Applications of trigonometric functions
Solve the missing parts of the triangle satisfying
a=125
b=150
alpha=55 degrees
There are two solutions
a/sin alpha = b/sin beta
125/sin 55 = 150/sin beta
sin beta = .819/125 * 150 = .9828
beta = 79.25 deg or 100.75 deg
gamma = 180 - (alpha+beta)
= 45.75 or 24.25
sin gamma = .716 or .411
c / sin gamma = 152.6
c = 152.6*.716 or 152.6*.411
= 109.26 or 62.72
To solve for the missing parts of the triangle, we can use the trigonometric functions sine, cosine, and tangent. Let's find the missing side and angle.
1. Use the Law of Cosines to find side c:
c^2 = a^2 + b^2 - 2ab*cos(alpha)
c^2 = 125^2 + 150^2 - 2(125)(150)*cos(55)
c^2 = 15625 + 22500 - 37500*cos(55)
c^2 = 38125 - 37500*cos(55)
c^2 = 38125 - 37500*0.57358 (rounded cosine value for 55 degrees)
c^2 ≈ 38125 - 21520.5
c^2 ≈ 16504.5
c ≈ √16504.5
c ≈ 128.39 (rounded to two decimal places)
So, we find that side c is approximately equal to 128.39.
2. Use the Law of Sines to find the other two angles:
sin(alpha) / a = sin(beta) / b
sin(beta) = (b * sin(alpha)) / a
sin(beta) = (150 * sin(55)) / 125
sin(beta) ≈ 1.02181 (rounded to five decimal places)
Since sine values range from -1 to 1, this is not a valid solution.
However, since there are two solutions, let's find the second solution by using the inverse sine function:
beta = arcsin(1.02181)
beta ≈ 46.407 degrees (rounded to three decimal places)
Therefore, the second solution is beta ≈ 46.407 degrees.
In summary, the missing parts of the triangle are:
- Side c ≈ 128.39 (rounded to two decimal places)
- Angle beta ≈ 46.407 degrees (rounded to three decimal places)
To solve the missing parts of the triangle, we can use the trigonometric functions sine, cosine, and tangent. Given that triangle ABC has sides a, b, and c, with angles alpha, beta, and gamma, we can use the following trigonometric equations:
1. Law of Sines: sin(alpha)/a = sin(beta)/b = sin(gamma)/c
2. Law of Cosines: c^2 = a^2 + b^2 - 2ab * cos(gamma)
3. Pythagorean identity: sin^2(x) + cos^2(x) = 1
In this particular problem, we are given a, b, and alpha, and we need to find the missing parts of the triangle. Let's go through the solution step-by-step.
1. Start by using the Law of Sines to find the remaining angles beta and gamma:
- sin(beta)/b = sin(alpha)/a
- sin(beta)/150 = sin(55)/125
- sin(beta) = (150 * sin(55))/125
- sin(beta) ≈ 0.8607
- beta ≈ sin^(-1)(0.8607)
- beta ≈ 59.65 degrees
Since there are two possible solutions, we have found one of them.
2. To find gamma, use the fact that the sum of all angles in a triangle is 180 degrees:
- gamma = 180 - alpha - beta
- gamma ≈ 180 - 55 - 59.65
- gamma ≈ 65.35 degrees
We have found the second solution by subtracting the angles from 180 degrees.
3. Now, use the Law of Cosines to find the length of side c:
- c^2 = a^2 + b^2 - 2ab * cos(gamma)
- c^2 = 125^2 + 150^2 - 2 * 125 * 150 * cos(65.35)
- c^2 ≈ 38127.73
- c ≈ √(38127.73)
- c ≈ 195.28 (approx.)
Therefore, the two possible solutions for the missing parts of the triangle are:
Solution 1:
- beta ≈ 59.65 degrees
- gamma ≈ 65.35 degrees
- c ≈ 195.28
Solution 2:
- beta ≈ 120.35 degrees (180 - 59.65)
- gamma ≈ 65.35 degrees
- c ≈ 195.28
These solutions satisfy the given values of a, b, and alpha.