Question 9
(a). \(\frac{T}{\sin 90^o}\) = \(\frac{120}{sin 135^o}\) and found T = 169.71N
(b) \(\frac{R}{\sin 135^o}\) = \(\frac{120}{\sin 135^o}\)
R = 120N
Theory
Question details
- Exam body
- WAEC
- Subject
- Further Mathematics
- Year
- 2019
- Question no.
- #9
- Type
- Theory