Question

plzzz solve it??!??​

Answer 1

Given:-

BC= 15cm

$\rm{sin\ B=\dfrac{4}{5}}$

To Find:-

i) Measurement of AB and AC

ii) Measurement of CD and AD, if tan(∠ADC)=1

To Prove:-

$\longrightarrow\bf{tan^{2}\ B-\dfrac{1}{cos^{2}\ B}=-1}$

Solution:-

In ΔABC,

$\longrightarrow\rm{cos\ B=\sqrt{1-sin^{2}B}}$

$\longrightarrow\rm{cos\ B=\sqrt{1-\bigg(\dfrac{4}{5}\bigg)^{2}}}$

$\longrightarrow\rm{cos\ B=\sqrt{1-\dfrac{16}{25}}}$

$\longrightarrow\rm{cos\ B=\sqrt{\dfrac{25-16}{25}}}$

$\longrightarrow\rm{cos\ B=\sqrt{\dfrac{9}{25}}}$

$\longrightarrow\rm{cos\ B=\dfrac{3}{5}}$

Now,

$\longrightarrow\rm{cos\ B=\dfrac{BC}{AB}}$

$\longrightarrow\rm{\dfrac{3}{5}=\dfrac{15}{AB}}$

$\longrightarrow\rm{AB=\dfrac{15\times5}{3}}$

$\longrightarrow\rm\green{AB=25\ cm}$

Now,

$\longrightarrow\rm{sin\ B=\dfrac{AC}{AB}}$

$\longrightarrow\rm{\dfrac{4}{5}=\dfrac{AC}{25}}$

$\longrightarrow\rm{\dfrac{4\times25}{5}=AC}$

$\longrightarrow\rm{AC=\dfrac{4\times25}{5}}$

$\longrightarrow\rm\green{AC=20\ cm}$

━━━━━━━━━━━━━━━━━━━━━

Now, In ΔACD

tan(∠ADC)=1

That means,

∠ADC= 45°

Or

D= 45°

So,

$\longrightarrow\rm{tan\ D=\dfrac{AC}{CD}}$

$\longrightarrow\rm{tan\ 45^{\circ}=\dfrac{20}{CD}}$

$\longrightarrow\rm{1=\dfrac{20}{CD}}$

$\longrightarrow\rm\green{CD=20\ cm}$

Now,

$\longrightarrow\rm{sin\ D=\dfrac{AC}{AD}}$

$\longrightarrow\rm{sin\ 45^{\circ}=\dfrac{20}{AD}}$

$\longrightarrow\rm{\dfrac{1}{\sqrt{2}} =\dfrac{20}{AD}}$

$\longrightarrow\rm\green{AD=20\sqrt{2}\ cm}$

━━━━━━━━━━━━━━━━━━━━━

Now, we know that,

$\pink{\underline{\boxed{\bf{1+tan^{2}\theta=sec^{2}\theta}}}}$

$\purple{\underline{\boxed{\bf{sec \theta}=\dfrac{1}{cos\theta}}}}$

So, by using above properties

$\longrightarrow\rm{1+tan^{2}B=sec^{2}B}$

$\longrightarrow\rm{tan^{2}B-sec^{2}B=-1}$

$\longrightarrow\rm{tan^{2}B-\bigg(\dfrac{1}{cosB} \bigg)^{2}=-1}$

$\longrightarrow\rm{tan^{2}B-\dfrac{1}{cos^{2}B}=-1}$

$\bigstar\bf\red{Hence\ Proved}$

Answer 2

$\large \bf \pink{Given:-}$

BC= 15cm

$\rm{sin\ B=\dfrac{4}{5}}$

$\large \bf \pink{To \: Find:-}$

i) Measurement of AB and AC

ii) Measurement of CD and AD, if tan(∠ADC)=1

$\large \bf \pink{To \: Prove:-}$

$\longrightarrow\bf{tan^{2}\ B-\dfrac{1}{cos^{2}\ B}=-1}$

Solution:-

In ΔABC,

$\longrightarrow\rm{cos\ B=\sqrt{1-sin^{2}B}}$

$\longrightarrow\rm{cos\ B=\sqrt{1-\bigg(\dfrac{4}{5}\bigg)^{2}}}$

$\longrightarrow\rm{cos\ B=\sqrt{1-\dfrac{16}{25}}}$

$\longrightarrow\rm{cos\ B=\sqrt{\dfrac{25-16}{25}}}$

$\longrightarrow\rm{cos\ B=\sqrt{\dfrac{9}{25}}}$

$\longrightarrow\rm{cos\ B=\dfrac{3}{5}}$

Now,

$\longrightarrow\rm{cos\ B=\dfrac{BC}{AB}}$

$\longrightarrow\rm{\dfrac{3}{5}=\dfrac{15}{AB}}$

$\longrightarrow\rm{AB=\dfrac{15\times5}{3}}$

$\longrightarrow\rm\green{AB=25\ cm}$

Now,

$\longrightarrow\rm{sin\ B=\dfrac{AC}{AB}}$

$\longrightarrow\rm{\dfrac{4}{5}=\dfrac{AC}{25}}$

$\longrightarrow\rm{\dfrac{4\times25}{5}=AC}$

$\longrightarrow\rm{AC=\dfrac{4\times25}{5}}$

$\longrightarrow\rm\green{AC=20\ cm}$

━━━━━━━━━━━━━━━━━━━━━

Now, In ΔACD

tan(∠ADC)=1

That means,

∠ADC= 45°

Or

D= 45°

So,

$\longrightarrow\rm{tan\ D=\dfrac{AC}{CD}}$

$\longrightarrow\rm{tan\ 45^{\circ}=\dfrac{20}{CD}}$

$\longrightarrow\rm{1=\dfrac{20}{CD}}$

$\longrightarrow\rm\green{CD=20\ cm}$

Now,

$\longrightarrow\rm{sin\ D=\dfrac{AC}{AD}}$

$\longrightarrow\rm{sin\ 45^{\circ}=\dfrac{20}{AD}}$

$\longrightarrow\rm{\dfrac{1}{\sqrt{2}} =\dfrac{20}{AD}}$

$\longrightarrow\rm\green{AD=20\sqrt{2}\ cm}$

━━━━━━━━━━━━━━━━━━━━━

Now, we know that,

$\pink{\underline{\boxed{\bf{1+tan^{2}\theta=sec^{2}\theta}}}}$

$\purple{\underline{\boxed{\bf{sec \theta}=\dfrac{1}{cos\theta}}}}$

So, by using above properties

$\longrightarrow\rm{1+tan^{2}B=sec^{2}B}$

$\longrightarrow\rm{tan^{2}B-sec^{2}B=-1}$

$\longrightarrow\rm{tan^{2}B-\bigg(\dfrac{1}{cosB} \bigg)^{2}=-1}$

$\longrightarrow\rm{tan^{2}B-\dfrac{1}{cos^{2}B}=-1}$

$\bigstar\bf\blue{Hence\ Proved}$