In - ABC - sin C -cos C -sin 2 B - C -cos 2 B - C -2 -2- Then the - ABC isA- Acute angled isosceles triangleB- Right angled isosceles triangleC- Equilateral triangleD- Obtuse angled isosceles triangle

The correct option is B Right angled isosceles trianglesin C +cos C+sin (2B+C) cos (2B+C)=2√(2) ⇒√(2)sin(C+π4)+√(2)sin(2B+C π4)=2√(2) ⇒sin(C+π4)+sin(2B+C π4)=2 ...

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Standard XII

Mathematics

Sine Rule

In Δ ABC , si...

Question

In $Δ A B C, sin C + cos C + sin (2 B + C) - cos (2 B + C) = 2 \sqrt{2} .$ Then the $Δ A B C$ is

Acute angled isosceles triangle

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Right angled isosceles triangle

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Obtuse angled isosceles triangle

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Equilateral triangle

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Solution

The correct option is B Right angled isosceles triangle
$sin C + cos C + sin (2 B + C) - cos (2 B + C) = 2 \sqrt{2} \Rightarrow \sqrt{2} sin (C + \frac{π}{4}) + \sqrt{2} sin (2 B + C - \frac{π}{4}) = 2 \sqrt{2} \Rightarrow sin (C + \frac{π}{4}) + sin (2 B + C - \frac{π}{4}) = 2$

The above equation is valid only if
$sin (C + \frac{π}{4}) = 1$
and $sin (2 B + C - \frac{π}{4}) = 1$

$\Rightarrow C + \frac{π}{4} = \frac{π}{2} \Rightarrow C = \frac{π}{4}$
and $2 B + C - \frac{π}{4} = \frac{π}{2} \Rightarrow B = \frac{π}{4}$
$\Rightarrow A = \frac{π}{2}$

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In ΔABC, sinC+cosC+sin(2B+C)−cos(2B+C)=2√2. Then the ΔABC is

The correct option is B Right angled isosceles trianglesinC+cosC+sin(2B+C)−cos(2B+C)=2√2⇒√2sin(C+π4)+√2sin(2B+C−π4)=2√2⇒sin(C+π4)+sin(2B+C−π4)=2 The above equation is valid only if sin(C+π4)=1 and sin(2B+C−π4)=1 ⇒C+π4=π2⇒C=π4 and 2B+C−π4=π2⇒B=π4 ⇒A=π2

In $Δ A B C, sin C + cos C + sin (2 B + C) - cos (2 B + C) = 2 \sqrt{2} .$ Then the $Δ A B C$ is