Question

Please solve it I will mark as brainiest​

Answer 1

In A,

2x+4=90

2x=90-4

x=86/2

x=43

In D,

4x-5=90

4x=90+5

x=95/4

x=23.75

In B,

y+3=90

y=90-3

y=87

In C,

2y+10=90

2y=90-10

y=80/2

y=40

Answer 2

Step-by-step explanation:

In a cyclic quadrilateral, the opposite angles are supplementary.

$\begin{array}{rrrl} & &\angle P+\angle R & =180^{\circ} \\ \\ \Rightarrow & & (2 x+4)^{\circ}+(2 y+10)^{\circ} & =180^{\circ} \\ \\ \Rightarrow & & 2 x+2 y+14-180 =0 \\ \\ \Rightarrow & 2 x+2 y-166 =0 \\ \\ \Rightarrow & x+y-83 =0 \\ \\ & \text { Also } \angle Q + \angle S = 180 \degree& \ldots(1) \\ \\ \therefore & (y+3)^{\circ}+(4 x-5)^{\circ} =180^{\circ} \\ \\ \Rightarrow & y+4 x-2-180 =0 \\ \\ \Rightarrow & y+4 x-182 =0\end{array}$

From (1) and (2),

$\[ \begin{array}{ll} \rm a_{1}=1, &\rm b_{1}=1, &\rm c_{1}=-83 \\ \\ \rm \rm \: a_{2}=4, &\rm b_{2}=1, &\rm c_{2}=-182 \end{array} \]$

$\begin{array}{l} \\ \displaystyle \rm \therefore \frac{x}{-182+83}-\frac{y}{-332+182}=\frac{1}{1-4} \\ \\ \displaystyle \rm \Rightarrow \frac{x}{-99}=\frac{y}{-150}=\frac{1}{-3} \\ \\ \displaystyle \rm \Rightarrow x=\frac{1}{-3} \times(-99)=33 \\ \\ \displaystyle \rm\Rightarrow y=\frac{1}{-3} \times(-150)=50 \end{array}$

$\begin{aligned} \color{red}\rm \therefore \angle P &\color{red} \rm=(2 x+4)^{\circ}=[(2 \times 33)+4]^{\circ}=70^{\circ} \\ \\ \color{darkorange}\angle\rm Q &\color{darkorange}\rm=(y+3)^{\circ}=[50+3]^{\circ}=53^{\circ} \\ \\ \color{olive}\rm\angle R &\color{olive}\rm=(2 y+10)^{\circ}=[2 \times 50+10]^{\circ}=110^{\circ} \\ \\ \color{darkgreen}\rm \angle S &\color{darkgreen}\rm=(4 x-5)^{\circ}=\left[4 \times 33-5^{\circ}\right]=127^{\circ} \end{aligned}$