ABC ত্রিভুজে cosA + cosC = sinB হলে $$\angle$$A সমান-

Question

Accepted Answer

$$\begin{aligned}& \cos \mathrm{A}+\cos \mathrm{C}=\sin \mathrm{B} \& \Rightarrow 2 \cos \left(\frac{\mathrm{~A}+\mathrm{C}}{2}ight) \cos \left(\frac{\mathrm{A}-\mathrm{C}}{2}ight)=2 \sin \frac{\mathrm{~B}}{2} \cos \frac{\mathrm{~B}}{2} \& \Rightarrow \cos \left(\frac{\pi}{2}-\frac{\mathrm{B}}{2}ight) \cos \left(\frac{\mathrm{A}-\mathrm{C}}{2}ight)=\sin \frac{\mathrm{B}}{2} \cos \frac{\mathrm{~B}}{2} \& \Rightarrow \sin \frac{\mathrm{~B}}{2} \cos \left(\frac{\mathrm{~A}-\mathrm{C}}{2}ight)=\sin \frac{\mathrm{B}}{2} \cos \frac{\mathrm{~B}}{2} \& \Rightarrow \cos \left(\frac{\mathrm{~A}-\mathrm{C}}{2}ight)=\cos \frac{\mathrm{B}}{2} \Rightarrow \frac{\mathrm{~A}-\mathrm{C}}{2}=\frac{\mathrm{B}}{2} \& \Rightarrow \mathrm{~A}=\mathrm{B}+\mathrm{C}=\pi-\mathrm{A} \Rightarrow 2 \mathrm{~A}=\pi \Rightarrow \mathrm{A}= \& 90^{\circ} \& 	ext{ অপশন টেস্ট }: \mathrm{C}=90^{\circ} 	ext{ হলে, } \mathrm{A}+\mathrm{B}=90^{\circ} \& 	herefore \cos \mathrm{A}+\cos \mathrm{C}=\cos \left(90^{\circ}-\mathrm{B}ight)+\cos 90^{\circ}= \& \sin \mathrm{B}\end{aligned}$$

Admission DU A প্রশ্নব্যাংক MCQ — ABC ত্রিভুজে cosA + cosC = sinB হলে \[\angle\] A সমান -

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