solve traingle ABC if A=63 degrees 10minutes, b=18 and a=17. Mark all solutions clearly.
SO
A=63
a=17
B=?
b=18
C=?
c=?
using law of sines
sine 63/17=sinB/18
SinB= sin63x18/17
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B=70.6 degrees '
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to find C
C= 180-63-70.6
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C= 46.4 degrees '
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using law of cosines
c^2=17^2+18^2-2(17)(18)(cos(46.4))
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c=13.8 '
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You ignored the 10 minutes of the 63º angle
so A = 63.1667º which would slightly alter your angles.
However I tested your solution for the angles if A = 63 and the angles are correct for that version
your value of c is correct for A=63
why did you use the cosine law?
The Sine Law would have been easier.
To solve triangle ABC, we are given:
Angle A = 63 degrees 10 minutes
Side b = 18
Side a = 17
To find the solution, we will use the Law of Cosines, which states that for any triangle with sides a, b, and c, and angle C opposite side c:
c^2 = a^2 + b^2 - 2ab * cos(C)
Let's start by finding angle B using the Law of Cosines:
b^2 = a^2 + c^2 - 2ac * cos(B)
Plugging in the given values:
(18)^2 = (17)^2 + c^2 - 2(17)(c) * cos(B)
324 = 289 + c^2 - 34c * cos(B)
0 = c^2 - 34c * cos(B) + 35
Now, let's try to find the value of c using a quadratic equation. The quadratic equation is:
c = (-b ± √(b^2 - 4ac)) / 2a
Substituting the values into the quadratic equation:
c = (-(-34) ± √((-34)^2 - 4(1)(35))) / (2 * 1)
c = (34 ± √(1156 - 140)) / 2
c = (34 ± √1016) / 2
c = (34 ± 31.906) / 2
Now we have two possible values for c:
c1 = (34 + 31.906) / 2 = 65.906 / 2 = 32.953
c2 = (34 - 31.906) / 2 = 2.094 / 2 = 1.047
Since we can't have a negative length for a side, we discard c2 = 1.047.
Now that we have the value of c, we can find angle B using the Law of Cosines:
b^2 = a^2 + c^2 - 2ac * cos(B)
(18)^2 = (17)^2 + (32.953)^2 - 2(17)(32.953) * cos(B)
324 = 289 + 1083.490 + (-1116.631) * cos(B)
0 = -805.510 - (-1116.631) * cos(B)
1116.631 * cos(B) = 805.510
cos(B) = 805.510 / 1116.631
cos(B) ≈ 0.721
Now, we can find angle B by taking the inverse cosine of 0.721:
B ≈ cos^(-1)(0.721)
Using a calculator, we find:
B ≈ 44.5 degrees
Now, to find angle C, we can use the fact that the sum of the angles in a triangle is 180 degrees:
C = 180 - A - B
C = 180 - 63° 10' - 44.5°
To perform the calculation, we need to convert the given angle A from degrees and minutes to decimal degrees:
A = 63° + (10'/60')
A ≈ 63.167 degrees
C = 180 - 63.167 - 44.5
C ≈ 180 - 63.167 - 44.5
C ≈ 72.333 degrees
So, the possible solution for triangle ABC is:
A = 63° 10'
B ≈ 44.5°
C ≈ 72.333°
a = 17
b = 18
c ≈ 32.953
To solve the triangle ABC, we can use the law of sines, which states that the ratio of the length of a side to the sine of the opposite angle is the same for all three sides of a triangle.
In this case, we are given angle A as 63 degrees 10 minutes (63° 10'), side b as 18, and side a as 17.
Here are the steps to solve the triangle:
1. Convert the given angle A from degrees and minutes to decimal degrees:
A = 63° 10' = 63 + (10/60) = 63.167 degrees
2. Apply the law of sines to find the measure of angle B:
sin(B) / b = sin(A) / a
sin(B) / 18 = sin(63.167) / 17
We can cross-multiply and solve for sin(B):
sin(B) = (18 * sin(63.167)) / 17
Taking the inverse sine of sin(B) will give us the measure of angle B:
B = arcsin[(18 * sin(63.167)) / 17]
After calculating arcsin, we get B ≈ 86.546 degrees
3. To find the measure of angle C, we can use the fact that the sum of angles in a triangle is always 180 degrees:
C = 180 - A - B
= 180 - 63.167 - 86.546
After calculation, we get C ≈ 30.287 degrees
Now we have solved the triangle ABC.
Angle A = 63° 10'
Angle B ≈ 86.546 degrees
Angle C ≈ 30.287 degrees
Side a = 17
Side b = 18