Question

1. b, d and e are points on the circumference of a circle. ac is a tangent to the circle at point b. angle dbe = 2x° angle cbe = x + 15° work out the size of angle abd.

Explanation:

Step1: Use the tangent - chord angle theorem

The angle between a tangent and a chord is equal to the inscribed angle subtended by the same chord. So, $\angle ABD=\angle BED$. In $\triangle BDE$, since $BD = BE$, $\angle BDE=\angle BED$. And the sum of angles in $\triangle BDE$ is $180^{\circ}$, so $\angle BDE+\angle BED+\angle DBE = 180^{\circ}$, that is $2\angle BED + 2x^{\circ}=180^{\circ}$, then $\angle BED = 90^{\circ}-x^{\circ}$. Also, $\angle ABD+\angle DBE+\angle CBE = 180^{\circ}$ (a straight - line angle), and $\angle ABD=\angle BED = 90^{\circ}-x^{\circ}$. So, $(90 - x)+2x+(x + 15)=180$.

Step2: Solve the equation for $x$

Simplify the left - hand side of the equation: $90 - x+2x+x + 15=90+15+( - x+2x+x)=105 + 2x$. Then the equation $105 + 2x=180$, subtract 105 from both sides: $2x=180 - 105=75$, so $x = 37.5$.

Step3: Calculate $\angle ABD$

Since $\angle ABD = 90^{\circ}-x^{\circ}$, substitute $x = 37.5$ into it. $\angle ABD=90 - 37.5=52.5^{\circ}$.

Answer:

$52.5^{\circ}$