3.3 Right-Triangle Trigonometry and Applications

3.3 Right-Triangle Trigonometry and Applications

Right-triangle trigonometry converts an angle into a ratio of sides, which is how examiners turn an unreachable height into arithmetic. Senior-high-school trigonometry works almost entirely from the special angles, so answers stay exact: $8\sqrt{3}$, $15(\sqrt{3}+1)$, $\frac{7\pi}{6}$. If a decimal is needed, use $\sqrt{2}\approx 1.414$ and $\sqrt{3}\approx 1.732$.

1. The six ratios

Label the triangle from the angle you are working with. The hypotenuse never moves — it is always across from the right angle — but opposite and adjacent swap the moment you switch to the other acute angle.

        B
        |\
        | \
      a |  \  c        right angle at C
        |   \         theta is measured at A
        |    \
        |_____\
        C  b  A

        opposite = a      adjacent = b      hypotenuse = c

SOH-CAH-TOA: $\sin\theta = \frac{\text{opp}}{\text{hyp}},\qquad \cos\theta = \frac{\text{adj}}{\text{hyp}},\qquad \tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{\sin\theta}{\cos\theta}$

The three reciprocals pair up in a deliberately confusing order — cosecant goes with sine, secant with cosine: $\csc\theta = \frac{1}{\sin\theta} = \frac{\text{hyp}}{\text{opp}},\qquad \sec\theta = \frac{1}{\cos\theta} = \frac{\text{hyp}}{\text{adj}},\qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\text{adj}}{\text{opp}}$

Cofunction rule: $\sin\theta = \cos(90^\circ - \theta)$ and $\tan\theta = \cot(90^\circ-\theta)$. That is why $\sin 30^\circ$ and $\cos 60^\circ$ are the same number.

2. The exact-value table

You never have to memorize this cold. Draw a $45$–$45$–$90$ triangle with sides $1, 1, \sqrt{2}$ and a $30$–$60$–$90$ triangle with sides $1, \sqrt{3}, 2$, then read the ratios straight off. Note the pattern in the sine column as the angle grows: $\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$.

3. Pythagorean identities

$\sin^2\theta + \cos^2\theta = 1,\qquad 1 + \tan^2\theta = \sec^2\theta,\qquad 1 + \cot^2\theta = \csc^2\theta$

The second and third are just the first divided through by $\cos^2\theta$ and $\sin^2\theta$. Use them to jump between ratios without finding the angle: if $\theta$ is acute and $\sin\theta = \frac{3}{5}$, then $\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$, so $\cos\theta = \frac{4}{5}$ (positive, because $\theta$ is acute) and $\tan\theta = \frac{3}{4}$. The faster route is to recognize the $3$-$4$-$5$ triangle directly.

4. Finding an unknown side or angle

When the unknown sits in the denominator, cross-multiply rather than guess: $\sin 30^\circ = \frac{12}{x}$ means $x = \frac{12}{\sin 30^\circ} = \frac{12}{1/2} = 24$, not $6$. Because only special angles appear, "finding the angle" means recognizing a value: $\tan\theta = 1 \Rightarrow \theta = 45^\circ$; $\tan\theta = \sqrt{3} \Rightarrow \theta = 60^\circ$; $\sin\theta = \frac{1}{2} \Rightarrow \theta = 30^\circ$.

5. Angles of elevation and depression

• Angle of elevation: measured upward from the observer's horizontal line of sight to the object.
• Angle of depression: measured downward from the observer's horizontal line of sight.

   observer A ========================   horizontal through A
            | \
            |  \      <- angle of depression, measured at A
          h |   \        between the horizontal and the line of sight
            |    \
            +-----B =====================   ground: horizontal through B
                x      <- angle of elevation, measured at B
                          (equal to the depression angle at A)

Since the two horizontal lines are parallel, the angle of elevation from the ground point up to the observer equals the angle of depression from the observer down to that point — they are alternate interior angles. Depression angles are therefore usable inside the triangle without any extra step, provided you place them at the bottom vertex.

