Category Archives: Finding an angle

Clock (Geometry Question)

At what time after 2pm will the minute hand over take the hour hand?

Let the distance the minute hand moves be x^\circ. Then the distance the hour hand moves is x-60^\circ (The angle between the hands at 2pm is 60^\circ)

The rate the hour hand moves is R_H=\frac{30}{60}=\frac{1}{2}^\circ per minute, and the rate the minute hand moves is R_M=\frac{360}{60}=6^\circ per minute.

The time is the distance divided by the rate,

    \begin{equation*}\frac{x}{6}=\frac{x-60}{\frac{1}{2}}\end{equation}

    \begin{equation*}\frac{x}{2}=6x-360\end{equation}

    \begin{equation*}360=\frac{11x}{2}\end{equation}

    \begin{equation*}x=65\frac{5}{11}^\circ\end{equation}

As the minute hand moves 65\frac{5}{11}^\circ, the change in time is 65\frac{5}{11}^\circ \div 6=10 minutes and 54 seconds. Hence the time is 2:11pm

Leave a Comment

Filed under Finding an angle, Geometry, Puzzles

Circle Geometry Question

In the above diagram O is the centre of the larger circle. A, B,D and E are points on the circumference of the larger circle. A, C, E and 0 are points on the circumference of the smaller circle. Show that \angle{CAB}=\angle{ABC}. AB, AC and BC are straight lines.

AO=OB (radii of the larger circle)

At a line from O to E (it is also a radius of the larger circle)

Let \angle{CAB}=\alpha.

ACEO is a cyclic quadrilateral.

Hence, \angle{CED}=180-\alpha (AECO is a cyclic quadrilateral)

As CB is a straight line \angle{OEB}=180-(180-\alpha)=\alpha.

\Delta OEB is an isosceles triangle.

Therefore, \angle{ABC}=\alpha

Therefore \angle{ABC}=\angle{CAB}

Leave a Comment

Filed under Circle Theorems, Finding an angle, Geometry, Year 11 Specialist Mathematics

Geometry Problem

This problem is from Geometry Snacks by Ed Southall and Vincent Pantaloni – it’s a great book.

Two squares are constructed such that three vertices are collinear as shown. Find the value of the marked angle.

I started by marking in the right angles. And I added the diagonal of the larger square (pink line).

Because there are right angles at O and P, we know there is a circle, which has the diagonal of the square as its diameter (see second image below).

\angle{RSP} is 45^{\circ} (Angle between the diagonal and side of a square)

PORS is a cyclic quadrilateral.

In cyclic quadrilaterals opposite angles are supplementary.

Hence, \angle{ROP}=180^{\circ}-45^{\circ}=135^{\circ}

As \angle{ROS}=90^{\circ}, \angle{SOP} must be 45^{\circ}

Leave a Comment

Filed under Finding an angle, Geometry