## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 244: 79

#### Answer

The sum of the three angles of a triangle add to $180^{\circ}$, and angle $ABD = 150^{\circ}$. Therefore the sum of angle $BAD$ and angle $BDA$ add to $30^{\circ}$ Since triangle $ABD$ is an isosceles triangle and $AB = BD$, then angle $BAD$ is equal to angle $BDA$. Therefore, angle $BDA = 15^{\circ}$, and angle $BAD = 15^{\circ}$

#### 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. The sum of the three angles of a triangle add to $180^{\circ}$, and angle $ABD = 150^{\circ}$. Therefore the sum of angle $BAD$ and angle $BDA$ add to $30^{\circ}$ Since triangle $ABD$ is an isosceles triangle and $AB = BD$, then angle $BAD$ is equal to angle $BDA$. Therefore, angle $BDA = 15^{\circ}$, and angle $BAD = 15^{\circ}$

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