Case study inverse trigonometric 1 chapter 2 class 12

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Csae study Chapter 2 (Inverse Trigonometric functio)

Case study 1:-  The government of India is planning to fix a hoarding board at the face of a building on yhe road of bisy market for awaeeness on COVID-19 protocol. Ram , Robert and Rahim are the three engineers who are working on this project. “A” is considered to be person viewing the hoarding board 20 metres away from the building , standing at the edge of a pathwaynearby . Ram , Robert and Rahim suggested to the firm to place the hoarding board at three different locations namely C, D and E. “C” is at the height of  10 metres from the ground level. for the viewer A, the angle of elevation of “D” is double the angle of elevation of “C” the angle of elevation of “E” is triple the angle of elevation of “C” for the same viewer.(Case study inverse trigonometric 1 )

Based on the above information answer the following question:

(i) Find the measure of ∠DAB.

(ii) Find the measure of ∠EAB.

Solution :(i)Let ∠DAB = 2α

In ΔABC

\tan \alpha = \frac{BC}{AB} =\frac{10}{20} = \frac{1}{2}

\therefore \tan 2\alpha = \frac{2\tan\alpha}{1-\tan^2\alpha}

= \frac{2\times \frac{1}{2}}{1-\frac{1}{4}}

\tan 2\alpha = \frac{1}{3/4}= \frac{4}{3}

\Rightarrow 2\alpha = \tan^{-1}(4/3)

⇒ ∠DAB = \tan^{-1}(4/3)

(ii) Let ∠EAB = 3α

We have

\tan\alpha = \frac{1}{2}

\therefore \tan 3\alpha = \frac{3\tan\alpha-\tan^3\alpha}{1-3\tan^2\alpha}

= \frac{3\times \frac{1}{2}-(\frac{1}{2})^3}{1-3\times (\frac{1}{2})^2)}

\tan 3\alpha = \frac{\frac{3}{2}-\frac{1}{8}}{1-\frac{3}{4}}

= \frac{11/8}{1/4}= \frac{11}{8}\times 4=\frac{11}{2}

3\alpha =\tan^{-1}(11/2)

⇒ ∠EAB  = \tan^{-1}(11/2).

 

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