Question

From the figure given below, calculate all the six t-ratios for both acute angles.

Accepted Answer

From the figure,

Perpendicular, $B C = 2 cm$

Base, $A C = 3 cm$

By Pythagoras theorem,
${HYPOTENUS}^{2} = {BASE}^{2} + {PERPENDICULAR}^{2}$
$A C^{2} = B C^{2} + A B^{2} \Rightarrow A B^{2} = 3^{2} - 2^{2} = 9 - 4 = 5 \Rightarrow A B = \sqrt{5}$

Then,

$S in A = \frac{PERPENDICULAR}{HYPOTENUS E} = \frac{B C}{A C} = \frac{2}{3} C osec A = \frac{HYPOTENUS E}{PERPENDICULAR} = \frac{A C}{B C} = \frac{3}{2} C os A = \frac{A B}{A C} = \frac{BASE}{HYPOTENUS E} = \frac{\sqrt{5}}{3} S ec A = \frac{HYPOTENUS E}{BASE} = \frac{A C}{A B} = \frac{3}{\sqrt{5}} TanA = \frac{PERPENDICULAR}{BASE} = \frac{B C}{A B} = \frac{2}{\sqrt{5}} CotA = \frac{BASE}{PERPENDICULAR} = \frac{A B}{B C} = \frac{\sqrt{5}}{2}$

And,

$S in C = \frac{PERPENDICULAR}{HYPOTENUS E} = \frac{A B}{A C} = \frac{\sqrt{5}}{3} C osec C = \frac{HYPOTENUS E}{PERPENDICULAR} = \frac{A C}{A B} = \frac{3}{\sqrt{5}} C os C = \frac{BASE}{HYPOTENUS E} = \frac{B C}{A C} = \frac{2}{3}, S ec C = \frac{HYPOTENUS E}{BASE} = \frac{A C}{B C} = \frac{3}{2} TanC = \frac{PERPENDICULAR}{BASE} = \frac{A B}{B C} = \frac{\sqrt{5}}{2} CotC = \frac{BASE}{PERPENDICULAR} = \frac{B C}{A B} = \frac{2}{\sqrt{5}}$

M L Aggarwal Solutions for Exercise 3: Chapter Test

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Exercise 3: Chapter Test

Questions from M L Aggarwal Solutions for Exercise 3: Chapter Test with Hints & Solutions

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