Question
Find the unknown angles in the given figure:
✓
Answer
In the figure,
$a=b$ (Angles opposite to equal sides)
But $a+b+80^{\circ}=180^{\circ}$ (Angles of a triangle)
$ \Rightarrow \mathrm{a}+\mathrm{a}+80^{\circ}=180^{\circ} $
$\Rightarrow 2 \mathrm{a}+80^{\circ}=180^{\circ}$
$\Rightarrow 2 \mathrm{a}=180^{\circ}-80^{\circ}=100^{\circ}$
$\Rightarrow \mathrm{a}=\frac{100^{\circ}}{2}=50^{\circ}$
$ \therefore \mathrm{b}=\mathrm{a}=50^{\circ} $
$x=a+80^{\circ}$ (Exterior angle of a triangle is equal to sum of its opposite interior angles)
$ =50^{\circ}+80^{\circ}=130^{\circ} $
Hence $\mathrm{a}=50^{\circ}, \mathrm{b}=50^{\circ}$ and $\mathrm{x}=130^{\circ}$
$a=b$ (Angles opposite to equal sides)
But $a+b+80^{\circ}=180^{\circ}$ (Angles of a triangle)
$ \Rightarrow \mathrm{a}+\mathrm{a}+80^{\circ}=180^{\circ} $
$\Rightarrow 2 \mathrm{a}+80^{\circ}=180^{\circ}$
$\Rightarrow 2 \mathrm{a}=180^{\circ}-80^{\circ}=100^{\circ}$
$\Rightarrow \mathrm{a}=\frac{100^{\circ}}{2}=50^{\circ}$
$ \therefore \mathrm{b}=\mathrm{a}=50^{\circ} $
$x=a+80^{\circ}$ (Exterior angle of a triangle is equal to sum of its opposite interior angles)
$ =50^{\circ}+80^{\circ}=130^{\circ} $
Hence $\mathrm{a}=50^{\circ}, \mathrm{b}=50^{\circ}$ and $\mathrm{x}=130^{\circ}$
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→Q.1. Today he managed to reach the railway crossing in 3 hours. Find his speed in m/sec:
(a) 2 m/sec$\quad$$\quad$ (b) 2.5 m/sec $\quad$$\quad$(c) 3 m/sec $\quad$$\quad$(d) 4.5 m/sec
Q.2. If tomorrow he wishes to cover this distance in 2 hours, by how much does he need to increase his speed?
(a) 1.25 m/s$\quad$$\quad$ $\quad$$\quad$(b) 1.5 m/s$\quad$$\quad$ (c) 1.75 m/s $\quad$$\quad$(d) 2.25 m/s
Q.3. Today, he has to stop at the railway crossing for a train to pass. What time will the train 180 m long, running at 54 km per hour take to pass him:
(a) 9 sec$\quad$$\quad$(b) 10 sec $\quad$$\quad$(c) 12 sec $\quad$$\quad$(d) 15 sec
Q.4. What time will this train take to pass a platform 300 m long at the speed of 10 m/s?
(a) 30 sec $\quad$$\quad$(b) 32 sec $\quad$$\quad$(c) 36 sec$\quad$$\quad$ (d) 35 sec