MCQ
In $\triangle\text{ABC}$ if $\angle\text{A} = \text{x}^\circ,\angle\text{B} = {\text{3x}}^\circ$ and $\angle{\text{C}}=\text{y}^\circ$ if 3y - 5x = 30, then $\angle\text{B} = $
- A$30^\circ$
- B$60^\circ$
- C$45^\circ$
- ✓$90^\circ $
✓
Answer
Correct option: D.
$90^\circ $
Here, $A = x, B = 3x, C = y.$
$180 = 4x + y ... (i)\ ($Sum of angles of a triangle $= x + 3x + y)$
$180 - 4x = y ... (ii)$
Also $, 3y - 5x = 30 ... (iii)$
Substituting the value of $(ii)$ in $(iii)$
$3(180 - 4x) - 5x = 30$
$540 - 12x - 5x = 30$
$-17x = 30 - 540$
$17x = 510$
$x = 30 ... (iv)$
But, $\angle\text{B} = 3\text{x} ... ($Given$)$
$\therefore\angle\text{B} = {3}\times{30} = {90}^\circ$
$180 = 4x + y ... (i)\ ($Sum of angles of a triangle $= x + 3x + y)$
$180 - 4x = y ... (ii)$
Also $, 3y - 5x = 30 ... (iii)$
Substituting the value of $(ii)$ in $(iii)$
$3(180 - 4x) - 5x = 30$
$540 - 12x - 5x = 30$
$-17x = 30 - 540$
$17x = 510$
$x = 30 ... (iv)$
But, $\angle\text{B} = 3\text{x} ... ($Given$)$
$\therefore\angle\text{B} = {3}\times{30} = {90}^\circ$
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