Question 19

Let us find the missing angles step-by-step using the given information and properties of triangles.

Step 1 — Angles in triangle CUR

We know that C is a corner of the rectangle. So, angle C is 90 degrees. The diagram shows that segment CU and segment CR have a single tick mark. This means CU = CR. Since two sides are equal, triangle CUR is an isosceles triangle. The angles opposite to the equal sides are also equal. So, ∠CUR = ∠CRU. The sum of angles in any triangle is 180 degrees. Let us call ∠CUR and ∠CRU 'x'.

$x + x + 90^\circ = 180^\circ$

$2x = 180^\circ - 90^\circ$

$2x = 90^\circ$

$x = \frac{90^\circ}{2}$

$\boxed{x = 45^\circ}$

So, ∠CUR = 45° and ∠CRU = 45°.

Step 2 — Angles in triangle AUP

Let P be the point where the line from A meets the line from U. The diagram shows an angle of 56 degrees at A, which is ∠UAP. Segment AU has a single tick mark. We are given that AU = UP. So, UP also has a single tick mark. Since AU = UP, triangle AUP is an isosceles triangle. The angles opposite to the equal sides are equal. So, ∠UPA = ∠UAP. Thus, ∠UPA = 56°. The sum of angles in triangle AUP is 180 degrees.

$\angle AUP = 180^\circ - (\angle UAP + \angle UPA)$

$\angle AUP = 180^\circ - (56^\circ + 56^\circ)$

$\angle AUP = 180^\circ - 112^\circ$

$\boxed{\angle AUP = 68^\circ}$

Step 3 — Angles related to points V, R, D, and N

Let N be the point where the line from R meets the line from V. The diagram shows an angle of 68 degrees at V, which is ∠RVN. Segment VR has a single tick mark. We are given that VR = VN. So, VN also has a single tick mark. In triangle RVN, since VR = VN, it is an isosceles triangle. The angles opposite to the equal sides are equal. So, ∠VNR = ∠VRN. The sum of angles in triangle RVN is 180 degrees. Let us call ∠VNR and ∠VRN 'a'.

$a + a + 68^\circ = 180^\circ$

$2a = 180^\circ - 68^\circ$

$2a = 112^\circ$

$a = \frac{112^\circ}{2}$

$\boxed{a = 56^\circ}$

So, ∠VNR = 56° and ∠VRN = 56°. Points R, V, D are on the straight line CD. So, ∠RVD is a straight angle, which is 180 degrees. ∠RVN and ∠DVN form a linear pair.

$\angle RVN + \angle DVN = 180^\circ$

$68^\circ + \angle DVN = 180^\circ$

$\angle DVN = 180^\circ - 68^\circ$

$\boxed{\angle DVN = 112^\circ}$

Segment VD has a single tick mark. We are given that VN = VD. So, VN also has a single tick mark. This means VN = VR = VD. All segments with a single tick mark are equal. In triangle VND, since VN = VD, it is an isosceles triangle. The angles opposite to the equal sides are equal. So, ∠VND = ∠VDN. The sum of angles in triangle VND is 180 degrees. Let us call ∠VND and ∠VDN 'c'.

$c + c + 112^\circ = 180^\circ$

$2c = 180^\circ - 112^\circ$

$2c = 68^\circ$

$c = \frac{68^\circ}{2}$

$\boxed{c = 34^\circ}$

So, ∠VND = 34° and ∠VDN = 34°.

Step 4 — Angles in triangle BOF

Let O be the central point where the 90-degree angle is marked. We are given that triangle BOF is an equilateral triangle. This means all its sides are equal: OB = OF = BF. The diagram shows BF has double tick marks. So, OB and OF also have double tick marks. In an equilateral triangle, all angles are equal to 60 degrees.

$\boxed{\angle FOB = \angle FBO = \angle OFB = 60^\circ}$

Step 5 — Angles around point B and L

B is a corner of the rectangle, so angle B is 90 degrees. We found ∠FBO = 60° in Step 4. ∠OBL is the remaining part of ∠B.

