2 Marks Questions

Question 12 Marks

The angle of elevation of the top of a tower from a point on the ground, which is $30\ m$ away from the foot of the tower, is $30^\circ$. Find the height of the tower.

Answer

In right triangle $ABC$,
$\tan 30 ^ { \circ } = \frac { A B } { B C } \Rightarrow \frac { 1 } { \sqrt { 3 } } - \frac { A B } { 30 }$

$A B = \frac { 30 } { \sqrt { 3 } } \Rightarrow A B = 10 \sqrt { 3 } \mathrm { m }$
Hence, the height of the tower is $10\sqrt3$ m.

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Question 22 Marks

A circus artist is climbing a $20\ m$ long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is $30^\circ$.

Answer

In right triangle $ABC$,

$\sin 30 ^ { \circ } = \frac { A B } { A C } \Rightarrow \frac { 1 } { 2 } = \frac { A B } { 20 } \Rightarrow A B = 10 \mathrm { m }$
Hence, the height of the pole is $10\ m$.

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Question 32 Marks

The shadow of a tower standing on a level ground is found to be $40\ m$ longer when the Sun's altitude is $30^\circ$, than when it is $60^\circ$. Find the height of the tower.

Answer

In $\triangle A B C,$ $\tan 60 ^ { \circ } = \frac { A B } { B C }$
$\Rightarrow \quad A B = \sqrt { 3 } B C$ .....(i)
In $\triangle A B D,$
$\tan 30 ^ { \circ } = \frac { A B } { B C + 40 }$
$\frac{ 1}{√3}=\frac{AB}{BC+40}=\frac{√3BC}{BC+40}$
$3BC = BC + 40$
$BC = 20$, Hence from $(i)$ we get
$AB = 20\sqrt3 = 20$ $\times$ $1.73 = 34.6$ meter

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Maths 2 Marks Questions

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