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 statement in step 17?

$sin(m ang BAE) = (AE)/(AB)$
$sin(m ang BAE) = (BE)/(AB)$
$sin(m ang BAE) = (BE)/(BC)$
$sin(m ang BAE) = (BE)/(AE)$

Question Info

Trigonometry Question

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