Question 14 Marks
Read the text carefully and answer the questions:
The houses of Ajay and Sooraj are at $100 m$ distance and the height of their houses is the same as approx $150 m$ . One big tower was situated near their house.
Once both friends decided to measure the height of the tower.
They measure the angle of elevation of the top of the tower from the roof of their houses.
The angle of elevation of ajay's house to the tower and sooraj's house to the tower are $45^{\circ}$ and $30^{\circ}$ respectively as shown in the figure.
$(a)$ Find the height of the tower.
$(b)$ What is the distance between the tower and the house of Sooraj?
OR
Find the distance between top of tower and top of Ajay's house?
$(c)$ Find the distance between top of the tower and top of Sooraj's house?
The houses of Ajay and Sooraj are at $100 m$ distance and the height of their houses is the same as approx $150 m$ . One big tower was situated near their house.
Once both friends decided to measure the height of the tower.
They measure the angle of elevation of the top of the tower from the roof of their houses.
The angle of elevation of ajay's house to the tower and sooraj's house to the tower are $45^{\circ}$ and $30^{\circ}$ respectively as shown in the figure.
$(a)$ Find the height of the tower.
$(b)$ What is the distance between the tower and the house of Sooraj?
OR
Find the distance between top of tower and top of Ajay's house?
$(c)$ Find the distance between top of the tower and top of Sooraj's house?
Answer
View full question & answer→Read the text carefully and answer the questions:
The houses of Ajay and Sooraj are at $100 m$ distance and the height of their houses is the same as approx $150 m$. One big tower was situated near their house.
Once both friends decided to measure the height of the tower.
They meaof the top of the tower from the roof of their houses.
The angle of elevation of ajay's house to the tower and sooraj's house to the sure the angle of elevation tower are $45^{\circ}$ and $30^{\circ}$ respectively as shown in the figure.
$(i)$ The above figure can be redrawn as shown below:
Let $P Q=y$
$\text { In } \triangle P Q A,$
$\tan 45=\frac{P Q}{Q A}=\frac{y}{x}$
$1=\frac{y}{x}$
$x=y \ldots \text { (i) }$
$\text { In } \triangle P Q B,$
$\tan 30=\frac{P Q}{Q B}=\frac{P Q}{x+100}=\frac{y}{x+100}=\frac{x}{x+100}$
$\frac{1}{\sqrt{3}}=\frac{x}{x+100}$
$x \sqrt{3}=x+100$
$x=\frac{100}{\sqrt{3}-1}=136.61 m$
From the figure, height of tower $h=P Q+Q R$
$=x+150=136.61+150$
$h=286.61 m$
$(ii)$ The above figure can be redrawn as shown below:
Distance of Sooraj's house from tower $= QA + AB =x+100$
$=136.61+100=236.61 m$
OR
The above figure can be redrawn as shown below:
Distance between top of the tower and top of Ajay's house is $PA$
$\text { In } \triangle PQA$
$\sin 45^{\circ}=\frac{P Q}{P A}$
$\Rightarrow P A=\frac{P Q}{\sin 45^{\circ}}$
$\Rightarrow P A=\frac{y}{\frac{1}{\sqrt{2}}}=\sqrt{2} \times 136.61$
$\Rightarrow P A=193.20 m$
$(iii)$ The above figure can be redrawn as shown below:
Distance between top of tower and Top of Sooraj's house is $PB$
$\text { In } \triangle PQB$
$\sin 30^{\circ}=\frac{P Q}{P B}$
$\Rightarrow P B=\frac{P Q}{\sin 30^{\circ}}$
$\Rightarrow PB=\frac{y}{\frac{1}{2}}=2 \times 136.61$
$\Rightarrow PB=273.20 m$
The houses of Ajay and Sooraj are at $100 m$ distance and the height of their houses is the same as approx $150 m$. One big tower was situated near their house.
Once both friends decided to measure the height of the tower.
They meaof the top of the tower from the roof of their houses.
The angle of elevation of ajay's house to the tower and sooraj's house to the sure the angle of elevation tower are $45^{\circ}$ and $30^{\circ}$ respectively as shown in the figure.
$(i)$ The above figure can be redrawn as shown below:
Let $P Q=y$
$\text { In } \triangle P Q A,$
$\tan 45=\frac{P Q}{Q A}=\frac{y}{x}$
$1=\frac{y}{x}$
$x=y \ldots \text { (i) }$
$\text { In } \triangle P Q B,$
$\tan 30=\frac{P Q}{Q B}=\frac{P Q}{x+100}=\frac{y}{x+100}=\frac{x}{x+100}$
$\frac{1}{\sqrt{3}}=\frac{x}{x+100}$
$x \sqrt{3}=x+100$
$x=\frac{100}{\sqrt{3}-1}=136.61 m$
From the figure, height of tower $h=P Q+Q R$
$=x+150=136.61+150$
$h=286.61 m$
$(ii)$ The above figure can be redrawn as shown below:
Distance of Sooraj's house from tower $= QA + AB =x+100$
$=136.61+100=236.61 m$
OR
The above figure can be redrawn as shown below:
Distance between top of the tower and top of Ajay's house is $PA$
$\text { In } \triangle PQA$
$\sin 45^{\circ}=\frac{P Q}{P A}$
$\Rightarrow P A=\frac{P Q}{\sin 45^{\circ}}$
$\Rightarrow P A=\frac{y}{\frac{1}{\sqrt{2}}}=\sqrt{2} \times 136.61$
$\Rightarrow P A=193.20 m$
$(iii)$ The above figure can be redrawn as shown below:
Distance between top of tower and Top of Sooraj's house is $PB$
$\text { In } \triangle PQB$
$\sin 30^{\circ}=\frac{P Q}{P B}$
$\Rightarrow P B=\frac{P Q}{\sin 30^{\circ}}$
$\Rightarrow PB=\frac{y}{\frac{1}{2}}=2 \times 136.61$
$\Rightarrow PB=273.20 m$
