Section 4.2 – Law of Sines & Surveying Applications

Administrative Announcements

  • Answers for Test #2 are posted on D2L.
  • Section covered today: 4.2 – Law of Sines (start on p. 272 of the black book).
  • Homework 4.2 due next Thursday.
  • Absolute last day to submit any homework (Chs 1‒5) for credit: May 4 ("May the 4th be with you").
    • Work submitted after May 4 (May 5,6,7…) will be deleted and receive no credit.
    • Ch. 5 homework (5.1–5.3) can be turned in early—even before the lectures—if you wish.
  • Always keep the calculator in degree mode when working with degree‐measure angles.

Right Triangles vs. Oblique Triangles

  • Right triangle (one $90^{\circ}$ angle)
    • Pythagorean theorem: a2+b2=c2a^2+b^2=c^2
  • Oblique triangle (no $90^{\circ}$ angle)
    • Pythagorean theorem does not apply.
    • Primary tools: Law of Sines (today) & Law of Cosines (next).

The Law of Sines (Section 4.2)

sin⁡Aa=sin⁡Bb=sin⁡Cc\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}

  • Capital letters ($A,B,C$) = angles.
  • Lower-case letters ($a,b,c$) = sides opposite the corresponding angles.
  • Common procedural pattern
    1. Identify a side–angle pair that is completely known.
    2. Set up a proportion with the unknown side or angle.
    3. Cross-multiply and solve.
    4. When solving for an angle, take the inverse trigonometric function (e.g., $A=\sin^{-1}(\;\cdot\;)$).
    5. Use A+B+C=180∘A+B+C=180^{\circ} to obtain the third angle if needed.
    6. Rounding conventions (used in class):
    • Sine values: 4 decimal places.
    • Final angles: 2 decimal places.
    • Final sides: 2 (or 3) significant decimals, depending on context.
Calculator Tips
  • Degree mode ⇔ radian mode mistakes are a common source of error.
  • If an inverse sine produces a value outside the expected range ($0^{\circ}$–$180^{\circ}$), re-check mode & inputs.

Example 1 (p. 275) – Small Oblique Triangle

Known:  side b = 3, side c = 2, included angle B = 40°
Find:  angle C, angle A, side a
  1. Set up to solve for $C$ first (only one unknown):
    sin⁡C2=sin⁡40∘3\frac{\sin C}{2}=\frac{\sin 40^{\circ}}{3}
    sin⁡C=2sin⁡40∘3=0.4285\sin C=\frac{2\sin 40^{\circ}}{3}=0.4285
    C=sin⁡−1(0.4285)=25.37∘C=\sin^{-1}(0.4285)=25.37^{\circ}
  2. Use the angle-sum rule:
    A=180∘−40∘−25.37∘=114.63∘A=180^{\circ}-40^{\circ}-25.37^{\circ}=114.63^{\circ}
  3. Solve for side $a$:
    \frac{a}{\sin 114.63^{\circ}}=\frac{3}{\sin 40^{\circ}}\quad\Rightarrow\quad
    a=\frac{3\sin 114.63^{\circ}}{\sin 40^{\circ}}=4.24
    Final results:
    $A=114.63^{\circ}$, $B=40^{\circ}$, $C=25.37^{\circ}$, $a=4.24$, $b=3$, $c=2$.

Key idea: choose the proportion containing only one unknown to avoid a dead-end.

Example 2 (p. 275, ex.#5) – Composite & Isosceles Structure

Sketch (values inside small right-side triangle):

