# Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 244: 78

Since $BDA = 15^{\circ}$, then $BAD = 15^{\circ}$ Since the sum of the three angles of a triangle add to $180^{\circ}$, then angle $ABD = 150^{\circ}$

#### Work Step by Step

$AB$ is a radius of the circle and $BD$ is a radius of the circle. Therefore, $AB = BD$. In the triangle $ABD$, since the two sides $AB$ and $BD$ are equal, the triangle $ABD$ is an isosceles triangle. Since triangle $ABD$ is an isosceles triangle and $AB = BD$, then angle $BAD$ is equal to angle $BDA$. Since $BDA = 15^{\circ}$, then $BAD = 15^{\circ}$ Since the sum of the three angles of a triangle add to $180^{\circ}$, then angle $ABD = 150^{\circ}$

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