Can you show us how to use law of cosines to show that theta equals inverse cosine of (a^2 + b^2 -484)/(2ab) where a^2=(7+xcos a)^2 + (28-xsin a)^2 and b^2=(7+xcos a)^2 + (xsin a -6)^2? Thanks
Lisa, saw that one last night,
confused about the angle 'a'
in xcosa and xsina.
secondly where does x come in ?
describe your triangle.
I have a triangle with sides a and b and Ø as its contained angle.
Sure! To show that θ equals the inverse cosine of (a^2 + b^2 - 484)/(2ab), we can begin by simplifying the given expressions for a^2 and b^2 as follows:
a^2 = (7 + xcos θ)^2 + (28 - xsin θ)^2
= 49 + 14xcos θ + x^2cos^2 θ + 784 - 56xsin θ + x^2sin^2 θ
b^2 = (7 + xcos θ)^2 + (xsin θ - 6)^2
= 49 + 14xcos θ + x^2cos^2 θ + x^2sin^2 θ - 12xsin θ + 36
Now, we can use the Law of Cosines, which states that for any triangle with sides of lengths a, b, and c, and angle θ opposite the side of length c:
c^2 = a^2 + b^2 - 2abcos θ
Comparing this with the given expression for b^2, we can see that c^2 = b^2, so we substitute:
b^2 = a^2 + b^2 - 2abcos θ
Now, we can simplify the equation by canceling out the common terms:
0 = a^2 - 2abcos θ
Next, we isolate the cosine term by moving the a^2 to the other side of the equation:
2abcos θ = a^2
Now, we can solve for cos θ:
cos θ = a^2 / (2ab)
Simplifying further:
cos θ = a / 2b
Finally, we can substitute the given expression for a:
cos θ = (7 + xcos θ) / (2b)
To solve for θ, we take the inverse cosine of both sides:
θ = cos^(-1)((7 + xcos θ) / (2b))
And there you have it! The expression for θ using the Law of Cosines.