Circles
Solve problems related to the theorem "Angle in a semicircle is a right angle"
Consider the circle given below.
Can we find the measure of ∠PQR?
Let us try and find out.
Observe that a ΔPQR is formed inside the circle. We know that the sum of all the angles in a triangle is 180°.
However, in ΔPQR, we only know the measure of ∠RPQ. Hence, to find the measure of ∠PQR, we must first know the measure of ∠QRP.
Observe that ∠QRP is an angle in a semi-circle.
There is an important theorem related to angle in a semi-circle. It states that
An angle in a semi-circle is a right angle. |
Using this theorem, we can say that ∠QRP = 90°
From ΔPQR, we obtain
∠PQR + ∠QRP + ∠RPQ = 180°
∠PQR + 90° + 32° = 180°
∠PQR = 180° − 90° − 32° = 58°
Thus, we find that the measure of ∠PQR is 58°.
In this manner we use the theorem related to angle in semi-circle to solve various problems.
Let us now solve a few more problems to understand this concept better.
Example 1
In the given figure, ΔPRS is an isosceles triangle in which PS = SR. Find the measure of ∠QRS if ∠QPS = 110° and ∠QPR = 65°.
Solution:
It is given that ∠QPS = 110° and ∠QPR = 65°. Therefore,
∠RPS = 110° − 65° = 45°
PS = SR [Given]
We also know that the sides opposite to equal angles are equal. Therefore,
∠PRS = ∠RPS
∴∠PRS = 45°
We know that the angle in a semi-circle is a right angle.
∴∠PQR = ∠PSR = 90°
Consider ΔPQR.
∠PQR + ∠PRQ + ∠QPR = 180°(Using angle sum property of triangles)
⇒ 90° + ∠PRQ + 65° = 180°
⇒ ∠PRQ = 25°
Consider ΔPRS.
∠PRS + ∠PSR + ∠RPS = 180°(Using angle sum property of triangles)
⇒ ∠PRS + 90° + 45° = 180°
⇒ ∠PRS = 45°
Thus, ∠QRS = ∠PRQ + ∠PRS = 25° + 45° = 70°
Consider the circle given below.
Can we find the measure of ∠PQR?
Let us try and find out.
Observe that a ΔPQR is formed inside the circle. We know that the sum of all the angles in a triangle is 180°.
However, in ΔPQR, we only know the measure of ∠RPQ. Hence, to find the measure of ∠PQR, we must first know the measure of ∠QRP.
Observe that ∠QRP is an angle in a semi-circle.
There is an important theorem related to angle in a semi-circle. It states that
An angle in a semi-circle is a right angle. |
Using this theorem, we can say that ∠QRP = 90°
From ΔPQR, we obtain
∠PQR + ∠QRP + ∠RPQ = 180°
∠PQR + 90° + 32° = 180°
∠PQR = 180° − 90° − 32° = 58°
Thus, we find that the measure of ∠PQR is 58°.
In this manner we use the theorem related to angle in semi-circle to solve various problems.
Let us now solve a few more problems to understand this concept better.
Example 1
In the given figure, ΔPRS is an isosceles triangle in which PS = SR. Find the measure of ∠QRS if ∠QPS = 110° and ∠QPR = 65°.
Solution:
It is given that ∠QPS = 110° and ∠QPR = 65°. Therefore,
∠RPS = 110° − 65° = 45°
PS = SR [Given]
We also know that the sides opposite to equal angles are equal. Therefore,
∠PRS = ∠RPS
∴∠PRS = 45°
We know that the angle in a semi-circle is a right angle.
∴∠PQR = ∠PSR = 90°
Consider ΔPQR.
∠PQR + ∠PRQ + ∠QPR = 180°(Using angle sum property of triangles)
⇒ 90° + ∠PRQ + 65° = 180°
⇒ ∠PRQ = 25°
Consider ΔPRS.
∠PRS + ∠PSR + ∠RPS = 180°(Using angle sum property of triangles)
⇒ ∠PRS + 90° + 45° = 180°
⇒ ∠PRS = 45°
Thus, ∠QRS = ∠PRQ + ∠PRS = 25° + 45° = 70°
We are quite familiar with an important property of angles subtended by an arc at the centre and anywhere on the circle.
It states that the angle made by an arc, at any point on the alternate arc, is equal to half the angle made by the same arc at the centre.
This property can be used to construct a variety of angles.
For example, let us construct an angle of measure
For this, we need to draw a circle of any radius, OB.
Since we need to draw an angle of measure , we should draw ∠BOC = 67° at the centre of the circle by using the property stated above.
Draw ∠BOC of measure 67° by taking BO as the base such that point C lies on the circle.
Now, take a point A anywhere on the circle on the alternate arc and join AB and AC.
∠BAC is the required angle of measure
The same concept can be applied for constructing a triangle when the measures of all the angles are given.
Let us understand this concept by taking an example.
Suppose we need to construct a triangle with angles 40°, 70° and 70°, within a circle of radius 3 cm.
For this, first we will draw a circle of radius OB = 3 cm.
One angle of a triangle is 40°, i.e. the angle at the circumference of the circle is 40°. In order to have such an angle, we need to draw an angle of measure 2 × 40° = 80°, at the centre of the circle.
For this, draw ∠BOC = 80° by taking OB as the base such that point C lies on the circle.
In the similar manner, we need to draw an angle of measure 2 × 70° = 140°, at the centre of the circle.
So, draw ∠AOC = 140° by taking OC as the base such that point A lies on the circle.
Join AB, AC and BC.
ΔABC is the required triangle with the given measurements.
Justification of construction:
We have drawn ∠BAC = 40° and ∠ABC = 70°.
We know that the sum of the measures of angles of a triangle is 180°.
∴ ∠ABC + ∠BCA + ∠CAB = 180°
⇒ 70° + ∠BCA + 40° = 180°
⇒ ∠BCA = 180° – 110°
⇒ ∠BCA = 70°, which is the required measure of the third angle of the triangle.
Hence, DABC is correctly drawn with the given measurements.
Let us construct some more angles.
Example 1:
Draw an angle of measure at one end of the diameter.
Solution:
We know that the angle made by an arc at any point on the alternate arc is equal to half the angle made by the same arc at the centre of the circle.
The given angle can be drawn at one end of the diameter using the above stated property.
The steps of construction are as follows:
(1) Draw a circle of any radius with O as the centre and AB as the diameter.
(2) Draw ∠BOC = 55° by taking OB as the base such that point …
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