Maths 1 Marks Question

Let $AB$ and $CD$ be the two towers such that $AB = x$ and $CD = y.$
We have,
$\angle\text{AEB}=30^\circ,\angle\text{CED}=60^\circ$ and $BE = DE$
In $\triangle\text{ABE},$
$\tan30^\circ=\frac{\text{AB}}{\text{BE}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{\text{x}}{\text{BE}}$
$\Rightarrow\text{BE}=\text{x}{\sqrt{3}}$
Also, in $\triangle\text{CDE},$
$\Rightarrow\tan60^\circ=\frac{\text{CD}}{\text{DE}}$
$\Rightarrow\sqrt{3}=\frac{\text{y}}{\text{DE}}$
$\Rightarrow\text{DE}=\frac{\text{y}}{\sqrt{3}}$
As, $BE = DE$
$\Rightarrow\text{x}\sqrt{3}=\frac{\text{y}}{\sqrt{3}}$
$\Rightarrow\frac{\text{x}}{\text{y}}=\frac{1}{\sqrt{3}\times\sqrt{3}}$
$\Rightarrow\frac{\text{x}}{\text{y}}=\frac13$
$\therefore\ \text{x}:\text{y}=1:3$

Let $ABCD$ be the rectangle in which $\angle\text{BAD}=30^\circ$ and $AC = 8\ cm.$

In $\triangle\text{BAC},$
We have:
$\frac{\text{AB}}{\text{AC}}=\cos30^\circ=\frac{\sqrt{3}}{2}$
$\Rightarrow\frac{\text{BC}}{8}=\frac{1}{2}$
$\Rightarrow\text{BC}=\frac82=4\text{m}$
$\therefore$ Area of the rectangle $=(\text{AB}\times\text{BC})=(4\sqrt{3}\times4)=16\sqrt{3}\text{cm}^2$​​

Let $AB$ be the cliff and $CD$ be the tower.
We have,
$AB = 20m$
Also, $CE = AB = 20m$
Let $\angle\text{ACB}=\angle\text{CAE}=\angle\text{DAE}=\theta$
In $\triangle\text{ABC},$
$\tan\theta=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\tan\theta=\frac{20}{\text{BC}}$
$\Rightarrow\tan\theta=\frac{20}{\text{AE}}$ $($As, $BC = AE)$
$\Rightarrow\text{AE}=\frac{20}{\tan\theta}\dots(\text{i})$
Also, in $\triangle\text{ADE},$
$\tan\theta=\frac{\text{DE}}{\text{AE}}$
$\Rightarrow\tan\theta=\frac{\text{DE}}{\Big(\frac{20}{\tan\theta}\Big)}$ $[$Using $(i)]$
$\Rightarrow\tan\theta=\frac{\text{DE}\times\tan\theta}{20}$
$\Rightarrow\text{DE}=\frac{20\times\tan\theta}{\tan\theta}$
$\Rightarrow\text{DE}=20\text{m}$
Now, $\text{CD} = \text{DE} + \text{CE}$
$=20+20$
$\therefore\ \text{CD}=40\text{m}$
Disclaimer: The answer given in the textbook is incorrect, The same has been rectified above.

Let $AB$ be the pole and $BC$ be its shadow and $\theta$ be the sun's elevation.
We have,
$AB = 12m$ and $\text{BC}=4\sqrt{3}\text{m}$
In $\triangle\text{ABC},$
$\tan\theta=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\tan\theta=\frac{12}{4\sqrt{3}}$
$\Rightarrow\tan\theta=\frac{3}{\sqrt{3}}$
$\Rightarrow\tan\theta=\frac{3}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$
$\Rightarrow\tan\theta=\frac{3\sqrt{3}}{3}$
$\Rightarrow\tan\theta=\sqrt{3}$
$\Rightarrow\tan\theta=\tan60^\circ$
$\therefore\ \theta=60^\circ$

Let point $A$ be the position of the kite and $AC$ be its string.
We have,
$AB = 30m$, and $AC = 60m$
Let $\angle\text{ACB}=\theta$
In $\triangle\text{ABC},$
$\sin\theta=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\sin\theta=\frac{30}{60}$
$\Rightarrow\sin\theta=\frac{1}{2}$
$\Rightarrow\sin\theta=\sin30^\circ$
$\therefore\ \theta=30^\circ$

Let the sun's altitude be $\theta.$
We have,
$AB = 20m$ and $\text{BC}=20\sqrt{3}\text{m}$
In $\triangle\text{ABC},$
$\tan\theta=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\tan\theta=\frac{20}{20\sqrt{3}}$
$\Rightarrow\tan\theta=\frac{1}{\sqrt{3}}$
$\Rightarrow\tan\theta=\tan30^\circ$
$\therefore\ \theta=30^\circ$

Let $AB$ be the observer and $CD$ be the tower.


