Question

Construct a pair of tangents to a circle of radius 3 cm which are inclined to each other at
an angle of 60°​

Answer 1

Step-by-step explanation:

PQ and PR are the tangents to the given circle. If they are inclined at 60°,then

QPO = OPR = 30°

Hence, POQ = POR = 60° Consider ∆QSO,

QOS = 60°

OQ = OS (radius)

So, OQS = OSQ = 60°

∆QSO is an equilateral triangle.

So QS = SO = QO = radius

PQS = 90° - OQS = 90° - 60° = 30°

QPS = 30°

PS = SQ (isosceles triangle)

Hence PS = SQ = OS (radius)

Now the tangents to the given circle can be drawn as follows:

1. Draw a circle of 3 cm radius and with centre O.

2. Take a point S on circumference of this circle. Extend OS to P such that OS =PS.

3. Midpoint of OP is S. Draw a circle with radius OS and centre as S.

Let it intersect our circle at Q and R. Join PQ and PR. PQ and PR are requiredtangents.

OR

Following steps will be followed to draw the required triangle.

(1) Draw a line segment AB = 4 cm draw a ray SA making 900 with it.

(2) Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C.Join BC. ABC is required triangle.

(3) Draw a ray AX making an acute angle with AB, opposite to vertex C.

(4) Locate 5 points A1, A2, A3, A4, A5 on line segment AX.

(5) Join A5B. Draw a line through A3 parallel to A5B intersecting AB at B'.

(6) Through B' draw a line parallel to BC intersecting line segment AC at

C'. A'BC' is required triangle.

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