Trigonometry Question

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The questions below relate to the following proof.

For

$Delta ABC$ , which is acute, let the intersection of the altitude from vertex

$B$ and

$bar{AC}$ be point

$D$ (not labeled). Also, let

$m ang C = theta$ . Prove that

$AB^2 = AC^2 + BC^2 - 2 AC \ BC cos(theta)$ .

Acute Triangle Height v2

$" Statement "$	$" Reason "$
$1. bar{BD} " is an altitude"$	$1. "Given"$
$2. bar{BD} _\|_ bar{AC}$	$2. "Definition of an altitude"$
$3. ang ADB, \ ang BDC " are right angles"$	$3. ""$
$4. Delta ADB, \ Delta BDC " are right triangles"$	$4. "Definition of right triangles"$
$5.$	$5. "S""ine ratio in a right triangle"$
$6. BC sin(theta) = BD$	$6. "Multiplication Property of Equality"$
$7. "$	$7. "C""osine ratio in a right triangle"$
$8. BC cos(theta) = CD$	$8. "Multiplication Property of Equality"$
$9.$	$9. "Segment Addition Postulate"$
$10. AC = AD + BC cos(theta)$	$10. ""$
$11. AC - BC cos(theta) = AD$	$11. "Subtraction Property of Equality"$
$12. Delta ADB " is a right triangle"$	$12. "Earlier result"$
$13. AB^2 = AD^2 + BD^2$	$13. ""$
$14. AB^2 = (AC - BC cos(theta))^2 + (BC sin(theta))^2$	$14. "Substitution Property of Equality"$
$15. AB^2 = (AC - BC cos(theta))^2 + BC^2 sin^2(theta)$	$15. "Distributive Property of Exponents"$
$16. AB^2 = AC^2 - 2 AC \ BC cos(theta) +$ $\ \ \ BC^2 cos^2(theta) + BC^2 sin^2(theta)$	$16. ""$
$17. AB^2 = AC^2 - 2 AC \ BC cos(theta) +$ $\ \ \ BC^2 (cos^2(theta) + sin^2(theta))$	$17. "Distributive Property"$
$18. AB^2 = AC^2 - 2 AC \ BC cos(theta) + BC^2$	$18. ""$
$19. AB^2 = AC^2 + BC^2 - 2 AC \ BC cos(theta)$	$19. "Commutative Property of Addition" \ \ \ \ \$

Grade 11 Trigonometry CCSS: HSG-SRT.D.10

What is the missing reason in step 16?

Algebra (addition)
Quadratic Formula
Substitution Property of Equality
Distributive Property

Question Info

Trigonometry Question

Grade 11 Trigonometry CCSS: HSG-SRT.D.10