Trigonometry Question

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The questions below concern the following proof.

Let

$Delta ABC$ be an obtuse triangle, such that

$m ang BAC > 90°$ . Let

$m ang A$ refer to

$m ang BAC$ and

$m ang B$ refer to

$m ang ABC$ . Prove the law of sines,

$(AB) / sin(m ang C) = (BC) / sin(m ang A) = (AC) / sin(m ang B)$ .

Obtuse Triangle ABC v2

$" Statement "$	$" Reason "$
$1. "Construct altitude" \ \bar{AD} \ "such that point" \ D \ "lies on" \ bar{BC}$	$1. ""$
$2. \bar{AD} _\|_ \bar{BC}$	$2. "Definition of an altitude"$
$3. ang ADC, \ ang ADB " are right angles"$	$3. "Definition of perpendicular lines"$
$4. Delta ADC, \ Delta ADB " are right triangles"$	$4. "Definition of right triangles"$
$5.$	$5. "S""ine ratio in a right triangle"$
$6. AB sin(m ang B) = AD$	$6. "Multiplication Property of Equality"$
$7.$	$7. "S""ine ratio in a right triangle"$
$8. AC sin(m ang C) = AD$	$8. "Multiplication Property of Equality"$
$9. AB sin(m ang B) = AC sin(m ang C)$	$9. "Transitive Property of Equality"$
$10. AB = (AC sin(m ang C))/sin(m ang B)$	$10. "Division Property of Equality"$
$11. (AB)/sin(m ang C) = (AC)/sin(m ang B)$	$11. "Division Property of Equality"$
$12. "Extend " bar{AC} " to " \stackrel{leftrightarrow}{AC}$	$12. ""$
$13. "Construct altitude" \ \bar{BE} \ "such that point" \ E \ "lies on" \ \stackrel{leftrightarrow}{AC}$ $\ \ \ (E \ "lies on" \ \stackrel{leftrightarrow}{AC} \ "such that" \ AE + AC = CE)$	$13. ""$
$14. \bar{BE} _\|_ \bar{CE}$	$14. "Definition of an altitude"$
$15. ang BEC " is a right angle"$	$15. "Definition of perpendicular lines"$
$16. Delta AEB, \ Delta BEC " are right triangles"$	$16. "Definition of right triangles"$
$17.$	$17. "S""ine ratio in a right triangle"$
$18. AB sin(m ang BAE) = BE$	$18. "Multiplication Property of Equality" \ \ \ \$
$19.$	$19. "S""ine ratio in a right triangle"$
$20. BC sin(m ang C) = BE$	$20. "Multiplication Property of Equality"$
$21. BC sin(m ang C) = AB sin(m ang BAE)$	$21. "Transitive Property of Equality"$
$22. m ang BAE + m ang A = 180°$	$22. ""$
$23. m ang BAE = 180° - m ang A$	$23. "Subtraction Property of Equality"$
$24. BC sin(m ang C) = AB sin(180° - m ang A)$	$24. "Substitution Property of Equality"$
$25. BC sin(m ang C) = AB sin(m ang A)$	$25. "Trig Identity"$
$26. (BC sin(m ang C))/sin(m ang A) = AB$	$26. "Division Property of Equality"$
$27. (BC)/sin(m ang A) = (AB)/sin(m ang C)$	$27. "Division Property of Equality"$
$28. (BC)/sin(m ang A) = (AC)/sin(m ang B)$	$28. "Transitive Property of Equality"$
$29. (AB) / sin(m ang C) = (BC) / sin(m ang A) = (AC) / sin(m ang B)$	$29. "Combined results"$

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

What is the missing reason in step 13?

The altitude of an acute angle in an obtuse triangle intersects the extension of the opposite side of the triangle
The altitude of a triangle consists of infinitely many points
Every triangle has three altitudes
The orthocenter of an obtuse triangle lies outside the triangle

Question Info

Trigonometry Question

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