Question

In figure, PQ is tangent to the circle at A, DB is a diameter. \[\angle ADB = {30^ \circ }\] and \[\angle CBD = {60^ \circ }\]. Calculate
(i) \[\angle QAB\]
(ii) \[\angle PAD\]
(iii) \[\angle CDB\]

Answer 1

Hint: Here we calculate the remaining angles in the two triangles using the property of sum of angles in a triangle. We join the radius to the points A and C which will form two isosceles triangles. Given an angle opposite to equal sides we find the other angle which is opposite to the equal side. Use the property that radius makes the right angle at the point where tangent meets the radius on the circle to find the required angle.
* In a triangle, the sum of three angles is equal to \[{180^ \circ }\].
* An isosceles triangle has two sides equal to each other and the angles opposite to equal sides are also equal.
* Radius of the circle is perpendicular to the tangent at the point of contact.

Complete step-by-step solution:
We are given PQ is tangent to the circle at A.
We join the radius of the circle from center O to point A and to point C separately.

We know sum of angles in a triangle is equal to \[{180^ \circ }\]
(i) \[\angle QAB\]
In triangle ABD,
\[\angle DAB + \angle ABD + \angle ADB = {180^ \circ }\]
Since, we know \[\angle ADB = {30^ \circ },\angle DAB = {90^ \circ }\]
\[ \Rightarrow {90^ \circ } + \angle ABD + {30^ \circ } = {180^ \circ }\]
\[ \Rightarrow {120^ \circ } + \angle ABD = {180^ \circ }\]
Shift all the constant values in degrees to RHS of the equation.
\[ \Rightarrow \angle ABD = {180^ \circ } - {120^ \circ }\]
\[ \Rightarrow \angle ABD = {60^ \circ }\]
Now we know that\[OA = OB\] (radius of the circle)
Therefore, \[\vartriangle AOB\] is an isosceles triangle
We know the angles opposite to equal sides are equal in an isosceles triangle.
\[\angle OAB = \angle OBA = {60^ \circ }\]
Now we use the property that the radius of the circle makes a right angle with the tangent at the point of contact.
So, we have \[\angle OAQ = {90^ \circ }\]
Now we know \[\angle OAQ = \angle OAB + \angle QAB\]
Substitute the value of \[\angle OAQ = {90^ \circ }\] and \[\angle OAB = {60^ \circ }\]
\[ \Rightarrow {90^ \circ } = {60^ \circ } + \angle QAB\]
Shift all constant values in degrees at one side of the equation
\[ \Rightarrow \angle QAB = {90^ \circ } - {60^ \circ }\]
\[ \Rightarrow \angle QAB = {30^ \circ }\]
(ii) \[\angle PAD\]
We are given \[\angle ADB = {30^ \circ }\]
Now we know that\[OA = OD\] (radius of the circle)
Therefore, \[\vartriangle AOD\]is an isosceles triangle
We know the angles opposite to equal sides are equal in an isosceles triangle.
\[\angle OAD = \angle ODA = {30^ \circ }\]
Now we use the property that the radius of the circle makes a right angle with the tangent at the point of contact.
So, we have \[\angle OAP = {90^ \circ }\]
Now we know \[\angle OAP = \angle OAD + \angle PAD\]
Substitute the value of \[\angle OAP = {90^ \circ }\] and \[\angle OAD = {30^ \circ }\]
\[ \Rightarrow {90^ \circ } = {30^ \circ } + \angle PAD\]
Shift all constant values in degrees at one side of the equation
\[ \Rightarrow \angle PAD = {90^ \circ } - {30^ \circ }\]
\[ \Rightarrow \angle PAD = {60^ \circ }\]
(iii) \[\angle CDB\]
In triangle CBD,
\[\angle CDB + \angle CBD + \angle BCD = {180^ \circ }\]
Since, we know \[\angle CBD = {60^ \circ },\angle BCD = {90^ \circ }\]
\[ \Rightarrow {90^ \circ } + \angle CDB + {60^ \circ } = {180^ \circ }\]
\[ \Rightarrow {150^ \circ } + \angle CDB = {180^ \circ }\]
Shift all the constant values in degrees to RHS of the equation.
\[ \Rightarrow \angle CDB = {180^ \circ } - {150^ \circ }\]
\[ \Rightarrow \angle CDB = {30^ \circ }\]

Note: Many students get confused if the diagram does not show the angle made by the triangle on the circle as the right angle. Keep in mind the angle at the circumference in a semicircle is \[{90^ \circ }\]. Students are likely to make mistake as the angle on the circumference is \[{90^ \circ }\] and the angle made by radius with the tangent is also \[{90^ \circ }\], never write the name of the angle as single letter instead choose to write the three letter form of angle that shows the path from where to where we are taking the angle.