Question

In an equilateral triangle $ABC$ , match the following Trigonometric ratios to the values
$\begin{array}{cc}Trigonometricratios& Values\\ \left(i\right)tan{60}^{\circ }& \left(p\right)\frac{1}{\sqrt{3}}\\ \left(ii\right)cot{30}^{\circ }& \left(q\right)\frac{2}{\sqrt{3}}\\ \left(iii\right)cosec{30}^{\circ }& \left(r\right)2\\ \left(iv\right)sec{30}^{\circ }& \left(s\right)\sqrt{3}\end{array}$

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

Answer 1

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

Draw a perpendicular from $A$ on $BC$ at $D$ as shown in above figure.
Let each side of the equilateral triangle be $a$.

Using Pythagoras theorem in $△ABD$, we get
Length of perpendicular $AD$ =$\frac{\sqrt{3}a}{2}$

$\tan {60}^{\circ }=tanB=\frac{AD}{BD}=\sqrt{3}$

$\cot {30}^{\circ }=cot\left(\frac{A}{2}\right)=\frac{AD}{BD}=\sqrt{3}$

cosec${30}^{\circ }=cosec\left(\frac{A}{2}\right)=\frac{AB}{BD}=2$

$\sec {30}^{\circ }=sec\left(\frac{A}{2}\right)=\frac{AB}{AD}=\frac{2}{\sqrt{3}}$