Question

$\purple{ \large{Answer \: the \: question}}$
$\pink{ Don’t \: forget \: the \: correct \: sign,}$
$\pink{Math \: is \: so \: easy \: it’s \: divine.}$
​

Answer 1

$\bigstar \underline{\underline{\sf{DIAGRAM}}}$

$\setlength{\unitlength}{20} \begin{picture}(6,6)\put(5,1){\line(1,0){3.5}}\put(5,1){\line(0,1){3.5}}\put(5,4.5){\line(1,-1){3.5}} \put(5,1){\line(-1,0){6.95}}\put(5,4.5){\line(-2,-1){6.95}}\put(5.5,1){\line(0,1){0.4}}\put(4.5,1){\line(0,1){0.4}}\put(4.5,1.4){\line(1,0){1}}\put(4.75,0.5){ \bf C }\put(8.25,0.5){ \bf D }\put(-2,0.5){ \bf B }\put(5,4.75){ \bf A }\put(1,0.5){\vector(-1,0){2.5}}\put(2.4,0.5){\vector(1,0){2.5}}\put(1,0.4){ \tt 15 \: cm }\end{picture}$

$\sf{(i)\:\: sin\ B=\dfrac{4}{5} }$

We know,

◘ sin θ = Perpendicular/Hypotenuse

⇒ sin θ = p/h

⇒ sin B = p/h

⇒ p/h = 4/5

⇒ AC = 4x cm and AB = 5x cm

Applying Pythagoras theorem :

AB² = BC² + AC²

⇒ (5x)² = BC² + (4x)²

⇒ 25x² = BC² + 16x²

⇒ BC² = 25x² - 16x²

⇒ BC² = 9x²

⇒ BC = 3x

Now,

BC = 15 cm = 3x

⇒ x = 15/3

⇒ x = 5

◘ Substituting the value of x :-

AB = 5x = 25 cm

AC = 4x = 20 cm

______________________

tan ∠ADC = 1

⇒ AC/CD = 1

⇒ AC = CD

Let AC = CD = k.

Applying Pythagoras theorem in Δ ACD,

⇒ (AD)² = (AC)² + (CD)²

⇒ AD² = k² + k²

⇒ AD² = 2k²

⇒ AD = √2k

From (i), we got that AC = 20 cm

→ k = 20 cm

⇒ AD = √2 × 20

⇒ AD = 20√2 cm

CD = k = 20 cm

______________________

tan B = p/b = AC/BC

⇒ tan B = 20/15

⇒ tan B = 4/3

cos B = b/h = BC/AB

⇒ cos B = 15/25

⇒ cos B = 3/5

$\sf{tan^2B-\dfrac{1}{cos^2B} }\\\\\displaystyle{\sf{= \bigg(\frac{4}{3} \bigg)^2-\frac{1}{(3/5)^2} }}\\\\\displaystyle{\sf{= \frac{16}{9}-\frac{1}{9/25} }}\\\\\displaystyle{\sf{= \frac{16}{9}-\frac{25}{9} }}\\\\\displaystyle{\sf{= \frac{16-25}{9} }}\\\\\displaystyle{\sf{= \frac{-9}{9} }}\\\\\displaystyle{\sf{= 1=RHS}}$

Answer 2

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

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

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

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

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

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