Question

Solve the question in the above attachment ✔⬆​

Answer 1

we have, Tan A=5/12=p/b

so, p=5 and b=12

so by Pythagoras theorem,

h²=p²+b²=>h=√(5²+12²)=>h=√(25+144)=>h=√(169)=>h=13

so now we know other trigonometric identities will be,

Sin A=p/h=5/13 ; Cos A=b/h=12/13 ; Cosec A=h/p=13/5

Sec A=h/b=13/12 ; Cot A=b/p=12/5

The above identities are the amswers

Answer 2

$\huge\star{\underline{\mathtt{\red{A}\pink{n}\green{s}\blue{w}\purple{e}\orange{r}}}}$

$\large\sf\purple{All\:other\:trignometric\:ratios\:are:-}$

$\small\sf{sin\:A = \frac{13}{12}}$

$\small\sf{cos\:A = \frac{12}{13}}$

$\small\sf{cosec \: x = \frac{13}{5}}$

$\small\sf{sec \: x = \frac{13}{12}}$

$\small\sf{cot \: x = \frac{12}{5}}$

$\large\sf\orange{It\:is\:given\:that}$

$\small\sf{tan \: A = \frac{5}{12}}$ ........(1)

$\small\sf{tan \: A = \frac{perpendicular}{base}}$ ......(2)

$\large\sf\pink{from\:(1)\:and\:(2)\:we\:get}$

$\small\sf{perpendicular=5x}$

$\small\sf{base=12x}$

$\small\sf{hypotenuse = \sqrt{ {perpendicular}^{2} + {base}^{2} }}$

$\small\sf{hypotenuse = \sqrt{ {(5x)}^{2} + {(12x)}^{2} }}$

$\small\sf{hypotenuse = \sqrt{ {169x}^{2} = 13x}}$

$\small\sf{sin \: A = \frac{perpendicular}{hypotenuse} = \frac{5x}{13x} = \frac{12}{13}}$

$\small\sf{cos \: A = \frac{base}{hypotenuse} = \frac{12x}{13x} = \frac{12}{13}}$

$\small\sf{cosec \: x = \frac{1}{sin \: a} = \frac{13}{5}}$

$\small\sf{sec \: x = \frac{1}{cos \: a} = \frac{13}{12}}$

$\small\sf{cot \: x = \frac{1}{tan \: a} = \frac{12}{5}}$

$\large\sf\green{Therefore,}$

$\large\sf\blue{All\:other\:trignometric\:ratios\:are:-}$

$\small\sf{sin\:A = \frac{5}{13}}$

$\small\sf{cos\:A = \frac{12}{13}}$

$\small\sf{cosec \: x = \frac{13}{5}}$

$\small\sf{sec \: x = \frac{13}{12}}$

$\small\sf{cot \: x = \frac{12}{5}}$