In - ABC - r -1- r 1-7 and R -3 then - ABC isA- equilateralB- acute angled not equilateralC- obtuse angledD- right angled

The correct option is D right angledr_1 r=4R sin A/2 [ cos B/2 cos C/2 sin B/2sin C/2 ] ⇒ 6 = 4R sin A/2cos ( B+C/2 ) ⇒ 6 = 4R sin^2 A/2⇒ sin^2 A/2=1/2⇒∠ A=π/ ...

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Byju's Answer

Standard XII

Mathematics

Cos(A+B)Cos(A-B)

In Δ ABC , r ...

Question

In $Δ$ ABC, r = 1, $r_{1}$ = 7 and R = 3 then $Δ$ ABC is

equilateral

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acute angled (not equilateral)

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obtuse angled

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right angled

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Solution

The correct option is D right angled
$r_{1} - r = 4 R s i n \frac{A}{2} [c o s \frac{B}{2} c o s \frac{C}{2} - s i n \frac{B}{2} s i n \frac{C}{2}]$
$\Rightarrow 6 = 4 R s i n \frac{A}{2} c o s (\frac{B + C}{2})$
$\Rightarrow 6 = 4 R s i n^{2} \frac{A}{2} \Rightarrow s i n^{2} \frac{A}{2} = \frac{1}{2} \Rightarrow ∠ A = \frac{π}{2}$

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The correct option is D right angledr1−r=4R sin A2[cosB2 cos C2−sin B2sin C2] ⇒6=4R sinA2cos(B+C2) ⇒6=4R sin2A2⇒sin2 A2=12⇒∠A=π2

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