Math, asked by riz92, 4 months ago

plzzz solve it??!??​

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Answers

Answered by BrainlyEmpire
53

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}

Answered by BabeHeart
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}

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