Category Archives: Finding an angle

December 10, 2025 · 7:54 am

Geometry Problem

Problem from *50 Maths Problems with Solutions* – Ajmal KP

The blue shaded area is the area of triangles $APO$ and $AQO$ subtract the sector $POQ$ .

We can use Heron’s law to find the area of the triangle $\Delta{ABC}$

$\begin{equation*}A=\sqrt{s(s-a)(s-b)(s-c)}\end{equation}$

where $s=\frac{a+b+c}{2}$

$\begin{equation*}A=\sqrt{20(20-16)(20-10)(20-14)}=40\sqrt{3}\end{equation}$

We also know the area of triangle $\Delta{ABC}=sr$ where $r$ is the radius of the inscribed circle.

Hence, $40\sqrt{3}=20r$ and $r=2\sqrt{3}$

We know $AP=AQ, CQ=CR$ , and $BP=BR$ – tangents to a circle are congruent.

$\begin{equation*}14-x=6+x\end{equation}$

(1) $\begin{equation*}8=2x\end{equation*}$

(2) $\begin{equation*}x=4\end{equation*}$

Area $\Delta{AQO}=\frac{1}{2}10\times 2\sqrt{3}=10\sqrt{3}$

Area $\Delta{APO}=$ Area $\Delta{AQO}$

$\begin{equation*}tan(\theta)=\frac{10}{2\sqrt{3}}\end{equation}$

$\begin{equation*}\theta=70.9^{\circ}\end{equation}$

Area of sector $OPQ=\frac{2\times70.9}{360}\pi (2\sqrt{3})^2=14.8$

Blue area = $20\sqrt{3}-14.8=19.8cm^2$

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Filed under Algebra, Area, Finding an angle, Finding an area, Geometry, Heron's Law, Interesting Mathematics, Puzzles, Radius and Semi-Perimeter, Right Trigonometry, Solving Equations, Trigonometry

Tagged as geometry puzzle

November 16, 2025 · 5:54 am

Geometry Circle Question

In the diagram below, $A, B, C$ and $D$ lie on the circle with centre $O$ . If $\angle{DBC} = 41^{\circ}$ and $\angle{ACD} = 53^{\circ}$ , determine with reasoning $\angle{BAC}$ and $\angle{AOB}$

We know $OA=OB=OD$ – radii of the circle.

Which means, $\Delta{AOB}$ is isosceles and $\angle{OAB}=\angle{OBA}$ – equal angles isosceles triangle.

$\angle{AOD}=2\angle{ACD}$ – angle at the centre twice the angle at the circumference.

$\angle{AOB}=106^{\circ}$

This means $\angle{AOB}=74^{\circ}$ – angles on a straight line are supplementary

$\angle{OAD}=\angle{ODA}=37^{\circ}$ – equal angles isosceles triangle and the angle sum of a triangle.

$\angle{DBA}=\angle{DCA}=53^{\circ}$ – angle at the circumference subtended by the same arc are congruent.

$\angle{CAD}=\angle{CBD}=41^{\circ}$ – angles at the circumference subtended by the same arc are congruent.

$\angle{OAB}=53^{\circ}$ – equal angle isosceles triangle

Hence $\angle{BAC}=12^{\circ}$

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

Tagged as circle theorems, geometry, Year 11 Specialist Mathematics

September 6, 2025 · 9:50 am

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 $2:11$ pm

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Filed under Finding an angle, Geometry, Puzzles

Tagged as geometry clock question

March 9, 2025 · 11:26 am

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 $\angle{ABC}=\angle{CAB}$

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

September 7, 2024 · 9:31 am

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).