
Using a ruler and compass only.
(i) Construct a $\vartriangle ABC$ with the following data.
AB = 3.5 cm, BC = 6 cm and $\angle ABC=120{}^\circ $
(ii) In the same diagram, draw a circle with BC as diameter. Find a point P on the circumference of the circle which is equidistant from AB and BC.
(iii) Measure $\angle BCP$ .
Answer
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Hint: In this problem, we are to find the figure with given instructions and also to find an angle. So, to start with we will draw the base which is of 6 cm and following that we will draw the angle of 120 degrees. Now, considering BC as a diameter we will draw a circle using the bisector. Now, we draw the angle bisector of the angle ABC and thus we get the equidistant points.
Complete step-by-step solution:
According to the question, we are to draw a triangle with the given conditions.
We will start with drawing a line segment BC which is equal to 6 cm.
Now, it is also said that, $\angle ABC=120{}^\circ $,
So, we get,
Again, it is said, AB = 3.5 cm,
Joining AB and AC will give us our desired triangle.
Now, for the second part,
We will start with drawing a bisector of BC.
Now, to complete the circle,
Again, we are to find a point on the circle which is equidistant from AB and BC.
So, we draw a bisector of the angle $\angle ABC=120{}^\circ $which is intersecting the circle at P.
Joining C and P and measuring the angle we get,
$(iii)\angle BCP=30{}^\circ $
Note: In this problem, we are to measure the angle we get in the last step. Sometimes, it might be the case that once we get the figure and we try to measure the angle, we will get the angles are not in whole numbers or in integers all the time. Then, we will try to consider it as a calculation problem and we will choose the nearest integer to get the solution.
Complete step-by-step solution:
According to the question, we are to draw a triangle with the given conditions.
We will start with drawing a line segment BC which is equal to 6 cm.
Now, it is also said that, $\angle ABC=120{}^\circ $,
So, we get,
Again, it is said, AB = 3.5 cm,
Joining AB and AC will give us our desired triangle.
Now, for the second part,
We will start with drawing a bisector of BC.
Now, to complete the circle,
Again, we are to find a point on the circle which is equidistant from AB and BC.
So, we draw a bisector of the angle $\angle ABC=120{}^\circ $which is intersecting the circle at P.
Joining C and P and measuring the angle we get,
$(iii)\angle BCP=30{}^\circ $
Note: In this problem, we are to measure the angle we get in the last step. Sometimes, it might be the case that once we get the figure and we try to measure the angle, we will get the angles are not in whole numbers or in integers all the time. Then, we will try to consider it as a calculation problem and we will choose the nearest integer to get the solution.
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