Laravel

Chapters For Class X- CBSE


Example: The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.

Solution : In Fig. 9.8, AB is the tower and BC is the length of the shadow when the Sun’s altitude is 60°, i.e., the angle of elevation of the top of the tower from the tip of the shadow is 60° and DB is the length of the shadow, when the angle of elevation is 30°.

Now, let AB be h m and BC be x m. According to the question, DB is 40 m longer than BC.
So,                                  DB =  (40 + $x$) m
Now, we have two right triangles ABC and ABD.
In ∆ ABC,                 tan 60° = ${AB}/{BC}$
						  √3= $h/x$
						  $h=√3x$ ........................(1)
In ∆ ABD,                tan 30° =${AB}/{BD}$
                         $1/√3$	=$h/{x+40}$...............(2)	
						 Putting the value of $h$ in equation  (2)
						 $1/√3$	=${√3x}/{x+40}$
						 $3x=x+40$
						 $x=20$
			So, $h=20√3$
Ans:Therefore, the height of the tower is 20√3m.
Example: The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively.Find the height of the multi-storeyed building and the distance between the two buildings.

Solution: In Fig. 9.9, PC denotes the multi- storyed building and AB denotes the 8 m tall building. We are interested to determine the height of the multi-storeyed building, i.e., PC and the distance between the two buildings, i.e., AC.

Look at the figure carefully. Observe that PB is a transversal to the parallel lines PQ and BD.

Therefore, ∠ QPB and ∠ PBD are alternate angles, and so are equal.

So ∠ PBD = 30°. Similarly, ∠ PAC = 45°.

In right ∆ PBD, we have
		${PD}/{BD}=tan 30° =1/√3$; i.e $BD=PD √3$
		
In right ∆ PAC, we have,
		${PC}/{AC}=tan 45° = 1$
		i.e.,PC =  AC
		Also, PC =  PD + DC, therefore, PD + DC = AC.
		Since, AC = BD and DC = AB = 8 m, we get PD + 8 = BD =  PD√3
        i.e $PD={8}/{√3 - 1}$ =${8(√3 + 1)}/{3-1}$ =$4(√3 + 1)$	

So, the height of the multi-storeyed building is 4(√3 + 1) +8 = $4(√3 + 3) m$	
and the distance between the two buildings is also $4(√3 + 3) m$
Example: From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.

Solution : In Fig 9.10, A and B represent points on the bank on opposite sides of the river, so that AB is the width of the river . P is a point on the bridge at a height of 3 m, i.e., DP = 3 m. We are interested to determine the width of the river, which is the length of the side AB of the ∆ APB. 
	Now,                  AB =  AD + DB
	In right ∆ APD, ∠ A = 30°.
	So, tan 30° =${PD}/{AD}$
	$1/√3={3}/{AD}$
	$AD=3 √3 $
	
	Also, in right ∆ PBD, ∠ B = 45°. So, BD = PD = 3 m.
	Now, AB =  BD + AD = 3 + 3 √3  = 3 (1 +√3) m.
 
Therefore, the width of the river is 3 (√3+ 1)m .