MCQ
If angles $A, B, C $ to a $\triangle\text{ABC}$ from an increasing $AP,$ then $\sin B =$
- A$\frac{1}{2}$
- ✓$\frac{\sqrt{3}}{2}$
- C$1$
- D$\frac{1}{\sqrt{2}}$
✓
Answer
Correct option: B.
$\frac{\sqrt{3}}{2}$
Let the angles $A, B , C$ of $\triangle\text{ABC}$
$\angle\text{A}=(\text{a}-\text{d})$
$\angle\text{B}=\text{a}$
$\angle\text{C}=\text{a}+\text{d}$
from an increasing $A.P$
then sum of the all there angles of $\triangle\text{ABC}$
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow(\text{a} -\text{d})+\text{a}+(\text{a+d})=180^\circ$
$\Rightarrow\text{3a}=180^\circ $
$\Rightarrow\text{a}=60^\circ =\angle\text{B}$
then $\sin\text{b=}\sin\text{a}=\sin60^\circ\ ($from the table$)$
$=\frac{\sqrt{3}}{2}$
Hence the correct option is $(b)$
$\angle\text{A}=(\text{a}-\text{d})$
$\angle\text{B}=\text{a}$
$\angle\text{C}=\text{a}+\text{d}$
from an increasing $A.P$
then sum of the all there angles of $\triangle\text{ABC}$
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow(\text{a} -\text{d})+\text{a}+(\text{a+d})=180^\circ$
$\Rightarrow\text{3a}=180^\circ $
$\Rightarrow\text{a}=60^\circ =\angle\text{B}$
then $\sin\text{b=}\sin\text{a}=\sin60^\circ\ ($from the table$)$
$=\frac{\sqrt{3}}{2}$
Hence the correct option is $(b)$
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