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Author: nsharp1
No. Questions: 9
Created: Feb 17, 2021
Last Modified: 4 years ago

Prove Law of Sines, Obtuse

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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

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

A.

What is the missing reason in step 1?

Given
Altitudes in non-right triangles do not intersect the other vertices of the triangle
An altitude from the vertex of an obtuse angle in a triangle always intersects the opposite side
The altitude of a triangle consists of infinitely many points

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

B.

What is the missing statement in step 5?

$sin(m ang B) = (AD)/(AB)$
$sin(m ang B) = (AB)/(AD)$
$sin(m ang B) = (BD)/(AB)$
$sin(m ang B) = (BD)/(AD)$

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

C.

What is the missing statement in step 7?

$sin(m ang C) = (AB)/(BC)$
$sin(m ang C) = (CD)/(AC)$
$sin(m ang C) = (AD)/(CD)$
$sin(m ang C) = (AD)/(AC)$

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

D.

What is the missing reason in step 12?

Segment Addition Postulate
A line segment consists of infinitely many points
A line segment can be extended indefinitely as a line
There are infinitely many lines in a plane

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

E.

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

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

F.

What is the missing statement in step 17?

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

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

G.

What is the missing statement in step 19?

$sin(m ang C) = (BE)/(AB)$
$sin(m ang C) = (BE)/(BC)$
$sin(m ang C) = (BE)/(CE)$
$sin(m ang C) = (AC)/(BC)$

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

H.

What is the missing reason in step 22?

Given
Sum of non-right angles in a triangle is 180°
Sum of the angles on a straight line is 180°
Exterior Angle Theorem

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

I.

For which category of triangles is this proof valid?

All triangles.
Acute triangles.
Non-right triangles.
Obtuse triangles.