Angle at base = 35°
Two equal sides = 6,6 (creates an isosceles triangle on the right)
External side = 8 (across the big triangle)
  1. Find angle E using the big triangle:
    sin⁡E8=sin⁡35∘6\frac{\sin E}{8}=\frac{\sin 35^{\circ}}{6}
    E=sin⁡−1(8sin⁡35∘6)=49.89∘E=\sin^{-1}\left(\frac{8\sin35^{\circ}}{6}\right)=49.89^{\circ}
  2. Isosceles consequence: if two sides are 6 & 6, the opposite base angles are equal:
    Right-small triangle angles=49.9∘,49.9∘\text{Right-small triangle angles}=49.9^{\circ},49.9^{\circ}
  3. Remaining large-triangle angle:
    C=180∘−49.9∘−49.9∘=80.2∘C=180^{\circ}-49.9^{\circ}-49.9^{\circ}=80.2^{\circ}
  4. External straight-line supplement gives:
    A=180∘−49.9∘=130.1∘A=180^{\circ}-49.9^{\circ}=130.1^{\circ}
  5. Final angle in the left small triangle:
    B=180∘−130.1∘−35∘=14.9∘B=180^{\circ}-130.1^{\circ}-35^{\circ}=14.9^{\circ}
  6. Find side $c$ (across $C=80.2^{\circ}$) inside small right-side triangle:
    \frac{c}{\sin 80.2^{\circ}}=\frac{6}{\sin 49.9^{\circ}}
    \;\Rightarrow\; c=7.70
  7. Find side $b$ (across $B=14.9^{\circ}$) inside left small triangle:
    bsin⁡14.9∘=6sin⁡35∘  ⇒  b=2.70\frac{b}{\sin 14.9^{\circ}}=\frac{6}{\sin 35^{\circ}}\;\Rightarrow\; b=2.70

Take-away concepts

  • Distinguish which triangle you are working in (students & instructor corrected a mis-selection).
  • Recognize isosceles properties automatically to save computations.

Example 3 – Height of a Mountain (900 m Baseline)

Problem statement (paraphrased):

  • Two observation points, 900 m apart along a straight line toward the mountain.
  • Angle of elevation at the nearest point: 47∘47^{\circ}.
  • Angle of elevation at the farther point: 35∘35^{\circ}.
  • Find the mountain height $h$.
  1. Geometry of the ground triangle (observer 1 – observer 2 – peak):
    • Exterior angle at the peak:
      133∘=180∘−47∘133^{\circ}=180^{\circ}-47^{\circ}
    • Remaining interior angle at ground-far point:
      12∘=180∘−133∘−35∘12^{\circ}=180^{\circ}-133^{\circ}-35^{\circ}
  2. Law of Sines on ground triangle to get slanted distance $d$ (far point to peak):
    sin⁡12∘900=sin⁡133∘d  ⇒  d=3165.86 m\frac{\sin 12^{\circ}}{900}=\frac{\sin 133^{\circ}}{d}\;\Rightarrow\; d=3165.86\,\text{m}
  3. Right-triangle trigonometry (at farther point):
    sin⁡35∘=hd\sin 35^{\circ}=\frac{h}{d}
    h=d sin⁡35∘=3165.86×sin⁡35∘=1815.86 mh=d\,\sin 35^{\circ}=3165.86\times\sin 35^{\circ}=1815.86\,\text{m}

Observation: Combine Law of Sines (oblique sub-triangle) plus a final right-triangle step.

Example 4 – Second Mountain (Angles 25° & 15°, 1000 ft Baseline)

Given

  • Baseline QP=1000 ftQP=1000\,\text{ft} between two sight points.
  • Angle at $Q$ (closer to mountain): 25∘25^{\circ}.
  • Angle at $P$ (farther): 15∘15^{\circ}.

Goal

  1. Height of mountain ($h$).
  2. Distance $QP$ (already given, but we compute $PQ$’s projection!).
    Class actually computed $PQ$ (slant) & $BQ$ (horizontal to foot of mountain).

Step A – Determine third angle in the sight triangle:
50∘=25∘+15∘,∴  ∠B=180∘−50∘=130∘50^{\circ}=25^{\circ}+15^{\circ},\quad \therefore\; \angle B=180^{\circ}-50^{\circ}=130^{\circ}

Step B – Distance $QP$ (slanted) using Law of Sines:
sin⁡15∘QP=sin⁡50∘1000  ⇒  QP=1490.48 ft\frac{\sin 15^{\circ}}{QP}=\frac{\sin 50^{\circ}}{1000}\;\Rightarrow\; QP=1490.48\,\text{ft}

Step C – Form right triangle at $Q$ with angle 25∘25^{\circ} and hypotenuse $QP$:
h=QP sin⁡25∘=1490.48×sin⁡25∘=629.90 fth=QP\,\sin 25^{\circ}=1490.48\times\sin25^{\circ}=629.90\,\text{ft}

Alternate route shown in class: first find outside hypotenuse ($2433.76$ ft) then use $15^{\circ}$ angle; both approaches yield the same $h$.

Practice Problem – Two Lighthouses (Text p. 280, #52)

Data

  • Lighthouses P and Q are 3 mi3\,\text{mi} apart along shore (straight line).
  • From a ship at sea, the angles between line‐of‐sight to shore and the two lighthouse sight lines are 15∘15^{\circ} (to P) and 35∘35^{\circ} (to Q).

Definitions
$S$ = Ship, $PQ=3$ mi, $\angle SP!Q=15^{\circ}$, $\angle SQ!P=35^{\circ}$.

  1. Distance Ship → Lighthouse P ($SP$):
    ∠S=180∘−(15∘+35∘)=130∘\angle S=180^{\circ}-(15^{\circ}+35^{\circ})=130^{\circ}
    SPsin⁡35∘=3sin⁡130∘  ⇒  SP=3.21 mi\frac{SP}{\sin 35^{\circ}}=\frac{3}{\sin 130^{\circ}}\;\Rightarrow\; SP=3.21\,\text{mi}
  2. Distance Ship → Shore line (perpendicular from $S$):
    \cos 15^{\circ}=\frac{\text{adjacent (shore)}}{3.21}\;\Rightarrow\;
    \text{shore distance}=3.21\cos15^{\circ}=3.10\,\text{mi}
  3. Distance Ship → Lighthouse Q ($SQ$): form right triangle at $Q$:
    cos⁡35∘=3.10SQ  ⇒  SQ=3.10cos⁡35∘=3.78 mi\cos 35^{\circ}=\frac{3.10}{SQ}\;\Rightarrow\; SQ=\frac{3.10}{\cos35^{\circ}}=3.78\,\text{mi}
    Results verified in session chat (Mario’s numbers matched: 3.21 mi, 3.10 mi, 3.78 mi).

Conceptual & Procedural Highlights

  • Choose the correct triangle: When a diagram contains nested triangles, isolate the one that corresponds to your known data; otherwise you may pair wrong angles/sides.
  • Isosceles cue: equal sides ⇒ equal opposite angles. Exploiting this speeds up solutions.
  • Supplementary angles: Angles on a straight line sum to 180∘180^{\circ}. Used repeatedly for external‐internal relationships.
  • Angle sum in triangle A+B+C=180∘A+B+C=180^{\circ} – fastest way to locate the third angle once two are known.
  • Decimal practice:
    • Keep at least 4 decimals in intermediate sine/cosine calculations to reduce round-off error.
    • Final displayed answers: 2 decimals (angles) or 2–3 decimals (distances) according to instructions.
  • Inverse trig: Every time an angle is unknown but its sine/cosine number is known, apply $\sin^{-1},\cos^{-1}$, etc.
  • Degree vs. Radian mode: first debugging step whenever answers appear nonsensical.

Ethical / Practical Notes

  • Instructor reminders
    • Clarified that “extra eye height of observer” in mountain problems was ignored today; focus strictly on triangle geometry.
  • Mountains & lighthouses illustrate direct real-world surveying applications of trigonometry.

Quick Reference Formulas Used Today

  • Law of Sines: sin⁡Aa=sin⁡Bb=sin⁡Cc\displaystyle \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}
  • Pythagorean (right only): a2+b2=c2a^{2}+b^{2}=c^{2}
  • Right-triangle ratios:
    • sin⁡θ=opphyp\sin\theta=\dfrac{\text{opp}}{\text{hyp}}
    • cos⁡θ=adjhyp\cos\theta=\dfrac{\text{adj}}{\text{hyp}}
    • tan⁡θ=oppadj\tan\theta=\dfrac{\text{opp}}{\text{adj}}
  • Supplementary: θ<em>1+θ</em>2=180∘\theta<em>{1}+\theta</em>{2}=180^{\circ} (straight line)
  • Interior triangle sum: A+B+C=180∘A+B+C=180^{\circ}

What’s Next

  • Upcoming lecture: Section 4.3 – Law of Cosines.
  • Keep practicing Law of Sines via textbook exercises; ensure calculator fluency and diagram accuracy.