in triangle ABC b=6 c=8 and angle A=60degree find the area of a triangle.let AD be the bisector of A where D lies on BC and AD=K??
Show that AD=24root3/7...
To find the area of triangle ABC, we can use the formula A = (1/2) * base * height, where the base and height are known.
First, let's find the height of triangle ABC. Draw a perpendicular line from vertex A to BC, and let's call the point of intersection E. This perpendicular line represents the height of the triangle.
Since angle A is 60 degrees and AD is the bisector of angle A, triangle ADE is a right triangle with a 30-degree angle. Triangle ADE is a special 30-60-90 triangle, where the ratio of the sides is 1:√3:2. Since DE represents the height of triangle ABC, we can determine its length.
In triangle ADE, the shortest side is AD = K, the longer side is DE, and the hypotenuse is AE. Since we know that DE is √3 times the shortest side (AD), we have:
DE = √3 * AD
DE = √3 * K
Now, let's find the length of AE. In triangle ABC, we have AB = c = 8 and AC = b = 6. Using the Law of Cosines, we can find the length of AE:
AE^2 = AB^2 + BE^2 - 2 * AB * BE * cos(A)
AE^2 = 8^2 + BE^2 - 2 * 8 * BE * cos(60°)
AE^2 = 64 + BE^2 - 16 * BE * cos(60°)
Since BE = BC - EC, and BC = 8 and EC = b - BE, we substitute the known values:
AE^2 = 64 + BE^2 - 16 * BE * cos(60°)
AE^2 = 64 + BE^2 - 16 * BE * cos(60°)
AE^2 = 64 + (8 - BE)^2 - 16 * (8 - BE) * cos(60°)
Expanding the equation:
AE^2 = 64 + 64 - 16 * 8BE + BE^2 -16 * 8 + 16BE * 0.5
AE^2 = 128 + BE^2 - 128BE + BE^2 - 64 + 8BE
Combining like terms:
AE^2 = 2BE^2 - 120BE + 64
Next, we need to find BE. We can use the Law of Cosines again in triangle ABC to find the value of cos(60°):
cos(60°) = (b^2 + c^2 - a^2) / (2bc)
cos(60°) = (6^2 + 8^2 - a^2) / (2 * 6 * 8)
cos(60°) = (36 + 64 - a^2) / 96
Since angle A = 60°, we know that angle B + angle C = 180° - angle A = 120°. Thus, angle B and angle C are equal. Let's call them x:
2x = 120°
x = 60°
So, angle B = angle C = 60°.
Now, we can substitute cos(60°) with its value in the equation for AE:
AE^2 = 2BE^2 - 120BE + 64
(64 + BE^2 - 128BE + BE^2 - 64 + 8BE) = 2BE^2 - 120BE + 64
2BE^2 - 120BE + 64 = 2BE^2 - 120BE + 64
Now, let's solve for BE:
2BE^2 - 120BE + 64 = 0
BE^2 - 60BE + 32 = 0
We can solve this quadratic equation using the quadratic formula:
BE = (-b ± √(b^2 - 4ac)) / 2a
BE = (-(-60) ± √((-60)^2 - 4 * 1 * 32)) / (2 * 1)
BE = (60 ± √(3600 - 128)) / 2
BE = (60 ± √(3472)) / 2
BE = (60 ± 40.28) / 2
Now, we have two solutions for BE:
BE1 = (60 + 40.28) / 2 = 50.14
BE2 = (60 - 40.28) / 2 = 9.72
Since BE cannot be longer than the base BC, we take BE = 9.72.
Now that we have the length of DE and BE, we can find the height using DE = √3 * K:
9.72 = √3 * K
To solve for K, we divide both sides of the equation by √3:
K = 9.72 / √3
To rationalize the denominator, we multiply both the numerator and denominator of the right side by √3:
K = (9.72 / √3) * (√3 / √3)
K = (9.72 * √3) / (3)
K = (9.72 * √3) / 3
Simplifying further, we have:
K = 3.24 * √3
Therefore, AD = K = 3.24 * √3
To find the exact value of AD, we multiply 3.24 by the square root of 3:
AD = 3.24 * √3 ≈ 5.61
Therefore, AD ≈ 5.61.
Hence, the length of AD is approximately 5.61.