Geometric Twins | FIO

Question 19

Find the missing angles. As per the convention that we have been following, all line segments marked with a single '|' are equal to each other and those marked with a double '|' are equal to each other, etc.

Solution

Understand the Question

To find all the unknown angles in the given figure, we use fundamental geometric properties:

Isosceles Triangle Property: Angles opposite to equal sides (indicated by matching tick marks) are equal.
Angle Sum Property: The sum of interior angles in any triangle is $180^\circ$ .
Equilateral Triangle Property: All sides are equal and all angles are $60^\circ$ .
Linear Pair / Straight Line Angle Property: Angles forming a straight line sum to $180^\circ$ .
Corner Angles: The corners of a rectangle are $90^\circ$ .

Step 1 · Find Angles in Triangle CUR

In $\Delta \text{CUR}$ , $\angle \text{C} = 90^\circ$ (corner of rectangle) and $\text{CU} = \text{CR}$ (marked with single tick). Diagram 1

Since $\Delta \text{CUR}$ is an isosceles right triangle, the angles opposite to equal sides are equal: $\angle \text{CUR} = \angle \text{CRU} = x$

By the angle sum property:

\begin{aligned} x + x + 90^\circ &= 180^\circ \\ 2x &= 90^\circ \\ x &= 45^\circ \end{aligned}

Therefore, $\angle \text{CUR} = 45^\circ$ and $\angle \text{CRU} = 45^\circ$ .

Step 2 · Find Angles in Triangle AUP

In $\Delta \text{AUP}$ , $\angle \text{UAP} = 56^\circ$ and $\text{AU} = \text{UP}$ . Diagram 2

Since $\text{AU} = \text{UP}$ , $\Delta \text{AUP}$ is isosceles: $\angle \text{UPA} = \angle \text{UAP} = 56^\circ$

Using the angle sum property in $\Delta \text{AUP}$ :

\begin{aligned} \angle \text{AUP} &= 180^\circ - (\angle \text{UAP} + \angle \text{UPA}) \\ &= 180^\circ - (56^\circ + 56^\circ) \\ &= 180^\circ - 112^\circ \\ &= 68^\circ \end{aligned}

Step 3 · Find Angles Related to Points V, R, D, and N

Given $\angle \text{RVN} = 68^\circ$ and $\text{VR} = \text{VN} = \text{VD}$ . Diagram 3

In isosceles $\Delta \text{RVN}$ with $\text{VR} = \text{VN}$ , let $\angle \text{VNR} = \angle \text{VRN} = a$ :

\begin{aligned} a + a + 68^\circ &= 180^\circ \\ 2a &= 112^\circ \\ a &= 56^\circ \end{aligned}

So, $\angle \text{VNR} = 56^\circ$ and $\angle \text{VRN} = 56^\circ$ .

Since points $\text{R, V, D}$ lie on a straight line:

\begin{aligned} \angle \text{RVN} + \angle \text{DVN} &= 180^\circ \\ 68^\circ + \angle \text{DVN} &= 180^\circ \\ \angle \text{DVN} &= 112^\circ \end{aligned}

In isosceles $\Delta \text{VND}$ with $\text{VN} = \text{VD}$ , let $\angle \text{VND} = \angle \text{VDN} = c$ :

\begin{aligned} c + c + 112^\circ &= 180^\circ \\ 2c &= 68^\circ \\ c &= 34^\circ \end{aligned}

So, $\angle \text{VND} = 34^\circ$ and $\angle \text{VDN} = 34^\circ$ .

Step 4 · Find Angles in Triangle BOF

Triangle $\text{BOF}$ is an equilateral triangle with double tick marks $\text{OB} = \text{OF} = \text{BF}$ . Diagram 4

In an equilateral triangle, all interior angles are equal to $60^\circ$ : $\angle \text{FOB} = \angle \text{FBO} = \angle \text{OFB} = 60^\circ$

Step 5 · Find Angles around Point B and L

The corner angle $\angle \text{B} = 90^\circ$ and $\angle \text{FBO} = 60^\circ$ . Diagram 5

\begin{aligned} \angle \text{OBL} &= \angle \text{B} - \angle \text{FBO} \\ &= 90^\circ - 60^\circ \\ &= 30^\circ \end{aligned}

Since $\text{LO} \parallel \text{BF}$ and $\text{BO}$ is a transversal, alternate interior angles are equal: $\angle \text{LOB} = \angle \text{FBO} = 60^\circ$ Also, $\text{OL} \perp \text{AB} \implies \angle \text{OLB} = 90^\circ$ .

