SATHEE: Properties of Triangles 1 Question 14

Properties of Triangles 1 Question 14

14. In a $△ A B C$ with fixed base $B C$ , the vertex $A$ moves such that $\cos B + \cos C = 4 \sin^{2} \frac{A}{2}$ . If $a, b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A, B$ and $C$ respectively, then

(2009)

(a) $b + c = 4 a$

(b) $b + c = 2 a$

(d) locus of point $A$ is a pair of straight line

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Answer:

Correct Answer: 14. $90^{\circ}$

Solution:

Given, $\cos B + \cos C = 4 \sin^{2} \frac{A}{2}$

$\Rightarrow 2 \cos \frac{B + C}{2} \cos \frac{B - C}{2} = 4 \sin^{2} \frac{A}{2}$

$\Rightarrow 2 \sin \frac{A}{2} \cos \frac{B - C}{2} - 2 \sin \frac{A}{2} = 0$

$\Rightarrow \cos \frac{B - C}{2} - 2 \cos \frac{B + C}{2} = 0$ as $\sin \frac{A}{2} \neq 0$

$\Rightarrow - \cos \frac{B}{2} \cos \frac{C}{2} + 3 \sin \frac{B}{2} \sin \frac{C}{2} = 0$

$\Rightarrow \tan \frac{B}{2} \tan \frac{C}{2} = \frac{1}{3}$

$\Rightarrow \sqrt{\frac{(s - a) (s - c)}{s (s - b)} \cdot \frac{(s - b) (s - a)}{s (s - c)}} = \frac{1}{3}$

$\Rightarrow \frac{s - a}{s} = \frac{1}{3} \Rightarrow 2 s = 3 a$

$\Rightarrow b + c = 2 a$

$∴$ Locus of $A$ is an ellipse.

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Properties of Triangles 1 Question 14

14. In a $△ A B C$ with fixed base $B C$ , the vertex $A$ moves such that $\cos B + \cos C = 4 \sin^{2} \frac{A}{2}$ . If $a, b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A, B$ and $C$ respectively, then

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Properties of Triangles 1 Question 14

14. In a △ABC with fixed base BC, the vertex A moves such that cos⁡B+cos⁡C=4sin2⁡A2. If a,b and c denote the lengths of the sides of the triangle opposite to the angles A,B and C respectively, then

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चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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14. In a $△ A B C$ with fixed base $B C$ , the vertex $A$ moves such that $\cos B + \cos C = 4 \sin^{2} \frac{A}{2}$ . If $a, b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A, B$ and $C$ respectively, then