Higher January 2021 Paper 2R Q22
22 \(ABC\) is an isosceles triangle in a horizontal plane.
The point \(T\) is vertically above \(B\).
Diagram NOT accurately drawn
Angle \(ABC = 140°\)
\(AB = BC = 8\) cm
\(TB = 10\) cm
\(M\) is the midpoint of \(AC\).
Calculate the size of the angle between \(MT\) and the horizontal plane \(ABC\).
Give your answer correct to one decimal place.
(4)
| Scheme | Marks |
|---|---|
\(\sin\left(\dfrac{180 - 140}{2}\right) = \dfrac{MB}{8}\) oe or \(\cos\left(\dfrac{140}{2}\right) = \dfrac{MB}{8}\) oe or \(\dfrac{8}{\sin 20} = \dfrac{AC}{\sin 140}\) and \((MB^2) = 8^2 - \left(\dfrac{\text{“}15.035\text{”}}{2}\right)^2\) or \(AC = \sqrt{8^2 + 8^2 - 2 \times 8 \times 8 \times \cos 140}\;(= 15.035\ldots)\) and \((MB^2) = 8^2 - \left(\dfrac{\text{“}15.035\text{”}}{2}\right)^2\) | M1 |
\((MB =)\) 8 sin(“20”) (= 2.736) or \((MB =)\) 8 cos(“70”) (= 2.736) or \((MB) = \sqrt{8^2 - \left(\dfrac{\text{“}15.035\text{”}}{2}\right)^2}\) | M1 |
| \(\tan TMB = \dfrac{10}{\text{“}2.736\text{”}}\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 74.7 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for a correct expression with \(MB\) included, or an expression for \(MB^2\)
If using sine or cosine rule on the isosceles triangle \(ABC\), use of Pythagoras required to obtain an expression for \(MB^2\)
M1: dep 1st M1
A1: 74.65 to 74.75