CIE iGCSE Maths E4.6 Angles Exam Style Practice Questions- Paper 2
Question
Three towns, $A,B$ and $C$, are equidistant from each other.
The bearing of $C$ from $A$ is 104$^\circ.$
Calculate the bearing of $B$ from $C.$
▶️Answer/Explanation
$224$
Towns \( A \), \( B \), and \( C \) form an equilateral triangle , all angles are \( 60^\circ \) (since they’re equidistant).
The bearing of \( C \) from \( A \) is \( 104^\circ \).
$104^\circ + 60^\circ = 164^\circ$
Bearings are measured clockwise from north.
Since we know \( A \)’s bearing from \( C \) is \( 164^\circ \) and the triangle is equilateral
The bearing of \( B \) from \( C \) is \( 60^\circ \) counterclockwise from \( 164^\circ \)
$164^\circ + 60^\circ = 224^\circ$
Question
The diagram shows 5 kites that are congruent to kite $ABCD$
Each kite is joined to the next kite along one edge.
Angle $DAB=40^{\circ}$ and $DCE$ is a straight line.
Find the value of $x.$
▶️Answer/Explanation
$145$
Properties of a kite:
1. Two pair of adjacent sides are equal.
$
\begin{aligned}
& A C=A D \\
& B C=B D
\end{aligned}
$
2. One pair of opposite angles (obtuse) are equal.
$
\angle C=\angle D
$
3. Sum of interior angles of a kite is 360 degrees.
$DCE$ is straight line
It means at point $C$ Sum of all angles will be $180^{\circ}$ and it divided by 6 equal angles.
$\begin{aligned} & \angle A+\angle B+\angle C+\angle D=360^{\circ} \\ & 40^{\circ}+x+30+x=360 \\ & 2 x=360-40-30 \\ & x=\frac{290}{ 2}=145^{\circ}\end{aligned}$