MCQ
In a $\triangle A B C$, if $\angle C=50^{\circ}$ and $\angle A$ exceeds $\angle B$ by $44^{\circ}$, then $\angle A=$
- A$43^{\circ}$
- B$40^{\circ}$
- C$67^{\circ}$
- ✓$87^{\circ}$
✓
Answer
Correct option: D.
$87^{\circ}$
(d) : Let $x$ and $y$ be the measures of $\angle A$ and $\angle B$ respectively.
$\begin{array}{lr}
\text { Now, } \angle A+\angle B+\angle C=180^{\circ} & \text { [By angle sum property] } \\
\Rightarrow x+y+50^{\circ}=180^{\circ} & \text { [Given, } \left.\angle C=50^{\circ}\right] \\
\Rightarrow x+y=130^{\circ} & \ldots \text { (i) }
\end{array}$
Also, $\angle A-\angle B=44^{\circ} \Rightarrow \quad x-y=44^{\circ}$$\ldots(ii)$
Adding (i) and (ii), we get
$2 x=174^{\circ} \Rightarrow x=87^{\circ} \Rightarrow \angle A=87^{\circ}$
$\begin{array}{lr}
\text { Now, } \angle A+\angle B+\angle C=180^{\circ} & \text { [By angle sum property] } \\
\Rightarrow x+y+50^{\circ}=180^{\circ} & \text { [Given, } \left.\angle C=50^{\circ}\right] \\
\Rightarrow x+y=130^{\circ} & \ldots \text { (i) }
\end{array}$
Also, $\angle A-\angle B=44^{\circ} \Rightarrow \quad x-y=44^{\circ}$$\ldots(ii)$
Adding (i) and (ii), we get
$2 x=174^{\circ} \Rightarrow x=87^{\circ} \Rightarrow \angle A=87^{\circ}$
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