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<b>Statement I:</b> If A,B,C,D are angle...

Statement I: If `A,B,C,D` are angles of a cyclic quadrilateral then `sum sin A=0.`
Statement II: If `A, B, C, D` are angles of cyclic quadrilateral then `sum cos A=0.`

A

(a) Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B

(b) Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I

C

(c) Statement I is correct but Statement II is incorrect

D

(d) Statement II is correct but Statement I is incorrect

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the statements regarding the angles of a cyclic quadrilateral. ### Step-by-Step Solution: 1. **Understanding the Properties of Cyclic Quadrilaterals:** A cyclic quadrilateral is a four-sided figure where all vertices lie on a single circle. The key property of cyclic quadrilaterals is that the sum of the opposite angles is supplementary. This means: \[ A + C = 180^\circ \quad \text{and} \quad B + D = 180^\circ \] 2. **Analyzing Statement I:** The first statement claims that the sum of the sines of the angles is zero: \[ \sin A + \sin B + \sin C + \sin D = 0 \] Since we know that \(C = 180^\circ - A\) and \(D = 180^\circ - B\), we can express \(\sin C\) and \(\sin D\) as: \[ \sin C = \sin(180^\circ - A) = \sin A \quad \text{and} \quad \sin D = \sin(180^\circ - B) = \sin B \] Therefore, the sum becomes: \[ \sin A + \sin B + \sin A + \sin B = 2\sin A + 2\sin B \] Since both \(\sin A\) and \(\sin B\) are positive for angles in the range \(0^\circ < A, B < 180^\circ\), the sum \(2\sin A + 2\sin B\) is always greater than zero. Thus, **Statement I is incorrect**. 3. **Analyzing Statement II:** The second statement claims that the sum of the cosines of the angles is zero: \[ \cos A + \cos B + \cos C + \cos D = 0 \] Using the same substitutions for \(C\) and \(D\): \[ \cos C = \cos(180^\circ - A) = -\cos A \quad \text{and} \quad \cos D = \cos(180^\circ - B) = -\cos B \] Therefore, the sum becomes: \[ \cos A + \cos B - \cos A - \cos B = 0 \] This shows that **Statement II is correct**. 4. **Conclusion:** Based on the analysis: - Statement I is incorrect. - Statement II is correct. Thus, the correct answer is that **Statement 2 is correct but Statement 1 is incorrect**. ### Final Answer: The correct option is **d**: Statement 2 is correct but Statement 1 is incorrect.
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