Solve Triangles Using The Law Of Sines Khan Academy Answers

The moment you search for solve triangles using the law of sines Khan Academy answers, the goal is not merely to copy a final result. You want to understand how side lengths, opposite angles, inverse sine, and the possible ambiguous case work together so that you can solve similar problems independently.

Introduction to the Law of Sines

The law of sines is a relationship among the sides and angles of any non-right triangle. If a triangle has sides (a), (b), and (c) opposite angles (A), (B), and (C), respectively, then:

[ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} ]

This formula allows you to find a missing side or angle when the triangle information is arranged in matching side-angle pairs. Take this: if side (a) and angle (A) are known, they form one pair. Another known pair can then be used to solve for an unknown Which is the point..

The law of sines is especially useful when you are given:

• ASA, meaning two angles and the included side

• AAS, meaning two angles and a non-included side

• SSA, meaning two sides and a non-included angle (this arrangement triggers the ambiguous case, discussed below)

Because the three interior angles of any triangle sum to (180^\circ), ASA and AAS problems almost always begin by subtracting the two known angles from (180^\circ) to find the third angle. Once you have a complete angle-side pair, you can set up a proportion to find the remaining sides That's the part that actually makes a difference. And it works..

Step-by-Step: Solving for a Missing Side (AAS Example)

Suppose you are given (A = 40^\circ), (B = 70^\circ), and side (a = 12). Find side (b) Simple, but easy to overlook..

1. Find the third angle.
   (C = 180^\circ - 40^\circ - 70^\circ = 70^\circ).

2. Identify a complete pair.
   You know side (a = 12) and its opposite angle (A = 40^\circ).

3. Set up the Law of Sines proportion for the unknown.
   [ \frac{a}{\sin A} = \frac{b}{\sin B} \implies \frac{12}{\sin 40^\circ} = \frac{b}{\sin 70^\circ} ]

4. Solve algebraically.
   [ b = \frac{12 \cdot \sin 70^\circ}{\sin 40^\circ} ]

5. Calculate.
   Using a calculator in degree mode:
   (b \approx \frac{12 \cdot 0.9397}{0.6428} \approx 17.5).

Step-by-Step: Solving for a Missing Angle (SSA Example)

Suppose you are given (a = 10), (b = 15), and (A = 30^\circ). Find angle (B).

1. Set up the proportion using the known pair (a) and (A).
   [ \frac{\sin A}{a} = \frac{\sin B}{b} \implies \frac{\sin 30^\circ}{10} = \frac{\sin B}{15} ]

2. Isolate (\sin B).
   [ \sin B = \frac{15 \cdot \sin 30^\circ}{10} = \frac{15 \cdot 0.5}{10} = 0.75 ]

3. Apply the inverse sine function.
   (B = \sin^{-1}(0.75) \approx 48.6^\circ) Easy to understand, harder to ignore..

   Pause here. On Khan Academy, this is often where students stop, but the SSA configuration requires a critical second check.

The Ambiguous Case (SSA Deep Dive)

When given two sides and a non-included angle (SSA), the sine ratio (\sin B = k) can produce zero, one, or two valid triangles. Here's the thing — this happens because (\sin \theta = \sin(180^\circ - \theta)). If (B_1 = \sin^{-1}(k)) is a solution, then (B_2 = 180^\circ - B_1) is mathematically a second solution, provided it creates a valid triangle.

The Decision Protocol:

1. Calculate the height (h = b \sin A) (using the side adjacent to the known angle).
2. Compare side (a) (the side opposite the known angle) to (h) and (b):
   • (a < h): No triangle exists (side (a) is too short to reach the base). 0 solutions.
   • (a = h): Exactly one right triangle exists. 1 solution.
   • (h < a < b): Side (a) can swing to intersect the base in two places. 2 solutions (one acute (B), one obtuse (B)).
   • (a \geq b): Side (a) is long enough to reach the base only once (the obtuse supplement would make the angle sum exceed (180^\circ)). 1 solution.

Applying it to the example above ((a=10, b=15, A=30^\circ)):

• (h = 15 \sin 30^\circ = 7.5).
• Since (7.5 < 10 < 15) ((h < a < b)), two triangles exist.
• Triangle 1: (B_1 \approx 48.6^\circ), (C_1 \approx 101.4^\circ).
• Triangle 2: (B_2 = 180^\circ - 48.6^\circ = 131

° so

[ C_2 = 180^\circ - 30^\circ - 131.Worth adding: 4^\circ = 18. 6^\circ.

Both triangles are valid because their angle sums are (180^\circ). So the given information (a=10), (b=15), and (A=30^\circ) produces two possible triangles:

[ B_1 \approx 48.6^\circ,\quad C_1 \approx 1