6. Worked example 1 — flagpole and shadow

Problem. A flagpole on level ground casts a $24$ m shadow at the moment the sun's angle of elevation is $30^\circ$. A shorter pole beside it casts a $9$ m shadow at the same instant. Find both heights.

1. The shadow is the adjacent side and the pole is the opposite side, so use tangent: $\tan 30^\circ = \frac{h}{24}$.
2. $h = 24\tan 30^\circ = \frac{24}{\sqrt{3}} = \frac{24\sqrt{3}}{3} = 8\sqrt{3} \approx 13.86$ m.
3. For the shorter pole, $h' = 9\tan 30^\circ = \frac{9}{\sqrt{3}} = 3\sqrt{3} \approx 5.20$ m.
4. Sanity check by similarity: at one instant the sun's angle is the same for both, so heights are proportional to shadows, and $\frac{8\sqrt{3}}{3\sqrt{3}} = \frac{8}{3} = \frac{24}{9}$.

7. Worked example 2 — one height shared by two triangles

Problem. Standing on level ground along Taft Avenue, Maricel measures the angle of elevation to the top of a condominium as $30^\circ$. She walks $30$ m straight toward the building and the angle of elevation becomes $45^\circ$. How tall is the building above her eye level, and how tall is it in total if her eyes are $1.5$ m above the ground?

1. Let $h$ be the height above eye level and $x$ the horizontal distance from the nearer position to the base. The unknown $h$ is the shared side of both right triangles.
2. Near triangle: $\tan 45^\circ = \frac{h}{x} \Rightarrow 1 = \frac{h}{x} \Rightarrow x = h$.
3. Far triangle: $\tan 30^\circ = \frac{h}{x + 30}$, and substituting $x = h$ gives $\frac{1}{\sqrt{3}} = \frac{h}{h+30}$.
4. Cross-multiply: $h + 30 = h\sqrt{3} \Rightarrow 30 = h(\sqrt{3}-1) \Rightarrow h = \frac{30}{\sqrt{3}-1}$.
5. Rationalize: $h = \frac{30(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{30(\sqrt{3}+1)}{2} = 15(\sqrt{3}+1) \approx 15(2.732) = 40.98$ m.
6. Total height $\approx 40.98 + 1.5 = 42.5$ m. Forgetting to add eye height is the built-in trap.

8. Bearings and simple navigation

Two notations appear:

1. Quadrant bearing — $N,30^\circ,E$ means start at north and rotate $30^\circ$ toward east.
2. True (azimuth) bearing — $030^\circ$, always three digits, measured clockwise from north.

So $N,30^\circ,E$ and $030^\circ$ describe the same heading, while $S,40^\circ,W$ equals $220^\circ$. To resolve a leg of a journey, draw the north–south line and use the angle at the starting point: a boat sailing $40$ km on bearing $N,30^\circ,E$ travels $\text{east} = 40\sin 30^\circ = 20\ \text{km},\qquad \text{north} = 40\cos 30^\circ = 20\sqrt{3} \approx 34.6\ \text{km}.$ Notice that sine takes the east–west component here because the angle is measured from north, the reverse of the usual habit.

9. Degrees and radians

$\pi\ \text{rad} = 180^\circ \quad\Longrightarrow\quad \text{radians} = \text{degrees}\times\frac{\pi}{180},\qquad \text{degrees} = \text{radians}\times\frac{180}{\pi}$

10. Common traps

• Reversing opposite and adjacent after switching to the other acute angle.
• Measuring an angle of depression from the vertical tower instead of from the horizontal.
• Using sine when the two known quantities are the two legs, where tangent is required.
• Multiplying by the ratio when the unknown is a hypotenuse in the denominator and you should divide.
• Omitting the observer's eye height or the height of a cliff base.
• Treating $N,30^\circ,E$ as $30^\circ$ from east.
• Leaving an answer in degrees when the choices are in radians.