MATHEMATICS
MORE EXERCISES
Tangents to circles.

Question  1
Complete the following statement : The angle
between thetangent to a circle and the chord
through the point of contact . . . .

Consider the diagram above and then
complete the following statements :
1.1   ∠PQT = . . .
1.2   ∠PQV = . . .
1.3   ∠QVS = . . .
1.4   ∠QSV = . . . [ A 1. ]

Question  2

In the diagram ABC is a tangent to
circle O. ∠DBC = 34° and ∠ABE = 73°

2.1   Calculate ∠BOD. [ A 2.1 ]
2.2   Calculate ∠OBD. [ A 2.2 ]
2.3   Prove EB = ED. [ A 2.3 ]

Question  3

In the diagram AE and BF are tangents
to a circle.
∠DAE = 38°,  ∠DBC = 22°,  ∠CBF = 50°
ABCD.                                                           [ A 3. ]

Question  4

In the diagram APB, AQC and BRC are
tangents to a circle.
∠A = 50° and ∠C = 48°
Calculate the angles of ΔPQR. [ A 4. ]

Question  5

In the diagram ABC is a tangent
and BE ǁ CD.
Prove that ∠DCB = ∠BDE. [ A 5. ]

Question  6

In the diagram AF is a tangent
and AF ǁ BD.
Prove that ∠AEB = ∠ABD. [ A 6. ]

Question  7

In the diagram CD is a tangent and
∠D = ∠ACB = x.
Prove that AC is a diameter. [ A 7. ]

Question  8

In the diagram KL and KM are tangents.
Prove that ∠K = ∠LNM − ∠P. [ A 8. ]

$$\text{The angle between the tangent to a circle and } \\ \text{the chord through the point of contact is equal to the}\\ \text{angle in the opposite segment of the circle.} \\ \text{van die sirkel} \\ \text{} \\ \text{1.1 \angle PQT = \angle QVT } \\ \text{1.2 \angle PQV = \angle QSV} \\ \text{1.3 \angle QVS = \angle SQR} \\ \text{1.4 \angle QSV = \angle PQV}$$
[ Q 1. ]

#### 2.1

\begin{align*} \angle OBC &= 90\degree \\ \tag{tangent \perp radius} \\ \angle OBD &= 90\degree - \angle DBC \\ &= 90\degree - 34\degree \\ &= 56\degree \\ \angle OBD\ &= \angle ODB \\ \tag{OB = OD} \\ \angle BOD\ &= 180\degree - (\angle OBD + \angle ODB) \\ \tag{\sum int. \angle^{'s} of \DeltaBOD} \\ &= 180\degree - (56\degree + 56\degree) \\ &= 68\degree \\ \bold{OR} \\ \angle BED &= \angle DBC \\ \tag{tangent ABC, chord BD} \\ &= 34\degree \\ \tag{given} \\ \angle BOD &= 2 \times \angle BED \\ \tag{midpt. \angle = 2 \times \angle at circumference.} \\ &= 2 \times 34\degree \\ &= 68\degree \\ \end{align*}
[ Q 2.1 ]

#### 2.2

\begin{align*} \angle OBC &= 90\degree \\ \tag{tangent ABC \perp radius OB} \\ \angle OBD + \angle DBC amp;= 90\degree \\ \angle OBD &= 90\degree - 34\degree \\ \tag{\angle DBC = 34\degree . . . gegee} \\ &= 56\degree \\ \end{align*}
[ Q 2.2 ]

#### 2.3

\begin{align*} \angle BDE &= \angle ABE \\ \tag{tangent ABC, chord BE} \\ &= 73\degree \\ \angle OBE &= 90\degree \\ \tag{tangent ABC \perp radius OB} \\ \angle OBE + \angle ABE &= 90\degree \\ \angle OBE &= 90\degree - 73\degree \\ &= 17\degree \\ \angle EBD &= \angle EBO + \angle OBD \\ &= 17\degree + 56\degree \\ &= 73\degree \\ \angle EBD &= \angle EDB \\ \tag{both = 73\degree} \\ \therefore EB &= ED \\ \end{align*}
[ Q 2.3 ]

