Join a line from B to C.
\(∠ABC = ∠BDC = 5z°\) [Alternate segment theorem]
Similarly, \(∠BCA = ∠CDB = 5z°\)
Thus, in \(△\;ABC\),
\(∠ABC + ∠BCA + ∠BAC = 180°\)
\(z = 15°\)
\(5z + 5z + 2z = 180°\)
So, \(∠BDC = 5(15) = 75°\)
Thus, \(∠BOC = 180° – 75° = 105°\) (As DBOC is a cyclic quadrilateral)
\(∠BAC = 2(15) = 30°\)
Hence, \(∠BOC + ∠BAC = 105° + 30° = 135°\)
Hence, option B is the correct answer.
In the given figure, \( PQ \) and \( PR \) are tangents to the circle such that \( PQ = 7 \, \text{cm} \) and \( \angle RPQ = 60^\circ \).
The length of chord QR is:
In the given figure, a circle inscribed in \( \triangle ABC \) touches \( AB, BC, \) and \( CA \) at \( X, Z, \) and \( Y \) respectively.
If \( AB = 12 \, \text{cm}, AY = 8 \, \text{cm}, \) and \( CY = 6 \, \text{cm} \), then the length of \( BC \) is:
A | B | C | D | Average |
---|---|---|---|---|
3 | 4 | 4 | ? | 4 |
3 | ? | 5 | ? | 4 |
? | 3 | 3 | ? | 4 |
? | ? | ? | ? | 4.25 |
4 | 4 | 4 | 4.25 |