ABX and ACX are congruent triangles (you should know how to prove using RH congruency).

So BX = CX = ½BC = ½(8cm) = 4cm
(X is the midpoint of BC)

Next, look at the right-angled triangle CAX first.

Recall that ABC is an equilateral triangle so ∠AXC is 60°

∠CAX = 180° - 90° - 60° = 30°
(Angle sum of triangle = 180°)


Then, use Pythagoras' Theorem on it.

AX ² + XC ² = AC ²

AX ² + (4cm)² = (8cm)²

AX ² + 16cm² = 64cm²

AX ² = 48cm²

AX = √48 cm = √(16x3) cm = √16√3 cm = 4√3 cm


Next, we look at right-angled triangle AXY


Using trigo,

tan ∠XAY = opp/adj = XY / AX = (XC + CY) / AX

tan ∠XAY = (4cm + 4cm) / 4√3 cm = 2/√3 cm

∠XAY = tan-¹ (2/√3)

Since ∠XAY = ∠CAX + ∠CAY , (adjacent angles)

Then ∠CAY = <∠XAY - ∠CAX

= tan-¹ (2/√3) - 30°

b)

4cos² A = sin 2A

4cos² A = 2 sin A cos A

4cos² A - 2 sin A cos A = 0

2 cos² A - sin A cos A = 0

(cos A)(2cos A - sin A) = 0

cos A = 0 or 2 cos A - sin A = 0

cos A = 0 or 2 cos A = sin A

cos A = 0 or sin A / cos A = tan A = 2


A = cos-¹ (0)

(repeats every π radian, try sketching or plotting the graph to visualise it)

Or

A = tan-¹ (2)

(positive, so look at the first and third quadrant)


Basic angle for cos-¹ (0) = π/2 rad
Basic angle for tan-¹(2) ≈ 1.107 rad


Since -π < A < π,


A = (π/2 - 2π) rad, π/2 rad

Or

A ≈ (-π + 1.107) rad, 1.107 rad

(Recall that the tangent function is periodic and repeats every π radians)


A = -π/2 rad , π/2 rad

Or

A ≈ -2.034 rad, 1.107 rad
A ≈ -2.03 rad, 1.11 rad (3s.f)


Note that in case you're unsure which values satisfy the range, use the following as reference :


π rad ≈ 3.1415926543... rad