In ΔABC, ∠A=90°, then cos−1(Rr2+r3)=

Rr2+r3=R4RsinA2⋅cosB2⋅sinC2+4RsinA2⋅sinB2⋅cosC2=R4RsinA2(cosB2⋅sinC2+cosC2⋅sinB2)=R4RsinA2⋅sin(B+C2)=R4RsinA2⋅cosA2=R2RsinA, ∠A=90°=12(1)=12cos−1(Rr2+r3)=cos−1(12)=cos−1(cos60°)=60°

In ΔABC, ∠A=90°, then cos−1(Rr2+r3)=

Rr2+r3=R4RsinA2⋅cosB2⋅sinC2+4RsinA2⋅sinB2⋅cosC2=R4RsinA2(cosB2⋅sinC2+cosC2⋅sinB2)=R4RsinA2⋅sin(B+C2)=R4RsinA2⋅cosA2=R2RsinA, ∠A=90°=12(1)=12cos−1(Rr2+r3)=cos−1(12)=cos−1(cos60°)=60°

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In $Δ A B C, ∠ A = 90 °$ , then $\cos^{- 1} (\frac{R}{r_{2} + r_{3}}) =$

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$60 °$

$45 °$

$30 °$

$90 °$

answer is D.

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Detailed Solution

$\frac{R}{r_{2} + r_{3}} = \frac{R}{4 R \sin \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \sin \frac{C}{2} + 4 R \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \cos \frac{C}{2}}$