Step 6 · Find Angles in Triangle OPN

In $\Delta \text{OPN}$ , given $\angle \text{PON} = 56^\circ$ and $\angle \text{PNO} = 90^\circ$ . Diagram 6

By angle sum property in $\Delta \text{OPN}$ :

\begin{aligned} \angle \text{OPN} + \angle \text{PON} + \angle \text{PNO} &= 180^\circ \\ \angle \text{OPN} + 56^\circ + 90^\circ &= 180^\circ \\ \angle \text{OPN} + 146^\circ &= 180^\circ \\ \angle \text{OPN} &= 34^\circ \end{aligned}

Step 7 · Find Angles on Straight Line at P

Angles $\angle \text{APK}$ , $\angle \text{KPO}$ , and $\angle \text{OPN}$ lie on a straight line. Diagram 7

Given $\angle \text{APK} = 44^\circ$ and $\angle \text{OPN} = 34^\circ$ :

\begin{aligned} \angle \text{APK} + \angle \text{KPO} + \angle \text{OPN} &= 180^\circ \\ 44^\circ + \angle \text{KPO} + 34^\circ &= 180^\circ \\ 78^\circ + \angle \text{KPO} &= 180^\circ \\ \angle \text{KPO} &= 102^\circ \end{aligned}

Step 8 · Find Angles in Triangle KPO

In $\Delta \text{KPO}$ , $\angle \text{POK} = 30^\circ$ and $\angle \text{KPO} = 102^\circ$ . Diagram 8

By angle sum property in $\Delta \text{KPO}$ :

\begin{aligned} \angle \text{KPO} + \angle \text{POK} + \angle \text{PKO} &= 180^\circ \\ 102^\circ + 30^\circ + \angle \text{PKO} &= 180^\circ \\ 132^\circ + \angle \text{PKO} &= 180^\circ \\ \angle \text{PKO} &= 48^\circ \end{aligned}

Step 9 · Find Angles in Triangle KAP

In $\Delta \text{KAP}$ , given $\angle \text{KAP} = 34^\circ$ and $\angle \text{KPA} = 44^\circ$ . Diagram 9

By angle sum property in $\Delta \text{KAP}$ :

\begin{aligned} \angle \text{KAP} + \angle \text{KPA} + \angle \text{AKP} &= 180^\circ \\ 34^\circ + 44^\circ + \angle \text{AKP} &= 180^\circ \\ 78^\circ + \angle \text{AKP} &= 180^\circ \\ \angle \text{AKP} &= 102^\circ \end{aligned}

Step 10 · Find Angles on Straight Line at K

Angles $\angle \text{AKP}$ , $\angle \text{PKO}$ , and $\angle \text{OKL}$ lie on a straight line. Diagram 10

\begin{aligned} \angle \text{AKP} + \angle \text{PKO} + \angle \text{OKL} &= 180^\circ \\ 102^\circ + 48^\circ + \angle \text{OKL} &= 180^\circ \\ 150^\circ + \angle \text{OKL} &= 180^\circ \\ \angle \text{OKL} &= 30^\circ \end{aligned}

Step 11 · Find Angles in Triangle KOL

In $\Delta \text{KOL}$ , $\angle \text{OKL} = 30^\circ$ and $\angle \text{OLK} = 90^\circ$ . Diagram 11

By angle sum property in $\Delta \text{KOL}$ :

\begin{aligned} \angle \text{OKL} + \angle \text{OLK} + \angle \text{KOL} &= 180^\circ \\ 30^\circ + 90^\circ + \angle \text{KOL} &= 180^\circ \\ 120^\circ + \angle \text{KOL} &= 180^\circ \\ \angle \text{KOL} &= 60^\circ \end{aligned}

Step 12 · Verify Congruence of Triangles OKL and OBL

In $\Delta \text{OKL}$ and $\Delta \text{OBL}$ :

$\text{KL} = \text{LB}$ (single tick mark)
$\angle \text{OLK} = \angle \text{OLB} = 90^\circ$ ( $\text{OL} \perp \text{AB}$ )
$\text{OL} = \text{OL}$ (common side)

Therefore, $\Delta \text{OKL} \cong \Delta \text{OBL}$ by SAS congruence criterion.

By CPCT (Corresponding Parts of Congruent Triangles): $\angle \text{OKL} = \angle \text{OBL} = 30^\circ$ $\angle \text{KOL} = \angle \text{BOL} = 60^\circ$

Answer

(i) $\angle \text{CUR} = 45^\circ$ , (ii) $\angle \text{CRU} = 45^\circ$ , (iii) $\angle \text{VRN} = 56^\circ$ , (iv) $\angle \text{VNR} = 56^\circ$ , (v) $\angle \text{AUP} = 68^\circ$ , (vi) $\angle \text{FOB} = 60^\circ$ , (vii) $\angle \text{FBO} = 60^\circ$ , (viii) $\angle \text{OFB} = 60^\circ$ , (ix) $\angle \text{DVN} = 112^\circ$ , (x) $\angle \text{VND} = 34^\circ$ , (xi) $\angle \text{VDN} = 34^\circ$ , (xii) $\angle \text{OBL} = 30^\circ$ , (xiii) $\angle \text{LOB} = 60^\circ$ , (xiv) $\angle \text{OPN} = 34^\circ$ , (xv) $\angle \text{KPO} = 102^\circ$ , (xvi) $\angle \text{PKO} = 48^\circ$ , (xvii) $\angle \text{AKP} = 102^\circ$ , (xviii) $\angle \text{OKL} = 30^\circ$ , (xix) $\angle \text{KOL} = 60^\circ$

Common Mistakes

Tick Mark Confusion: Segments with a single tick mark are all equal in length to each other, but not necessarily equal to segments with double tick marks.
Isosceles Base Angles: Always identify the vertex angle first; the two base angles opposite the equal sides are equal to each other, not the vertex angle.
Angles on a Straight Line: When multiple adjacent angles lie on a straight line, their sum is $180^\circ$ , not $90^\circ$ or $360^\circ$ .

Question 19

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