Please solve 8c(i,ii)

Please solve 8c(i,ii) In AABC,AB AC • 15 - 18 (d) /_ACu, Hint. AD to BC, D is mica-point of so BD A D = 12 S the fig ( ) given 10 ABC is isosce wi th AB AC 5 BC- 6 cm. Find (ii) tan B sin C (iii) tan C —cot B, (b) In the figure (2) given A ABC is right-angled at B. that ZAC-B Side AB = 2 and Side BC z I unit, find the value Of tan20, (c) In the figure (3) given below, AD is perpendicular to BC. BD B z — and tan C I. (i) Calculate the lengths Of AD, AB, DC and AC. If t-ao 19 Giv 20 11 22 23 2 9 (it) Show that tanz B — c•os2 B Hint. (a) Draw AD perpendicular to BC, then BD If sin 9 = and 9 is acute angle, find DC = 3 cm and AD— 4 cm.

Dear Student,

Please find below the solution to the asked query:

8 ( c ) Given : Sin B  = $\frac{4}{5}$  , We know :  Sin $\mathrm{\theta } = \frac{\mathrm{Opposite}}{\mathrm{Hypotenuse}}$  ,So 

Sin B = $\frac{\mathrm{AD}}{\mathrm{AB}}$
We assume ratio coefficient of sides =  x , So  AD  = 4 x  and AB = 5 x 

Now we apply Pythagoras theorem we get :

AB² = AD² + BD² , Substitute given values we get :

(5x)² = (4x)² + BD² ,

25x² =  16x² + BD² ,

BD² = 9x²,

BD = 3 x 

But we know BD  = 15 cm ( Given ) , So

3 x = 15

x  = 5 , So

AD = 4x  = 4 ( 5 ) = 20  cm

And

AB = 5 x = 5 ( 5 ) = 25 cm

And

Given : tan C  = 1  , We know : tan $\mathrm{\theta } = \frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$  ,So 

tan C = $\frac{\mathrm{AD}}{\mathrm{DC}}$ , So

1 = $\frac{20}{\mathrm{DC}}$ ,

DC =  20 cm

Now we apply Pythagoras theorem in triangle ACD we get :

AC² = AD² + DC² , Substitute values we get :

AC² = 20² + 20² ,

AC² = 400 + 400 ,

AC² = 800 ,

AC = 20$\sqrt{2} =20\times 1.414 =\mathbf{28}\mathbf{.}\mathbf{28}\mathbf{ }\mathbf{cm}$

Therefore,

AD  = 20 cm , AB = 25 cm , DC = 20 cm and AC = 28.28 cm                                               ( Ans )


ii ) tan B  = $\frac{\mathrm{AD}}{\mathrm{BD}} = \frac{20}{15} =\mathbf{ }\frac{\mathbf{4}}{\mathbf{3}}$   And We know :  Cos $\mathrm{\theta } = \frac{\mathrm{Adjacent}}{\mathrm{Hypotenuse}}$  ,So Cos B = $\frac{\mathrm{BD}}{\mathrm{AB}} = \frac{15}{25} = \frac{\mathbf{3}}{\mathbf{5}}$

To Show : ${\tan }^2 \mathrm{B} - \frac{1}{{\mathrm{Cos}}^2 \mathrm{B}} = -1$ , We take L.H.S. and substitute values , As :

$$\begin{gathered} {\tan }^2 \mathrm{B} - \frac{1}{{\mathrm{Cos}}^2 \mathrm{B}} \\ \Rightarrow {\left(\frac{4}{3}\right)}^2 - \frac{1}{{\left({\displaystyle \frac{3}{5}}\right)}^2} \\ \Rightarrow {\left(\frac{4}{3}\right)}^2 - {\left(\frac{5}{3}\right)}^2 \\ \Rightarrow \frac{16}{9} - \frac{25}{9} \\ \Rightarrow \frac{- 9}{9} \\ \mathbf{\Rightarrow }\mathbf{-}\mathbf{1} \end{gathered}$$
Therefore,

L.H.S. = R.H.S.                                            ( Hence proved )



Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards