c
Substituting the given values:
sin A / 63 = sin 53° / 50
To find sin A, cross multiply and solve for sin A:
50 * sin A = 63 * sin 53°
sin A can be found by dividing both sides by 50:
sin A = (63 * sin 53°) / 50
Now, using the inverse sine function, we can find the value of angle A:
A = sin^(-1)((63 * sin 53°) / 50)
Step 2: Find angle B.
Since the sum of angles in any triangle is 180°, we have:
Angle B = 180° - Angle A - Angle C
Substituting the values, we get:
Angle B = 180° - Angle A - 53°
Step 3: Find side b using the Law of Sines.
We can use the relationship between angle B and side b:
sin B / b = sin A / a
Substituting the known values:
sin B / b = sin(A) / 63
To find sin B, we cross multiply and solve for sin B:
sin B = (b * sin A) / 63
Now, using the inverse sine function, we can find the value of angle B:
B = sin^(-1)((b * sin A) / 63)
Step 4: Find side b using the Law of Cosines.
The Law of Cosines states that the square of one side of a triangle is equal to the sum of the squares of the other two sides, minus twice their product, multiplied by the cosine of the included angle.
So, we have:
b^2 = a^2 + c^2 - 2ac * cos B
Substituting the given values:
b^2 = 63^2 + 50^2 - 2 * 63 * 50 * cos B
Solving for b, we take the square root:
b = sqrt(63^2 + 50^2 - 2 * 63 * 50 * cos B)
Now we have found angle A, angle B, and side b. We can substitute the values we obtained and round them to one decimal place:
to find A,
sinA/63 = sin53°/50
having C and A, B is easy since A+B+C=180.
get b using law of sines or law of cosines.