Case study (4 Marks)

Question 14 Marks

Read the following text carefully and answer the questions that follow: A bird is sitting on the top of a tree, which is $80\ m$ high. The angle of elevation of the bird, from a point on the ground is $45^{\circ}$. The bird flies away from the point of observation horizontally and remains at a constant height. After $2$ seconds, the angle of elevation of the bird from the point of observation becomes $30^{\circ}$. Find the speed of flying of the bird.

$i.$ Find the distance between observer and the bottom of the tree?
$ii$. Find the speed of the bird?
$iii$. Find the distance between second position of bird and observer?
OR
Find the distance between initial position of bird and observer?

Answer

$i.$ Given height of tree $= 80m, P$ is the initial position of bird and $Q$ is position of bird after $2 \sec$ the distance between observer and the bottom of the tree
$\text { In } \triangle ABP$
$\tan 45^{\circ}=\frac{B P}{A B}$
$\Rightarrow 1=\frac{80}{A B}$
$\Rightarrow A B=80 m$
$ii$. The speed of the bird
$\text { In } \triangle AQC$
$\tan 30^{\circ}=\frac{Q C}{A C}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{80}{A C}$
$\Rightarrow AC =80 \sqrt{3}\ m$
$AC - AB = BC$
$\Rightarrow B C=80 \sqrt{3}-80$
$=80(\sqrt{3}-1) \ m$
Speed of bird $=\frac{\text { Distance }}{\text { Time }}$
$\Rightarrow \frac{B C}{2}$
$=\frac{80(\sqrt{3}-1)}{2}=40(\sqrt{3}-1)$
$\Rightarrow$ Speed of the bird $=29.28\ m / \sec$
$iii$. The distance between second position of bird and observer.
$\text { In } \triangle AQC$
$\sin 30^{\circ}=\frac{Q C}{A Q}$
$\Rightarrow \frac{1}{2}=\frac{80}{A Q}$
$\Rightarrow AQ =160 \ m$
OR
The distance between initial position of bird and observer
In $\triangle ABP$
$\sin 45^{\circ}=\frac{B P}{A P}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{80}{A P}$
$\Rightarrow A P=80 \sqrt{2} \ m$

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Question 24 Marks

Read the following text carefully and answer the questions that follow:
Mary and John are very excited because they are going to go on a dive to see a sunken ship. The dive is quite shallow which is unusual because most sunken ship dives are found at depths that are too deep for two junior divers. However, this one is at 40 feet, so the two divers can go to see it.

They have the following map to chart their course. John wants to figure out exactly how far the boat will be from the sunken ship. Use the information in this lesson to help John figure out the following.
i. What are the coordinates of the boat and the sunken ship respectively?
ii. How much distance will Mary and John swim through the water from the boat to the sunken ship?
iii. If each square represents 160 cubic feet of water, how many cubic feet of water will Mary and John swim through from the boat to the sunken ship?
OR
If the distance between the points (x, -1) and (3, 2) is 5, then what is the value of x?

Answer

i. (4, 8) and (-3, 7)
ii. 8 units
iii. 1280 cubic feet
OR
7 or -1

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Question 34 Marks

Read the following text carefully and answer the questions that follow:
Elpis Technology is a laptop manufacturer. The company works for many branded laptop companies and also provides them with spare parts. Elpis Technology produced $6000$ units in $3^{rd}$ year and $7000$ units in the $7^{th}$ year.

Assuming that production increases uniformly by a fixed number every year.
$i.$ Find the production in the $1^{st}$ year.
$ii.$ Find the production in the $5^{th}$ year.
$iii$. Find the total production in $7$ years.
OR
Find in which year $10000$ units are produced?

Answer

$i$. Let production in a $1^{st}$ year be $'a\ '$ unit and increase in production $($every year$)$ be $'d\ '$ units.
Increase in production is constant, therefore unit produced every year forms an $AP$.
Now, $a_3=6000$
$a+2 d=6000 $
$\Rightarrow a=6000-2 d \ldots \text { (i) }$
and $ a_7=7000$
$ \Rightarrow a+6 d=7000$
$\Rightarrow(6000-2 d)+6 d=7000 $
$\Rightarrow 4 d=1000 \ [$using eq. $(i)]$
$\Rightarrow d =250$
When $d = 250,$ eq. $(i)$ becomes
$a = 6000 - 2(250) = 5500$
$\therefore$ Production in $1^{st}$ year $=5500$
$ii$. Production in fifth year
$a_5=a+4 d$
$=5500+4(250)=6500$
$iii$. Total production in $7$ years $=\frac{7}{2}(5500+7000)=43750$
OR
$a_n=1000$ units
$a_n=1000$
$\Rightarrow 10000=a+(n-1) d$
$\Rightarrow 1000=5500+250 n-250$
$\Rightarrow 10000-5500+250=250 n$
$\Rightarrow 4750=250 n$
$\Rightarrow n=\frac{4750}{250}=19$

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Maths Case study (4 Marks)

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