Use the Law of Sines to solve the triangle? 6.) A=60 degrees, a=9, c=10.

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Use the Law of Sines to solve the triangle? 6.) A=60 degrees, a=9, c=10.

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Answer 1 · 2017-01-01T19:51:14

Check for the Ambiguous Case and, if appropriate, use the Law of Sines to solve the triangle(s).

Explanation:

Here is a reference for The Ambiguous Case

$∠ A$ $angle A$ is acute. Compute value of h:

$h = (c) sin (A)$ $h = (c)sin(A)$

$h = (10) sin (60^{\circ})$ $h = (10)sin(60^@)$

$h \approx 8.66$ $h ~~ 8.66$

$h < a < c$ $h < a < c$ , therefore, two possible triangles exist, one triangle has $∠ C_{acute}$ $angle C_("acute")$ and the other triangle has $∠ C_{obtuse}$ $angle C_("obtuse")$

Use The Law of Sines to compute $∠ C_{acute}$ $angle C_("acute")$

$\frac{sin (C_{acute})}{c} = \frac{sin (A)}{a}$ $sin(C_("acute"))/c = sin(A)/a$

$sin (C_{acute}) = sin (A) \frac{c}{a}$ $sin(C_("acute")) = sin(A)c/a$

$C_{acute} = {sin}^{- 1} (sin (A) \frac{c}{a})$ $C_("acute") = sin^-1(sin(A)c/a)$

$C_{acute} = {sin}^{- 1} (sin (60^{\circ}) \frac{10}{9})$ $C_("acute") = sin^-1(sin(60^@)10/9)$

$C_{acute} \approx {74.2}^{\circ}$ $C_("acute") ~~ 74.2^@$

Find the measure for angle B by subtracting the other angles from $180^{\circ}$ $180^@$ :

$∠ B = 180^{\circ} - 60^{\circ} - {74.2}^{\circ}$ $angle B = 180^@ - 60^@ - 74.2^@$

$∠ B = {45.8}^{\circ}$ $angle B = 45.8^@$

Use the Law of Sines to compute the length of side b:

side $b = a \frac{sin (B)}{sin (A)}$ $b = asin(B)/sin(A)$

$b = 9 \frac{sin ({45.8}^{\circ})}{sin (60^{\circ})}$ $b = 9sin(45.8^@)/sin(60^@)$

$b \approx 7.45$ $b ~~ 7.45$

For the first triangle:

$a = 9, b \approx 7.45, c = 10, A = 60^{\circ}, B \approx {45.8}^{\circ}, and C \approx {74.2}^{\circ}$ $a = 9,b ~~ 7.45, c = 10, A = 60^@, B ~~ 45.8^@, and C ~~ 74.2^@$

Onward to the second triangle:

$∠ C_{obtuse} \approx 180^{\circ} - C_{acute}$ $angle C_("obtuse") ~~ 180^@ - C_("acute")$

$C_{obtuse} \approx 180^{\circ} - {74.2}^{\circ} \approx {105.8}^{\circ}$ $C_("obtuse") ~~ 180^@ - 74.2^@ ~~ 105.8^@$

Find the measure for angle B by subtracting the other angles from $180^{\circ}$ $180^@$ :

$∠ B = 180^{\circ} - 60^{\circ} - {105.8}^{\circ} \approx {14.2}^{\circ}$ $angle B = 180^@ - 60^@ - 105.8^@ ~~ 14.2^@$

Use the Law of Sines to compute the length of side b:

$b = 9 \frac{sin ({14.2}^{\circ})}{sin (60^{\circ})}$ $b = 9sin(14.2^@)/sin(60^@)$

$b \approx 2.55$ $b ~~ 2.55$

For the second triangle:

$a = 9, b \approx 2.55, c = 10, A = 60^{\circ}, B \approx {14.2}^{\circ}, and C \approx {105.8}^{\circ}$ $a = 9,b ~~ 2.55, c = 10, A = 60^@, B ~~ 14.2^@, and C ~~ 105.8^@$

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Use the Law of Sines to solve the triangle? 6.) A=60 degrees, a=9, c=10.

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