In triangle ABC, b=14, c=15, and cos a= 3/5.
Find
a)side a
b) Sin B
Help please?
In this notation, small letters are usually used to represent sides, and capital letters to indicate the angles at the vertex,
so yours : b=14, c=15, cos A = 3/5
direct application of the cosine law
a^2 = 14^2 + 15^2 - 2(14)(15)cos A
= ....
once you have a
use the Sine Law to find angle B
ok. I have side A.
How do you find angle B?
sin B/b = sin A/a
To find side a and sin B in triangle ABC, we can use the Law of Cosines and the Law of Sines.
a) To find side a, we can use the Law of Cosines, which states that in a triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of the magnitudes of those two sides multiplied by the cosine of the included angle.
In this case, we are given side b = 14, side c = 15, and cos a = 3/5. We can use the Law of Cosines to find side a:
a^2 = b^2 + c^2 - 2bc * cos a
a^2 = 14^2 + 15^2 - 2 * 14 * 15 * (3/5)
a^2 = 196 + 225 - 504
a^2 = 73
Taking the square root of both sides, we find:
a = √73
So side a is approximately √73.
b) To find sin B, we can use the Law of Sines, which states that in a triangle, the ratio of the length of a side to the sine of the opposite angle is constant. In other words:
a/sin A = b/sin B = c/sin C
In this case, we are given side b = 14 and side c = 15. We already calculated side a to be approximately √73.
Using the Law of Sines, we can find sin B:
a/sin A = b/sin B
√73 / sin A = 14 / sin B
Since sin A is not given, we need to find it. We can use the cosine formula for a right triangle:
cos A = adjacent/hypotenuse = b/c = 14/15
Using the Pythagorean identity sin^2 A + cos^2 A = 1, we can solve for sin A:
sin A = √(1 - cos^2 A)
sin A = √(1 - (14/15)^2)
sin A = √(1 - 196/225)
sin A = √(15/225)
sin A = √(1/15)
sin A = 1/√(15)
sin A = √15/15
Now that we know sin A, we can solve for sin B:
√73 / (√15/15) = 14 / sin B
Cross-multiplying, we get:
sin B = (14 * √15) / ( √73)
So sin B is approximately (14 * √15) / ( √73).
Therefore, side a is approximately √73, and sin B is approximately (14 * √15) / ( √73).