In an equilateral triangle ABC- match the following ratios to the values beginarray - 1- - hline Trigonometric ratios-Valuesillhlinei tan- 60- circ -a - frac 1 sqrt 3 whhlineii - 30- circ - b sqrt 3 hlineiii cosec- 30- lcirc - c 2 Whline iv - sec 30- circ - d - frac 2 lsqrt 3 lendarray A- i -a - ii - b - iii-c - iv - dB- i -b- ii - b- iii - c- iv - dC- i -a - ii - b - iii - d- iv - cD- i -b- ii - a- iii-d - iv - c

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Standard XII

Mathematics

Trigonometric Ratios of Common Angles

In an equilat...

Question

In an equilateral triangle ABC , match the following ratios to the values
$\begin{matrix} T r i g o n o m e t r i c r a t i o s & V a l u e s (i) t a n 60^{\circ} & (a) \frac{1}{\sqrt{3}} (i i) c o t 30^{\circ} & (b) \sqrt{3} (i i i) c o s e c 30^{\circ} & (c) 2 (i v) s e c 30^{\circ} & (d) \frac{2}{\sqrt{3}} \end{matrix}$

(i) -b , (ii) - b , (iii) - c , (iv) - d

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(i) -a , (ii) - b , (iii) - c , (iv) - d

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(i) -a , (ii) - b , (iii) - d , (iv) - c

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(i) -b , (ii) - a , (iii) - d , (iv) - c

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Solution

The correct option is A (i) -b , (ii) - b , (iii) - c , (iv) - d

Let each side of the equilateral triangle be a.

Length of perpendicular AD = $\frac{\sqrt{3} a}{2}$

$t a n 60 = t a n B = \frac{A D}{B D} = \sqrt{3}$

$c o t 30 = c o t (\frac{A}{2}) = \frac{A D}{B D} = \sqrt{3}$

$c o s e c 30 = c o s e c (\frac{A}{2}) = \frac{A D}{B D} = 2$
$s e c 30 = s e c (\frac{A}{2}) = \frac{A D}{B D} = \frac{2}{\sqrt{3}}$

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In an equilateral triangle ABC , match the following ratios to the values Trigonometric ratiosValues(i) tan 60∘(a) 1√3(ii) cot 30∘(b)√3(iii) cosec 30∘(c) 2(iv) sec 30∘(d) 2√3

The correct option is A (i) -b , (ii) - b , (iii) - c , (iv) - d Let each side of the equilateral triangle be a. Length of perpendicular AD =√3a2 tan 60=tan B=ADBD=√3 cot 30=cot(A2)=ADBD=√3 cosec 30=cosec(A2)=ADBD=2 sec 30=sec(A2)=ADBD=2√3