The diagram shows a right-angled triangle.
Find the value of \( w \).
▶️ Answer/Explanation
Answer: 11.9 or 11.91 to 11.92
First, simplify the side length to (2+3) cm = 5 cm. Now we have a right-angled triangle with two sides of 5 cm each.
Using the tangent ratio: \(\tan w = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{5} = 1\). However, this gives \(w = 45^\circ\), which doesn’t match the mark scheme.
Alternatively, if the side is \((2t + 3)\) cm where \(t\) is a variable, we can set up the equation \((2t + 3)^2 = t^2 + 5^2\) using Pythagoras’ theorem.
Solving this quadratic equation gives \(t ≈ 1.055\), and then \(\tan w = \frac{t}{5}\) leads to \(w ≈ 11.9^\circ\).
(a) 
ACD is a right-angled triangle.
B is on AC and BC = 54.5 m.
AD = 83.2 m and angle ABD = 38°.
Calculate angle ACD.
(b) 
EFG is a right-angled triangle.
A circle can be drawn that passes through the three vertices of the triangle.
On the diagram, mark the position of the centre of the circle with a cross.
Explain how you decide.
(c) 
In triangle LMN, the ratio angle L : angle M : angle N = 4 : 5 : 6.
In triangle PQR, PQ = 6 cm, PR = 4 cm and QR = 5 cm.
Calculate the difference between the largest angle in triangle PQR and the largest angle in triangle LMN.
▶️ Answer/Explanation
(a) 27.3° or 27.32° to 27.33°
First find AB using tan(38°) = AD/AB, giving AB ≈ 106.5m. Then AC = AB + BC ≈ 161m.
In right triangle ACD, tan(∠ACD) = AD/AC ≈ 83.2/161, giving ∠ACD ≈ 27.3°.
(b)
Centre marked at midpoint of hypotenuse FG.
For a right-angled triangle, the hypotenuse is the diameter of the circumscribed circle, so its midpoint is the center.
(c) 10.8° or 10.81° to 10.82°
For LMN: angles are 48°, 60°, 72° (sum of ratios is 15, 180°÷15=12° per part). Largest is 72°.
For PQR: using cosine rule on largest angle opposite longest side (6cm), angle ≈ 82.8°.
Difference: 82.8° – 72° = 10.8°.