Chapter 9.7, Problem 12PSB

a

To determine

To find: The co-ordinate of “ B ”.

Answer to Problem 12PSB

The co-ordinate of “ B ” is $(1,\sqrt{3})$ .

Explanation of Solution

Given information :

  $30°-60°-90°$ triangle

Coordinates of O as (0, 0) and A as (1, 0)

Formula used :

In $30°-60°-90°$ triangle, sides are in ratio $1:\sqrt{3}\text{ :2}$

Calculation :

In triangle, OAB

  $\begin{array}{l}\tan {60}^o=\frac{AB}{OA}\\ \sqrt{3}=\frac{AB}{1}\\ AB=\sqrt{3}\end{array}$

Therefore, the y co-ordinate of B is $\sqrt{3}$

Hence, the coordinate of B is $(1,\sqrt{3})$

b

To determine

To find: The slope of line “ OB ”.

Answer to Problem 12PSB

The slope of a given line is $\sqrt{3}$

Explanation of Solution

Given information :

  $30°-60°-90°$ triangle

Coordinates of O as (0, 0) and A as (1, 0)

Formula used :

In $30°-60°-90°$ triangle, sides are in ratio $1:\sqrt{3}\text{ :2}$

Calculation :

In triangle, OAB

It is $30°-60°-90°$ triangle and OA = 1 unit

  $\therefore$ AB = $\sqrt{3}$ and

Required ratio = $\frac{AB}{OA}=\frac{\sqrt{3}}{1}=\sqrt{3}$