Chapter 32.4 - Solved examples on Tangents and Chords

In the previous section we saw the relations between tangents and chords. In this section we will see some solved examples based on that discussion.

Solved example 32.11
In fig.32.33 (a) below, the circumcircle of ΔPQR is shown in yellow colour. ∠PQR = 80^o and ∠PRQ = 60^o. 

Fig.30.33

Red tangents are drawn at P, Q and R. The tangents intersect at 3 points to form ΔABC. Find the angles of ΔABC.
Solution:
• Given that ∠PQR = 80^oand ∠PRQ = 60^o. 
    ♦ So ∠RPQ = [180 -(80+60)] = 40^o
1. Consider the side PR. It is a chord of the yellow circle. 
• This chord subtends an angle of 80^o on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be 80^o
• That is., ∠BPR = ∠BRP = 80^o. 
• So, in ΔPRB, ∠B =  [180 -(80+80)] = 20^o. This is shown in fig.(b)
2. Consider the side QR. It is a chord of the yellow circle. 
• This chord subtends an angle of 40^o on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be 40^o
• That is., ∠QRA = ∠AQR = 40^o. 
• So, in ΔAQR, ∠A =  [180 -(40+40)] = 100^o.
3. Consider the side PQ. It is a chord of the yellow circle. 
• This chord subtends an angle of 60^o on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be 60^o
• That is., ∠CPQ = ∠CQP = 60^o. 
• So, in ΔCQP, ∠C =  [180 -(60+60)] = 60^o.

Solved example 32.12
In fig.32.34 (a) below, the circumcircle of ΔPQR is shown in yellow colour.  

Fig.32.34

Red tangents are drawn at P, Q and R. The tangents intersect at 3 points to form ΔABC. 
∠ABC = 40^o and ∠ACB = 60^o. Find the angles of ΔPQR.
Solution:
• Given that ∠ABC = 40^o and ∠ACB = 60^o.
    ♦ So ∠BAC = [180 -(40+60)] = 80^o
1. Consider ΔPRB. It is an isosceles triangle (∵ PB = RB. See Theorem 32.2)
• So base angles will be equal. We can write:
(2 × ∠BPR) + 40 = 180 ⟹ ∠BPR = ∠PRB = 70^o. This is shown in fig(b)
• Consider the side PR. It is a chord of the yellow circle. 
• The angles (at the two ends of that chord) on the tangent side is 70^o
• So this chord will subtend an angle of 70^o on the non-tangent side
• That is., ∠PQR = 70^o.
2. Consider ΔQRA. It is an isosceles triangle (∵ QA = RA. See Theorem 32.2)
• So base angles will be equal. We can write:
(2 × ∠AQR) + 80 = 180 ⟹ ∠AQR = ∠ARQ = 50^o. This is shown in fig(b)
• Consider the side QR. It is a chord of the yellow circle. 
• The angles (at the two ends of that chord) on the tangent side is 50^o
• So this chord will subtend an angle of 50^o on the non-tangent side
• That is., ∠QPR = 50^o.
3. Consider ΔQPC. It is an isosceles triangle (∵ QC = PC. See Theorem 32.2)
• So base angles will be equal. We can write:
(2 × ∠CQP) + 60 = 180 ⟹ ∠CQP = ∠QPC = 60^o. This is shown in fig(b)
• Consider the side QP. It is a chord of the yellow circle. 
• The angles (at the two ends of that chord) on the tangent side is 60^o
• So this chord will subtend an angle of 60^o on the non-tangent side
• That is., ∠QRP = 60^o.

Solved example 32.13
In the fig.32.35(a) given below, PQ, RS and TU are tangents to the circumcircle of ABC. Sort out the equal angles in the fig.

Fig.32.35

Solution:
1. Consider the side AC. It is a chord of the yellow circle. 
• This chord subtends ∠ABC on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be equal to ∠ABC
• That is., ∠PAC = ∠TCA = ∠ABC. These are shown in green color in fig(b).
2. Consider the side BC. It is a chord of the yellow circle. 
• This chord subtends ∠BAC on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be equal to ∠BAC
• That is., ∠UCB = ∠RBC = ∠BAC. These are shown in magenta color.
3. Consider the side AB. It is a chord of the yellow circle. 
• This chord subtends ∠ACB on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be equal to ∠ACB
• That is., ∠QAB = ∠SBA = ∠ACB. These are shown in cyan color.

Solved example 32.14
In the fig.32.36(a) below, O is the center of the circle and AB is the tangent to the circle through Q. Find ∠PQA

Fig.32.36

Solution:
1. Given that, the central angle of chord PQ is 100^o. So ∠PRQ = 50^o. [Theorem 27.4]
• This is shown in fig(b).
2. In ΔPQR, consider the side PQ. It is a chord of the yellow circle. 
• This chord subtends an angle of 50^o on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will also be 50^o
• That is., ∠PQA = ∠PRQ = 50^o.

Solved example 32.15
In the fig.32.37(a) below, O is the center of the circle and QR is a diameter of the circle through Q. AB is a tangent to the circle at P. If ∠BPQ = 50^o, find ∠PQR

Fig.32.37

Solution:
1. In ΔPQR, consider the side PQ. It is a chord of the yellow circle
• The tangent AB is at the end P of the chord
• Draw a tangent CD at the end Q also. This is shown in fig(b)
2. The angle at the end P on the tangent side is given to be 50^o
• So the angle at end Q on the tangent side will also be 50^o
• That is., ∠DQP = 50^o
3. But ∠DQO = 90^o (∵ the tangent CD is at the end Q of the diameter QR)
4. So we get: ∠PQR = (∠DQO - ∠DQP) = (90 - 50) = 40^o

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In the next section, we will see more details about Tangents.

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