@PastaIsVeryUnderrated
XDD , you can also solve it using this way , I think this one should be easier, since my handwriting looks like squashed tomatoes 🤣 🤣 🤣, am typing instead.
Given:
AD ∥ BC (AD is parallel to BC)
∠B = 90° (right angle)
AB = 3 cm
AD = 6 cm
BC = 10 cm
Sooooo, let's find the length of CD:
Since AD ∥ BC, we can use similar triangles:
CD/BC = AB/AD
CD/10 = 3/6
CD = 5 cm
Finding ∠DCB and ∠ACB:
Using the right triangle BCD:
tan(∠DCB) = BC/CD = 10/5 = 2
Using right triangle ABC:
tan(∠ACB) = BC/AB = 10/3
cos(∠ACB) = AB/AC
To find AC, use the Pythagorean theorem:
AC² = AB² + BC²
AC² = 3² + 10²
AC = √(109)
cos(∠ACB) = 3/√109
Calculate::
cos(∠DCB) = CD/BC = 5/10 = 1/2
Using the addition to solve tangent
tan(∠DCB - ∠ACB) = (tan(∠DCB) - tan(∠ACB)) / (1 + tan(∠DCB)tan(∠ACB))
Substitute the values::
tan(∠DCB - ∠ACB) = (2 - 10/3) / (1 + 2*10/3)
tan(∠DCB - ∠ACB) = (6/3 - 10/3) / (1 + 20/3)
tan(∠DCB - ∠ACB) = -4/3 / (23/3)
tan(∠DCB - ∠ACB) = -4/23
Calculating cos(∠DCB) - tan(∠ACB):
cos(∠DCB) - tan(∠ACB) = 1/2 - 10/3
cos(∠DCB) - tan(∠ACB) = 3/6 - 10/3
cos(∠DCB) - tan(∠ACB) = 1/2
Therefore:
cos(∠DCB) - tan(∠ACB) = 1/2
*Bows* Ahh Thank you
