Given the indicated parts of triangle ABC with γ = 90°, approximate the remaining parts. (Round your answers for a and c to one decimal place.)
β = 51°31', b = 250.0
α = ° '
a =
c =
Given the indicated parts of triangle ABC with γ = 90°, approximate the remaining parts. (Round your answers for b and c to one decimal place.)
α = 27°50', a = 530
β = ° '
b =
c =
Given the indicated parts of triangle ABC with γ = 90°, express the third part in terms of the first two.
β, c; b
b =
Given the indicated parts of triangle ABC with γ = 90°, express the third part in terms of the first two.
α, a; b
b =
1. tanB = b/a = 250 / a,
a = 250 / tanB = 250 / tan51.52 = 198.7
A = 90 - 51.52 = 38.5 deg.
c = b / cosA = 250 / cos38.5 = 319.4.
2. B = 90 - A = 90 - 27.83 = 62.17 deg.
tanA = a / b = 530 / b,
b = 530 / tanA = 530 / tan27.83 = 1004.
c = a / sinA = 530 / sin27.83 = 1135.3.
3. sinB = b/c,
b = csinB.
4. tanA = a/b,
b = a/tanA.
Why did the triangle go to the doctor? Because it had acute angles!
For the first question:
Using the sine function, we have:
sin(β) = (b/c) = (250.0/c) = sin(51°31') = 0.7820
This gives us c = b/sin(β) = 250.0/0.7820 = 319.2 (approximated to one decimal place).
Using the Pythagorean theorem, we have:
a^2 = c^2 - b^2 = 319.2^2 - 250.0^2 = 41085.44 - 62500 = -21414.56
Since the triangle is a right triangle (γ = 90°), we know that a is the hypotenuse and therefore cannot be negative. Hence, there is no solution for the value of a.
For the second question:
Using the sine function, we have:
sin(β) = (b/c) = (b/530) = sin(27°50') = 0.4784
This gives us b = sin(β) * a = 0.4784 * 530 = 253.43 (approximated to one decimal place).
Using the Pythagorean theorem, we have:
c^2 = a^2 + b^2 = 530^2 + 253.43^2 = 280900 + 64200.96 = 345100.96
Hence, c = √(345100.96) = 586.86 (approximated to one decimal place).
For the third question:
Given β and c, we want to express b in terms of β and c. We know that sin(β) = (b/c), so multiplying both sides by c gives us b = c*sin(β).
For the fourth question:
Given α and a, we want to express b in terms of α and a. We know that sin(α) = (a/b), so multiplying both sides by b gives us b = a*sin(α).
I hope that helps, or at least makes you smile!
To solve these problems, we can use the trigonometric ratios sine, cosine, and tangent.
For the first problem:
To find angle α, we can use the sine ratio: sin α = opposite/hypotenuse
sin α = a/b
sin α = a/250.0
To find side a, we can use the cosine ratio: cos α = adjacent/hypotenuse
cos α = a/250.0
a = 250.0 * cos α
To find side c, we can use the Pythagorean theorem: a^2 + c^2 = b^2
c^2 = b^2 - a^2
c = √(b^2 - a^2)
For the second problem:
To find angle β, we can use the sine ratio: sin β = opposite/hypotenuse
sin β = b/a
sin β = 250/530
To find side b, we can use the cosine ratio: cos β = adjacent/hypotenuse
cos β = b/a
b = a * cos β
To find side c, we can use the Pythagorean theorem: a^2 + b^2 = c^2
c^2 = a^2 + b^2
c = √(a^2 + b^2)
For the third problem:
To express side b in terms of angle β and side c, we can use the tangent ratio: tan β = opposite/adjacent
tan β = c/b
b = c/tan β
For the fourth problem:
To express side b in terms of angle α and side a, we can use the tangent ratio: tan α = opposite/adjacent
tan α = b/a
b = a * tan α
To solve for the remaining parts of the triangle, we can use the trigonometric relationships between the angles and sides of a right triangle.
Case 1:
Given: β = 51°31', b = 250.0
To find: α, a, c
To find α:
We know that the sum of angles in a triangle is 180°.
Since γ = 90° and β = 51°31', we can calculate α as follows:
α = 180° - 90° - 51°31'
α ≈ 38°29'
To find a:
We can use the trigonometric relationship in a right triangle:
sin(β) = a / b
Rearranging the formula, we get:
a = b * sin(β)
a ≈ 250.0 * sin(51°31')
a ≈ 190.9 (rounded to one decimal place)
To find c:
We can use the Pythagorean theorem in a right triangle:
c^2 = a^2 + b^2
c = sqrt(a^2 + b^2)
c ≈ sqrt(190.9^2 + 250.0^2)
c ≈ 312.4 (rounded to one decimal place)
Therefore, α ≈ 38°29', a ≈ 190.9, c ≈ 312.4.
Case 2:
Given: α = 27°50', a = 530
To find: β, b, c
To find β:
We know that the sum of angles in a triangle is 180°.
Since γ = 90° and α = 27°50', we can calculate β as follows:
β = 180° - 90° - 27°50'
β ≈ 62°10'
To find b:
We can use the trigonometric relationship in a right triangle:
sin(α) = b / a
Rearranging the formula, we get:
b = a * sin(α)
b ≈ 530 * sin(27°50')
b ≈ 242.5 (rounded to one decimal place)
To find c:
We can use the Pythagorean theorem in a right triangle:
c^2 = a^2 + b^2
c = sqrt(a^2 + b^2)
c ≈ sqrt(530^2 + 242.5^2)
c ≈ 582.1 (rounded to one decimal place)
Therefore, β ≈ 62°10', b ≈ 242.5, c ≈ 582.1.
Given β, c, and b, we need to express the third part b in terms of the first two.
We know the trigonometric relationship in a right triangle: sin(β) = b / c.
Rearranging the formula, we have:
b = c * sin(β).
Therefore, b = c * sin(β).
Given α, a, and b, we need to express the third part b in terms of the first two.
We know the trigonometric relationship in a right triangle: sin(α) = b / a.
Rearranging the formula, we have:
b = a * sin(α).
Therefore, b = a * sin(α).