$\angle OBL = \angle B - \angle FBO$

$\angle OBL = 90^\circ - 60^\circ$

$\boxed{\angle OBL = 30^\circ}$

L is a point on AB. OL is perpendicular to AB (marked with a right angle symbol). So, ∠OLB = 90°. We are given that LO is parallel to BF. BO is a transversal line. When two parallel lines are cut by a transversal, alternate interior angles are equal. So, ∠LOB = ∠FBO. Since ∠FBO = 60°, then ∠LOB = 60°.

Step 6 — Angles in triangle OPN

Let O be the central point. P and N are other points as defined in previous steps. We are given that ∠PON = 56° and ∠PNO = 90°. The sum of angles in triangle OPN is 180 degrees.

$\angle OPN + \angle PON + \angle PNO = 180^\circ$

$\angle OPN + 56^\circ + 90^\circ = 180^\circ$

$\angle OPN + 146^\circ = 180^\circ$

$\angle OPN = 180^\circ - 146^\circ$

$\boxed{\angle OPN = 34^\circ}$

Step 7 — Angles on a straight line at P

We are given that ∠APK, ∠KPO, and ∠OPN are angles on a straight line. This means their sum is 180 degrees. The diagram shows ∠APK = 44°. We found ∠OPN = 34° in Step 6.

$\angle APK + \angle KPO + \angle OPN = 180^\circ$

$44^\circ + \angle KPO + 34^\circ = 180^\circ$

$78^\circ + \angle KPO = 180^\circ$

$\angle KPO = 180^\circ - 78^\circ$

$\boxed{\angle KPO = 102^\circ}$

Step 8 — Angles in triangle KPO

Let O be the central point. P and K are other points. We are given that ∠POK = 30°. We found ∠KPO = 102° in Step 7. The sum of angles in triangle KPO is 180 degrees.

$\angle KPO + \angle POK + \angle PKO = 180^\circ$

$102^\circ + 30^\circ + \angle PKO = 180^\circ$

$132^\circ + \angle PKO = 180^\circ$

$\angle PKO = 180^\circ - 132^\circ$

$\boxed{\angle PKO = 48^\circ}$

Step 9 — Angles in triangle KAP

Let A be the bottom-left corner. P and K are other points. The diagram shows ∠KAP = 34°. We are given that ∠KPA = 44°. The sum of angles in triangle KAP is 180 degrees.

$\angle KAP + \angle KPA + \angle AKP = 180^\circ$

$34^\circ + 44^\circ + \angle AKP = 180^\circ$

$78^\circ + \angle AKP = 180^\circ$

$\angle AKP = 180^\circ - 78^\circ$

$\boxed{\angle AKP = 102^\circ}$

Step 10 — Angles on a straight line at K

We are given that ∠AKP, ∠PKO, and ∠OKL are angles on a straight line. This means their sum is 180 degrees. We found ∠AKP = 102° in Step 9. We found ∠PKO = 48° in Step 8.

$\angle AKP + \angle PKO + \angle OKL = 180^\circ$

$102^\circ + 48^\circ + \angle OKL = 180^\circ$

$150^\circ + \angle OKL = 180^\circ$

$\angle OKL = 180^\circ - 150^\circ$

$\boxed{\angle OKL = 30^\circ}$

Step 11 — Angles in triangle KOL

Let O be the central point. K and L are other points. We found ∠OKL = 30° in Step 10. L is a right angle (∠OLK = 90°). The sum of angles in triangle KOL is 180 degrees.

$\angle OKL + \angle OLK + \angle KOL = 180^\circ$

$30^\circ + 90^\circ + \angle KOL = 180^\circ$

$120^\circ + \angle KOL = 180^\circ$

$\angle KOL = 180^\circ - 120^\circ$

$\boxed{\angle KOL = 60^\circ}$

Step 12 — Congruence of triangles OKL and OBL

We are given that triangle OKL is congruent to triangle OBL by SAS condition. Let's check the conditions:

1. KL = LB: The diagram shows KL and LB both have single tick marks. So, they are equal.
2. ∠OLK = ∠OLB = 90°: L is on AB, and OL is perpendicular to AB. So, both angles are 90 degrees.
3. OL is common: OL is a side shared by both triangles. Since these three conditions are met, ΔOKL ≅ ΔOBL by SAS (Side-Angle-Side). This congruence confirms the angles we found: ∠OKL = ∠OBL = 30°. (Matches our calculations). ∠KOL = ∠BOL = 60°. (Matches our calculations).

Answer

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