Question

Given that angle 1= (x^2+4y), angle 2= (27y+7) and angle 3=65. which of these is a possible value of x?
A.4
B.5
C.6
D.7

Answer 1

first you must know that each two neighbouring angles, their sum equals 180 degree
so angle 2+angle 3 = 180
(27y+7) +65=180
27y+7=115
27y=108
y=108/27
then angle 1 and 2 are neighbouring
so (x^2+4y)+(115)=180
x^2+4y=65
x^2+4(108/27) = 65

Answer 2

From the figure it is obvious that:
angle 1 and angle 3 are VERTICALLY OPPOSITE ANGLES,
THUS
\[\huge \angle 1 \cong \angle 3\]
\[\huge \rightarrow (x^2+4y) \cong 65^0---(1)\]
Since angle 2 and angle 3 are linear pair angles, hence
\[\huge \angle 2 + \angle 3 = 180^0\]
\[\huge \rightarrow (27y+7) + 65^0 = 180^0\]
\[\huge \rightarrow 27y+72^0 = 180^0\]
\[\huge \rightarrow 27y = 180^0- 72^0\]
\[\huge \rightarrow 27y = 108^0\]
\[\huge \rightarrow y = \frac{108^0}{27} \rightarrow y =4^0\]
Substituting y= 4 in eq (1) we find
\[\huge \rightarrow (x^2+4 \times 4) \cong 65^0\]
\[\huge \rightarrow (x^2+16) \cong 65^0\]
\[\huge \rightarrow x^2\cong 65^0 - 16^0\rightarrow x^2 = 49^0\]
\[\huge x= 7 \]( Taking square root both sides)
@ltp1212

Answer 3

Hence option D.7 is the correct answer.

Answer 4

@ltp1212 Are you there?