Want to see correct answers?
Login or join for free!

Question Group Info

This question group is public and is used in 1 test.

Author: nsharp1
No. Questions: 9
Created: Feb 15, 2021
Last Modified: 4 years ago

Law of Cosines Proof, Obtuse

View group questions.

To print this group, add it to a test.

The questions below relate to the following proof.

For

$Delta ABC$ , where

$m ang BAC > 90°$ , let the intersection of the altitude from vertex

$B$ and the extension of

$bar{AC}$ be point

$D$ (not labeled). Let

$theta = m ang BAC$ and

$alpha = m ang BAD$ . Prove that

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

Obtuse Triangle Height v2

Obtuse Triangle Height v2

$" Statement "$	$" Reason "$
$1. bar{BD} " is an altitude"$	$1. "Given"$
$2. bar{BD} _\|_ bar{DC}$	$2. "Definition of an altitude"$
$3. ang BDC " is a right angle"$	$3. "Definition of perpendicular lines"$
$4. Delta BDA, \ Delta BDC " are right triangles"$	$4. "Definition of right triangles"$
$5.$	$5. "C""osine ratio in a right triangle"$
$6. AB cos(alpha) = AD$	$6. "Multiplication Property of Equality"$
$7.$	$7. "S""ine ratio in a right triangle"$
$8. AB sin(alpha) = BD$	$8. "Multiplication Property of Equality"$
$9. BC^2 = BD^2 + CD^2$	$9. ""$
$10.$	$10. "Segment Addition Postulate"$
$11. BC^2 = BD^2 + (AD + AC)^2$	$11. "Substitution Property of Equality"$
$12. BC^2 = (AB sin(alpha))^2 + (ABcos(alpha) + AC)^2$	$12. ""$
$13. BC^2 = AB^2 sin^2(alpha) + (ABcos(alpha) + AC)^2$	$13. "Distributive Property of Exponents"$
$14. BC^2 = AB^2 sin^2(alpha) + AB^2cos^2(alpha) +$ $\ \ \ 2 AB \ AC cos(alpha) + AC^2$	$14. "Distributive Property"$
$15. BC^2 = AB^2 (sin^2(alpha) + cos^2(alpha)) +$ $\ \ \ 2 AB \ AC cos(alpha) + AC^2$	$15. "Distributive Property"$
$16. BC^2 = AB^2 + 2 AB \ AC cos(alpha) + AC^2$	$16. ""$
$17. BC^2 = AB^2 + AC^2 + 2 AB \ AC cos(alpha)$	$17. "Commutative Property of Addition"$
$18. alpha + theta = 180°$	$18. ""$
$19. alpha = 180° - theta$	$19. "Subtraction Property of Equality"$
$20. BC^2 = AB^2 + AC^2 + 2 AB \ AC cos(180° - theta)$	$20. "Substitution Property of Equality"$
$21. BC^2 = AB^2 + AC^2 - 2 AB \ AC cos(theta)$	$21. "Trigonometric Identity"$

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

A.

What is the missing statement in step 5?

$cos(alpha) = (BD)/(AB)$
$cos(alpha) = (AC)/(AB)$
$cos(alpha) = (AD)/(AB)$
$cos(alpha) = (BD)/(AD)$

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

B.

What is the missing statement in step 7?

$sin(alpha) = (BD)/(AB)$
$sin(alpha) = (BD)/(AD)$
$sin(alpha) = (AD)/(AB)$
$sin(alpha) = (AC)/(BC)$

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

C.

What is the missing reason in step 9?

Sum of Squares Formula
Pythagorean Identity
Pythagorean Theorem
Substitution Property of Equality

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

D.

What is the missing statement in step 10?

$CD = AB + AC$
$CD = AD + AC$
$AC = DB + AB$
$BC = AD + AC$

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

E.

What is the missing reason in step 12?

Pythagorean Theorem
Trig ratios in a right triangle
Substitution Property of Equality
Pythagorean Identity

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

F.

What is the missing reason in step 16?

Pythagorean Identity
Pythagorean Theorem
Algebra (subtraction)
Subtraction Property of Equality

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

G.

What is the missing reason in step 18?

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

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

H.

Is this proof also valid for the law of cosines formulas that include the other two angles of the triangle, namely

$AB^2 = BC^2 + AC^2 - 2 BC \ AC cos(m ang BCA)$ and

$AC^2 = AB^2 + BC^2 - 2 AB \ BC cos(m ang ABC) ?$

Yes. If altitudes were constructed from vertices $A$ or $C$ , the proofs for these two formulas would be identical to the one given.
Yes. Since the formula including the angle with the largest measure has been proven, it must necessarily hold for the formulas including the other angles in the triangle.
No. The most straightforward proof for these two formulas would be similar to the proof given for an acute triangle (and they would include constructing an altitude from vertex $A$ only).
No. The proof for either of these formulas would require additional trigonometric identities because of the obtuse angle.

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

I.

Given that a proof was also given for the law of cosines formula for an acute angle in a triangle, which type(s) of triangles are now covered by these proofs? Choose all that apply.

Acute
Obtuse
Scalene
Right