 Hello and welcome to the session. Let's work out the following question. It says draw a pair of tendons to a circle of radius 5 centimeter which are inclined to each other at an angle of 60 degrees. So let's now move on to the solution and let's do the construction. The first step of construction is to take a point O, draw a circle of radius Oa which is 5 centimeter. So we have drawn a circle with radius 5 centimeter and Oa is the radius which is 5 centimeter. Now the next step is produce such that Oa to 5 centimeter. So we have produced Oa to be where A B is 5 centimeter which is equal to Oa that is 5 centimeter. Now the next step is with A as center and radius AO equal to AB equal to 5 centimeter draw a circle setting the first circle points B and Q. Here we have drawn a circle with radius AB equal to Oa equal to 5 centimeter and the circle intersects the first circle at the points B and Q. Now the next step is join BP to get the required tendons. BP and BQ are the required tendons which are inclined at each other at an angle of 60 degrees that is angle PBQ is 60 degrees. Now we'll justify that angle PBQ is 60 degrees. Now for justification join AP and OP. So we have joined AP and OP and here we see that OP is equal to Oa since the radii of the given circle. Also Oa is equal to OP since is the radius of the circle with center A and circle with center A has radius 5 centimeter. So this is 5 centimeter this is also 5 centimeter and this is also 5 centimeter. So in triangle Oa P we have Oa is equal to OP is equal to 5 centimeter because they are the radii of the given circle but also P is equal to 5 centimeter because it's the radius of the circle center A. Therefore Oa is equal to OP is equal to AP. Therefore triangle Oa P is an equilator triangle since all the sides are equal. Therefore angle degrees right. So angle BAP angle BAP that is this angle will be 120 degrees since the linear pair therefore angle BAP is 120 degrees. Now equal to AP equal to AP since it's the radial circle with center A it has radius 5 centimeter. Angle BAP is 120 degrees. Therefore is equal to angle because these are the angles opposite to the equal sides and we know that angles opposite the equal sides are equal. Now in triangle APB plus angle BAP is equal to 180 degrees angle APB plus angle B is 180 degrees as we know that the sum of angles of triangle is 180 degrees. Now ABB is equal to APB angle BAP is 120 degrees. So this implies twice of angle ABP is equal to 180 degrees minus 120 degrees. So this implies twice of angle ABP is equal to 60 degrees and this implies angle ABP is equal to 30 degrees. So angle ABP is 30 degrees and angle ABP is equal to angle ABP. So this is also 30 degrees. Now similarly we'll join OQ and OA and then we'll consider the triangle BAQ. So we have joined OQ and AQ and again this will be a equilateral triangle that is triangle AQO will be equilateral triangle and this angle would be 120 degrees and then considering the triangle we'll get that angle ABQ would be 30 degrees. So similarly join OQ and AQ consider triangle BAQ or ABQ that is this triangle will get equal to 30 degrees. Now angle ABP that is PBA plus angle QBA that is ABQ is equal to 30 degrees plus 30 degrees that is 60 degrees and angle PBA angle PBA plus angle QBA is angle PBQ. So both the tangents are inclined to each other at an angle of 60 degrees hence proved. It's a question and a session. Bye for now. Take care. Have a good day.