MCQ
$\cos35^\circ+\cos85^\circ+\cos155^\circ=$
- A$0$
- B$\frac{1}{\sqrt3}$
- C$\frac{1}{\sqrt2}$
- D$\cos275^\circ$
Solution:
$\cos35^\circ+\cos85^\circ+\cos155^\circ$
$=\ 2\cos\Big(\frac{35^\circ+85^\circ}{2}\Big)\cos\Big(\frac{35^\circ-85^\circ}{2}\Big)+\cos155^\circ$ $\Big[\because\ \cos\text{A}+\cos\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=\ 2\cos60^\circ\cos(-25^\circ)+\cos155^\circ$
$=\ 2\times\frac{1}{2}\cos25^\circ+\cos155^\circ$
$=\ \cos25^\circ+\cos155^\circ$
$=\ 2\cos\Big(\frac{25^\circ+155^\circ}{2}\Big)\cos\Big(\frac{25^\circ-155^\circ}{2}\Big)$
$=\ 2\cos90^\circ\cos65^\circ$
$=\ 0$
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The equations of the sides AB, BC and CA of $\triangle \text{ABC}$ are y - x = 2, x + 2y = 1 and 3x + y + 5 = 0 respectively. The equation of the altitude through B is: