# Draw a circle of radius $3cm$

Draw triangle \[ABC\] with this circle as circumcircle and angle ${50^ \circ },{60^ \circ }$ and ${70^ \circ }$.

Construct triangle$PQR$, outside the circle, by drawing tangents to the circle at the points $A,B$and$C$.

Find all angles of triangle$PQR$

Answer

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Hint: Use the necessary tools to attempt construction problems. Follow the construction steps.

Step #1: Draw a circle with centre $O$ having radius \[3cm\].

Step #2: Take three points $A,B$and \[C\] in the circle and join \[A\] to the centre of the circle $O$.

Step #3: If \[\angle BAC\] is to be\[{50^ \circ }\], should be\[{100^ \circ }\].

Step #4: If \[\angle ABC\] is to be\[{60^ \circ }\], should be\[{120^ \circ }\].

Step #5: If \[\angle ACB\] is to be\[{70^ \circ }\], should be\[{140^ \circ }\].

Step #6: Join the points B and C such that \[m\angle AOC = {100^ \circ }\] and\[m\angle BOC = {120^ \circ }\].

Thus, \[\Delta ABC\]is the required triangle.

Step #1: Extend \[OA\] and draw perpendicular to it through\[A\].

Step #2: Extend \[OB\] and draw perpendicular to it through\[B\].

Step #3: In the same way draw a perpendicular from point \[C\] through\[OC\].

Let the points of intersection of these perpendicular be \[P,Q\] and\[R\], so we get the required \[\Delta PQR\].

\[m\angle AOC = {120^ \circ }\]

\[ \Rightarrow m\angle P = {180^ \circ } - {120^ \circ } = {60^ \circ }\] ……(opposite angles of a quadrilateral are supplementary)

In the same way,

\[ \Rightarrow m\angle Q = {180^ \circ } - {140^ \circ } = {40^ \circ }\]

And, \[ \Rightarrow m\angle R = {180^ \circ } - {100^ \circ } = {80^ \circ }\]

Note: Use the necessary tools to attempt construction problems. Mistakes can be made in reading the angles wrongly on the instruments or incorrect dimensioning in the diagram.

Steps of Construction:

Step #1: Draw a circle with centre $O$ having radius \[3cm\].

Step #2: Take three points $A,B$and \[C\] in the circle and join \[A\] to the centre of the circle $O$.

Step #3: If \[\angle BAC\] is to be\[{50^ \circ }\], should be\[{100^ \circ }\].

Step #4: If \[\angle ABC\] is to be\[{60^ \circ }\], should be\[{120^ \circ }\].

Step #5: If \[\angle ACB\] is to be\[{70^ \circ }\], should be\[{140^ \circ }\].

Step #6: Join the points B and C such that \[m\angle AOC = {100^ \circ }\] and\[m\angle BOC = {120^ \circ }\].

Thus, \[\Delta ABC\]is the required triangle.

Steps of Construction:

Step #1: Extend \[OA\] and draw perpendicular to it through\[A\].

Step #2: Extend \[OB\] and draw perpendicular to it through\[B\].

Step #3: In the same way draw a perpendicular from point \[C\] through\[OC\].

Let the points of intersection of these perpendicular be \[P,Q\] and\[R\], so we get the required \[\Delta PQR\].

In the quadrilateral \[PAOC\],

\[m\angle AOC = {120^ \circ }\]

\[ \Rightarrow m\angle P = {180^ \circ } - {120^ \circ } = {60^ \circ }\] ……(opposite angles of a quadrilateral are supplementary)

In the same way,

\[ \Rightarrow m\angle Q = {180^ \circ } - {140^ \circ } = {40^ \circ }\]

And, \[ \Rightarrow m\angle R = {180^ \circ } - {100^ \circ } = {80^ \circ }\]

Note: Use the necessary tools to attempt construction problems. Mistakes can be made in reading the angles wrongly on the instruments or incorrect dimensioning in the diagram.

Last updated date: 23rd Sep 2023

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