Question

please answer it......​

Answer 1

Q1. Since the triangle is right angled at B, AB and BC are perpendicular sides and AC is the hypotenuse. So,

$\longrightarrow\sf{AC=\sqrt{AB^2+BC^2}}$

$\longrightarrow\sf{AC=\sqrt{12^2+5^2}}$

$\longrightarrow\sf{AC=\sqrt{144+25}}$

$\longrightarrow\sf{AC=\sqrt{169}}$

$\longrightarrow\sf{AC=13\ cm}$

Then,

$\sf{(i)\quad\!\!\sin A=\dfrac{BC}{AC}}$

$\longrightarrow\sf{\underline{\underline{\sin A=\dfrac{5}{13}}}}$

And,

$\longrightarrow\sf{\tan A=\dfrac{BC}{AB}}$

$\longrightarrow\sf{\underline{\underline{\tan A=\dfrac{5}{12}}}}$

$\sf{(ii)\quad\!\!\!\sin C=\dfrac{AB}{AC}}$

$\longrightarrow\sf{\underline{\underline{\sin C=\dfrac{12}{13}}}}$

And,

$\longrightarrow\sf{\cot C=\dfrac{BC}{AB}}$

$\longrightarrow\sf{\underline{\underline{\cot C=\dfrac{5}{12}}}}$

Q2. Given,

$\displaystyle\longrightarrow\sf{\cot\theta=\dfrac{20}{21}}$

First we assume $\theta$ is an acute angle. Then $\theta$ can be the angle of a right triangle of perpendicular sides $\sf{20\ cm}$ which is adjacent to $\theta,$ and $\sf{21\ cm}$ which is opposite to $\theta.$

Hence the length of the hypotenuse of the triangle is,

$\displaystyle\longrightarrow\sf{\sqrt{20^2+21^2}\ cm}$

$\displaystyle\longrightarrow\sf{\sqrt{400+441}\ cm}$

$\displaystyle\longrightarrow\sf{\sqrt{841}\ cm}$

$\displaystyle\longrightarrow\sf{29\ cm}$

So we have,

• $\sf{Opposite\ Side\ of\ \theta=21\ cm}$

• $\sf{Adjacent\ Side\ of\ \theta=20\ cm}$

• $\sf{Hypotenuse=29\ cm}$

Then,

$\longrightarrow\sf{\underline{\underline{\sin\theta=\pm\dfrac{21}{29}}}}$

$\longrightarrow\sf{\underline{\underline{\cos\theta=\pm\dfrac{20}{29}}}}$

$\longrightarrow\sf{\underline{\underline{\tan\theta=\dfrac{21}{20}}}}$

$\longrightarrow\underline{\underline{\mathrm{cosec\,\theta}=\pm\sf{\dfrac{29}{21}}}}$

$\longrightarrow\sf{\underline{\underline{\sec\theta=\pm\dfrac{29}{20}}}}$

Negative sign is only when $\theta$ belongs to third quadrant.

Q3. Given,

$\longrightarrow\sf{\cos A=\dfrac{12}{13}}$

A is assumed to be an acute angle.

Then,

$\longrightarrow\sf{\sin^2A=1-\cos^2A}$

$\longrightarrow\sf{\sin^2A=1-\left(\dfrac{12}{13}\right)^2}$

$\longrightarrow\sf{\sin^2A=1-\dfrac{144}{169}}$

$\longrightarrow\sf{\sin^2A=\dfrac{169-144}{169}}$

$\longrightarrow\sf{\sin^2A=\dfrac{25}{169}}$

$\longrightarrow\sf{\sin A=\dfrac{5}{13}}$

And,

$\longrightarrow\sf{\tan A=\dfrac{\sin A}{\cos A}}$

$\longrightarrow\sf{\tan A=\dfrac{\left(\dfrac{5}{13}\right)}{\left(\dfrac{12}{13}\right)}}$

$\longrightarrow\sf{\tan A=\dfrac{5}{12}}$

Hence,

$\longrightarrow\sf{\sin A(1-\tan A)=\dfrac{5}{13}\left(1-\dfrac{5}{12}\right)}$

$\longrightarrow\sf{\sin A(1-\tan A)=\dfrac{5}{13}\times\dfrac{12-5}{12}}$

$\longrightarrow\sf{\sin A(1-\tan A)=\dfrac{5}{13}\times\dfrac{7}{12}}$

$\longrightarrow\sf{\sin A(1-\tan A)=\dfrac{35}{156}}$

Thus Verified!