Question
In a $\triangle\text{ABC},$ if $\cos\text{C}=\frac{\sin\text{A}}{2\sin\text{B}},$ prove that the triangle is isosceles.
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| (i) | $((\text{A}'\cup\text{B}')-\text{A})'$ | (a) | $\text{A} - \text{B}$ |
| (ii) | $[\text{B}'\cup(\text{B}'-\text{A})]'$ | (b) | $\text{A}$ |
| (iii) | $(\text{A} - \text{B}) - (\text{B} - \text{C})$ | (c) | $\text{B}$ |
| (iv) | $(\text{A}-\text{B})\cap(\text{C}-\text{B})$ | (d) | $(\text{A}\times\text{B})\cap(\text{A}\times\text{C})$ |
| (v) | $\text{A}\times(\text{B}\cap\text{C})$ | (e) | $(\text{A}\times\text{B})\cup(\text{A}\times\text{C})$ |
| (vi) | $\text{A}\times(\text{B}\cup\text{C})$ | (f) | $(\text{A}\cap\text{C})-\text{B}$ |