Draw $\text{BE}\perp\text{CD},$ let $CD = h$ metres.
Then, $\text{AB}=1.5\text{m},\text{BE}=\text{AC}=28.5\text{m}$ and $\angle\text{EBD}=45^\circ.$
$DE = (CD - EC) = (CD - AB) = (h - 1.5)m$.
In right $\triangle\text{BED},$
We have:
${\frac{\text{DE}}{\text{BE}}}=\tan45^\circ=1$
$\Rightarrow\frac{(\text{h}-1.5)}{28.5}=1$
$\Rightarrow\text{h}-1.5=28.5$
$\Rightarrow\text{h}=28.5+1.5=30\text{m}$
Hence the height of the tower is $30m$.

Let $AB$ be the tower and point $C$ be the position of the car.
We have,
$AB = 150m$, and $\angle\text{ACB}=30^\circ$
In $\triangle\text{ABC},$
$\tan30^\circ=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{150}{\text{BC}}$
$\therefore\ \text{BC}=150\sqrt{3}\text{m}$

Let $AB$ be the wall and $AC$ be the ladder
We have,
$BC = 2m$, and $\angle\text{ACB}=60^\circ$
In $\triangle\text{ABC},$
$\cos60^\circ=\frac{\text{BC}}{\text{AC}}$
$\Rightarrow\frac{1}{2}=\frac{2}{\text{AC}}$
$\therefore\ \text{AC}=4\text{m}$

Let $AB$ be the pole and $BC$ be its shadow.


Let $AB = h$ and $BC = x$ such that $\text{x}=\sqrt{3}\text{h}$ (given) and $\theta$ be the angle of elevation.
From $\triangle\text{ABC},$ we have:
$\frac{\text{AB}}{\text{BC}}=\tan\theta$
$\Rightarrow\frac{\text{h}}{\text{x}}=\frac{\text{h}}{\sqrt{3}\text{h}}=\tan\theta$
$\Rightarrow\tan\theta=\frac{1}{\sqrt{3}\text{h}}$
$\Rightarrow\theta=30^\circ$

Let $AB$ be the pole and $AC$ and $AD$ be its shadows.
We have:
$\angle\text{ACB}=30^\circ,\angle\text{ADB}=60^\circ$ and AB = 15m.

In $\triangle\text{ACB},$
We have:
$\frac{\text{AC}}{\text{AB}}=\cot30^\circ=\sqrt{3}$
$\Rightarrow\frac{\text{AC}}{15}={\sqrt{3}}\Rightarrow\text{AC}=15\sqrt{3}\text{m}$
Now, in $\triangle\text{ADB},$ we have:
$\frac{\text{AD}}{\text{AB}}=\cot60^\circ=\frac{1}{\sqrt{3}}$
$\Rightarrow\frac{\text{AD}}{15}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{AD}=\frac{15}{\sqrt{3}}=\frac{15\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}$
$\Rightarrow\frac{15\sqrt{3}}{3}=5\sqrt{3}\text{m}$
$\therefore$ Difference between the lengths of the shadows $=\text{AC}-\text{AD}=15\sqrt{3}-5\sqrt{3}=10\sqrt{3}\text{m}$

Let $AB$ be the lamp-post; $CD$ be the girl and $DE$ be her shadow.
We have,
$CD = 1.5m, AD =3m, DE = 4.5m$
Let $\angle\text{E}=\theta$
In $\triangle\text{CDE},$
$\tan\theta=\frac{\text{CD}}{\text{DE}}$
$\Rightarrow\tan\theta=\frac{1.5}{4.5}$
$\Rightarrow\tan\theta=\frac{1}{3}\dots(\text{i})$
Now, in $\triangle\text{ABE},$
$\tan\theta=\frac{\text{AB}}{\text{AE}}$
$\Rightarrow\frac{1}{3}=\frac{\text{AB}}{\text{AD}+\text{DE}}$ $[$Using $(i)]$
$\Rightarrow\frac{1}{3}=\frac{\text{AB}}{3+4.5}$
$\Rightarrow\text{AB}=\frac{7.5}{3}$
$\therefore\ \text{AB}=2.5\text{m}$

Let $AB$ represents the vertical pole and $BC$ represents the shadow on the ground and $\theta$ represents angle of elevation the sun.
In $\triangle\text{ABC},$
$\tan\theta=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\tan\theta=\frac{\text{x}}{\text{x}}$ $($As, the height of the pole, $AB =$ the shadow, $BC = x)$
$\Rightarrow\tan\theta=1$
$\Rightarrow\tan\theta=\tan45^\circ$
$\therefore\theta=45^\circ$

Here, $AO$ be the pole; $BO$ be its shadow and $\theta$ be the angle of elevation of the sun.
Let $BO = x$
Then, $\text{AO}=\text{x}\sqrt{3}$
In $\triangle\text{AOB},$
$\tan\theta=\frac{\text{AO}}{\text{BO}}$
$\Rightarrow\tan\theta=\frac{\text{x}\sqrt{3}}{\text{x}}$
$\Rightarrow\tan\theta=\sqrt{3}$
$\Rightarrow\tan\theta=\tan60^\circ$
$\therefore\theta=60^\circ$