Trigonometry Question

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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

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

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

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.

Question Info

Trigonometry Question

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