Relation Between Altitude Base and Hypotenuse


 
 
Concept Explanation
 

Relation Between Altitude Base and Hypotenuse

Relation Between Altitude Base and Hypotenuse:

Let us refer to fig. again. If you want to find the height CD of the minar without actually measuring it, what information do you need? You would need to know the following:

(i) The distance DE at which the student is standing fromthe foot of the minar.

(ii) the angle of elevation, large angle BAC, of the top of the minar.

(iii) the height AE of the student.

Assuming that the above three conditions are known, how can we determine the height of the minar?

In the figure, CD = CB+ BD. Here, BD = AE, which is the height of the student.

To find BC, we eill use trigonometric ratios of large angle BAC;or;angle A

In large Delta ABC, the side BC is the opposite side in relation to the known large angle A. Now, which of the trigonometric ratios can we use? Which one of them has the two values that we have and the one we need to determine? Our search narrows down to using either tan A or cot A, as these ratios involve AB and BC.

Therefore,  large tan;A=frac{BC}{AB} or large cot;A=frac{AB}{BC}, which on solving would given us BC.

By adding AE to BC, you will get the height of the minar.

Example: An electrician has to repair an electric fault on a pole of height 5 m.She needs to reach a point 1.3m below the top of the pole to undertake the repair work (see fig.) What should be the length of the ladder that she should use which, when inclined at an angle of large 60^{circ} to the horizontal, would enable her to reach the required position ? Also, how far from the foot of the pole should she place the foot of the ladder? (You may take large sqrt{3}=1.73)

SOLUTION: In fig. the electrician is requried to reach the point B on the pole AD.

SO,     BD = AD - AB = (5 - 1.3)m = 3.7 m.

Here, BC represents the ladder. We need to find its length, i.e.., the hypotenuse of the right triangle BDC.

Now, can you think which trigonometric ratio should we consider?

It should be sin large 60^{circ}.

So,          large frac{BD}{BC}=sin 60^{circ};;or;;frac{3.7}{BC}=frac{sqrt{3}}{2}

Therefore,     large BC=frac{3.7times2}{sqrt{3}}=4.28m

i.e., the length of the laddershould be 4.28 m.

Now,         large frac{DC}{BD}=cot;60^{circ}=frac{1}{sqrt{3}}

i.e.,          large DC=frac{3.7}{sqrt{3}}=2.14;m  (approx)

Therefore, she should place the foot of the ladder at a distance of 2.14 m from the pole.

 
 
 


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