\begin{align*} \angle ABD &= \angle DAE \\ \tag{tangent AE, chord AD} \\ &= 38\degree \\ \\ \angle BDC &= \angle CBF \\ \tag{tangent BF, chord BC} \\ &= 50\degree \\ \angle BAD &= \angle DBF \\ \tag{tangent BF, chord BD} \\ &= (50\degree + 22\degree) \\ &= 72\degree \\ \angle BDA &= 180\degree - (\angle ABD + \angle BAD) \\ \tag{\sum int. \angle^{'s} of \DeltaABD} \\ &= 189\degree - (38\degree + 72\degree) \\ &= 70\degree \\ \angle BCD &= 180\degree - (\angle CBD + \angle BDC) \\ \tag{\sum int. \angle^{'s} of \DeltaBCD} \\ &= 180\degree - (22\degree + 50\degree) \\ &= 108\degree \\ \angle BAD &= 72\degree \\ \angle ABC &= \angle ABD + \angle DBC \\ &= 38\degree + 22\degree \\ &= 60\degree \\ \angle BCD &= 108\degree \\ \angle ADC &= \angle BDC + \angle ADB \\ &= 50\degree + 70\degree \\ &= 120\degree \\ \end{align*}
[ Q 3. ]
\begin{align*} AP &= AQ \\ \tag{tangents from A} \\ \therefore \angle A + \angle APQ + \angle AQP &= 180\degree \\ \tag{\sum int. \angle^{'s} of \DeltaAQP} \\ 50\degree + \angle APQ + \angle AQP &= 180\degree \\ 2 \times \angle APQ &= 180\degree - 50\degree \\ &= 130\degree \\ \therefore \angle APQ &= 65\degree \\ \angle PRQ &= \angle APQ \\ \tag{tangent APB, chord PQ} \\ &= 65 \degree \\ \angle CRQ &= \angle CQR \\ \tag{tangents from C} \\ \angle C + \angle CRQ + \angle CQR &=180\degree \\ \tag{\sum int. \angle'^{s} of \DeltaQRC} \\ 48\degree + 2 \times \angle CRQ &= 180\degree \\ 2 \times \angle CRQ &= 180\degree - 48\degree \\ 2 \times \angle CRQ &= 132\degree \\ \angle CRQ &= 66\degree \\ \angle RPQ &= \angle QRC \\ \tag{tangent BRC, chord RQ} \\ &= 66\degree \\ \angle RPQ + \angle PQR + \angle PRQ &= 180\degree \\ \tag{\sum int. \angle^{'s} of \DeltaPQR} \\ 66\degree + \angle PQR + 65\degree &= 180\degree \\ \angle PQR &= 180\degree - 131\degree \\ &= 49\degree \\ \end{align*} ∠RPQ = 66°, ∠PQR = 49° and ∠PRQ 65°
[ Q 4. ]
\begin{align*} \angle C &= \angle ABE \\ \tag{corresp. \angle'^{s}, BE ǁ CD} \\ \angle BDE &= \angle ABE \\ \tag{tangent AB, chord BE} \\ \therefore \angle DCB &= \angle BDE \\ \tag{both = \angle ABE} \\ \bold{OR} \\ \end{align*} $$\text{Try the algebraic method} \\ \text{This method is very useful when the relation} \\ \text{between angles have to be proved.} \\$$ \text{Let ∠ABE = x} \\ \begin{align*} \\ \angle C &= \angle ABE \\ \tag{corresp. \angle'^{s}, BE ǁ CD} \\ \angle C &= x \\ \angle BDE &= \angle ABE \\ \tag{tangent ABC, chord BE} \\ \angle BDE = x \\ \therefore \angle DCB &= \angle BDE \\ \tag{both = x} \\ \end{align*} \\
[ Q 5. ]

\text{Use the algebraic method again.} \\ \text{Let ∠FAD = x and ∠EBD = y} \\ \begin{align*} \angle D &= \angle FAD \\ \tag{alt. \angle'^{s}, FA ǁ BCD} \\ &= x \\ \angle ABE &= \angle FAD \\ \tag{tangent FA, chord AE} \\amp; &= x \\ \angle AEB &= \angle EBD + \angle ABE \\ \tag{exterior angle of \DeltaEBD} \\ &= y + x \\ \angle ABD &= \angle EBD + \angle ABE \\ &= y + x \\ \angle AEB &= \angle ABD \\ \tag{both = x + y} \\ \end{align*}
[Q 6. ]

\begin{align*} \text{Use the algebraic method again.} \\ \text{Let ∠BCD = y} \\ \angle A &= \angle BCD \\ \tag{tangent CD, chord BC} \\ &= y \\ \angle CBD &= \angle ACB + \angle CAB \\ \tag{exterior angle of \DeltaCAB} \\ &= x + y \\ \angle CBA &= \angle D + \angle BCD \\ \tag{exterior angle of \DeltaCBD} \\ &= x + y \\ \therefore \angle CBA &= \angle CBD \\ \tag{both = x + y} \\ \angle CBA + \angle CBD &= 180\degree \\ \tag{ABD is a straight line} \\ 2 \times \angle CBA &= 180\degree \\ \tag{∠CBA = ∠CBD} \\ \text{∴ AC is a diameter . . . ∠ in semi-sircle} \\ \end{align*}
[ Q 7. ]

\begin{align*} \text{Let ∠KLN = x and ∠NLM = y} \\ \therefore \angle KLM &= x + y \\ KL &= KM \\ \tag{tangents from the same point} \\ \angle KLM &= \angle KML \\ \tag{KL = KM} \\ &= x + y \\ \angle KLM &= \angle P \\ \tag{tangent KL, chord LM} \\ &= x + y \\ \angle K + \angle KLM + \angle KML &= 180\degree \\ \tag{\sum int. \angle^{s} of \DeltaKLM} \\ \angle K &= 180\degree - (\angle KLM + \angle KML) \\ &= 180\degree - 2(x+ y) \\ \text{KL is produced to Q.} \\ \angle QLM &= \angle N \\ \tag{tangent KLQ, chord LM} \\ \angle QLM + \angle KLM &= 180\degree \\ \tag{KLQ is a straight line} \\ \angle QLM + (x + y) &= 180\degree \\ \angle QLM &= 180\degree - (x + y) \\ \angle N &= 180\degree - (x + y) \\ \angle K &= 180\degree - 2(x + y) \\ &= 180\degree - (x + y) - (x + y) \\ &= \angle LNM - \angle P \\ \end{align*}
[ Q 8. ]