Question

In an equilateral triangle ABC , match the following ratios to the values

$\begin{array}{cc}\text{Trigonometric Ratios}& \text{Values}\\ \left(i\right)tan{60}^o& \left(a\right)\frac{1}{\sqrt{3}}\\ \left(ii\right)cot{30}^o& \left(b\right)\sqrt{3}\\ \left(iii\right)cosec{30}^o& \left(c\right)2\\ \left(iv\right)sec{30}^o& \left(d\right)\frac{2}{\sqrt{3}}\end{array}$

• A. (i) -a , (ii) - a , (iii) - c , (iv) - d
• B. (i) -d , (ii) - a , (iii) - b , (iv) - d
• C. (i) -a , (ii) - b , (iii) - d , (iv) - c
• D. (i) -b , (ii) - b , (iii) - c , (iv) - d

Answer 1

The correct option is D

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

Let each side of the equilateral triangle be a.
Length of perpendicular $AD=\frac{\sqrt{3}a}{2}$
$tan{60}^o=tanB=\frac{AD}{BD}=\sqrt{3}$
$cot{30}^o=cot\left(\frac{A}{2}\right)=\frac{AD}{BD}=\sqrt{3}$
$cosec{30}^o=cosec\left(\frac{A}{2}\right)=\frac{AB}{BD}=2$
$sec{30}^o=sec\left(\frac{A}{2}\right)=\frac{AB}{AD}=\frac{2}{\sqrt{3}}$