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Unit 8: Geometry and Trigonometry

Topic 5

Right Triangle Trigonometry

Trigonometry on the SAT is limited to right triangles and the three basic ratios: sine, cosine, and tangent. These are defined relative to a specific acute angle in a right triangle. If θ\theta is one of the acute angles, then:
sin⁡(θ)=oppositehypotenuse,cos⁡(θ)=adjacenthypotenuse,tan⁡(θ)=oppositeadjacent\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
The mnemonic SOH-CAH-TOA helps you remember these: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent. ``Opposite'' and ``adjacent'' are always defined relative to the angle you are working with, not the triangle as a whole.
An important relationship connects sine and cosine of complementary angles (two angles that add up to 90∘90^\circ). In a right triangle with acute angles α\alpha and β\beta:
sin⁡(α)=cos⁡(β)andcos⁡(α)=sin⁡(β)\sin(\alpha) = \cos(\beta) \quad \text{and} \quad \cos(\alpha) = \sin(\beta)
This is because the side opposite α\alpha is the side adjacent to β\beta, and vice versa. This means sin⁡(x∘)=cos⁡(90∘−x∘)\sin(x^\circ) = \cos(90^\circ - x^\circ) for any acute angle xx. The SAT tests this relationship directly, often by asking you to recognize that sin⁡(32∘)=cos⁡(58∘)\sin(32^\circ) = \cos(58^\circ) or similar.
The Pythagorean identity states that for any angle θ\theta:
sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1
This follows directly from the Pythagorean Theorem applied to the unit definitions: if the sides are aa, bb, and cc (hypotenuse), then sin⁡(θ)=ac\sin(\theta) = \frac{a}{c} and cos⁡(θ)=bc\cos(\theta) = \frac{b}{c}, so
sin⁡2(θ)+cos⁡2(θ)=a2c2+b2c2=a2+b2c2=c2c2=1.\sin^2(\theta) + \cos^2(\theta) = \frac{a^2}{c^2} + \frac{b^2}{c^2} = \frac{a^2 + b^2}{c^2} = \frac{c^2}{c^2} = 1.
On the SAT, this identity is most often used when you are given sin⁡(θ)\sin(\theta) or cos⁡(θ)\cos(\theta) and need to find the other. For example, if sin⁡(θ)=35\sin(\theta) = \frac{3}{5}, then cos⁡2(θ)=1−925=1625\cos^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25}, so cos⁡(θ)=45\cos(\theta) = \frac{4}{5} (taking the positive root, since θ\theta is an acute angle in a right triangle).
To solve for a missing side using trigonometry, pick the trig ratio that involves the side you know and the side you want. To solve for a missing angle, use the inverse trig function on your calculator: if sin⁡(θ)=0.6\sin(\theta) = 0.6, then θ=sin⁡−1(0.6)≈36.87∘\theta = \sin^{-1}(0.6) \approx 36.87^\circ. However, the SAT usually keeps angles at values you can handle exactly (30, 45, 60) or asks you to leave the answer as a trig expression.
θoppositeadjacenthypotenuse

Core Skills

  • Identify the opposite side, adjacent side, and hypotenuse relative to a given angle and set up the correct trig ratio.
  • Use SOH-CAH-TOA to find a missing side length in a right triangle.
  • Apply the complementary angle relationship: sin⁡(x∘)=cos⁡(90∘−x∘)\sin(x^\circ) = \cos(90^\circ - x^\circ).
  • Use the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 to find one trig value given the other.
  • Solve word problems involving angles of elevation and depression using right triangle trigonometry.

Example 1: Setting Up a Trig Ratio

In a right triangle, the side opposite angle AA has length 5 and the hypotenuse has length 13. What is cos⁡(A)\cos(A)? Step 1: We need the adjacent side. Use the Pythagorean Theorem.
adjacent2+52=132  ⟹  adjacent2=169−25=144  ⟹  adjacent=12\text{adjacent}^2 + 5^2 = 13^2 \implies \text{adjacent}^2 = 169 - 25 = 144 \implies \text{adjacent} = 12
Step 2: Apply the definition.
cos⁡(A)=adjacenthypotenuse=1213\cos(A) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13}
1213\boxed{\dfrac{12}{13}}

Example 2: Finding a Missing Side

In a right triangle, one acute angle measures 30∘30^\circ and the hypotenuse is 20. What is the length of the side opposite the 30∘30^\circ angle? Step 1: The side we want is opposite the given angle, and we know the hypotenuse. Use sine.
sin⁡(30∘)=opposite20\sin(30^\circ) = \frac{\text{opposite}}{20}
Step 2: Since sin⁡(30∘)=12\sin(30^\circ) = \frac{1}{2}:
12=opposite20  ⟹  opposite=10\frac{1}{2} = \frac{\text{opposite}}{20} \implies \text{opposite} = 10
The side opposite the 30∘30^\circ angle is 10\boxed{10}.

Example 3: Complementary Angle Relationship

If sin⁡(3x+10)∘=cos⁡(2x+30)∘\sin(3x + 10)^\circ = \cos(2x + 30)^\circ, what is the value of xx? Step 1: Since sin⁡(α)=cos⁡(β)\sin(\alpha) = \cos(\beta) when α+β=90∘\alpha + \beta = 90^\circ:
(3x+10)+(2x+30)=90(3x + 10) + (2x + 30) = 90
Step 2: Solve.
5x+40=90  ⟹  5x=50  ⟹  x=105x + 40 = 90 \implies 5x = 50 \implies x = 10
The value of xx is 10\boxed{10}.

Example 4: Pythagorean Identity

If sin⁡(θ)=725\sin(\theta) = \dfrac{7}{25} and θ\theta is an acute angle, what is cos⁡(θ)\cos(\theta)? Step 1: Apply the Pythagorean identity.
sin⁡2(θ)+cos⁡2(θ)=1  ⟹  (725)2+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 \implies \left(\frac{7}{25}\right)^2 + \cos^2(\theta) = 1
49625+cos⁡2(θ)=1  ⟹  cos⁡2(θ)=576625\frac{49}{625} + \cos^2(\theta) = 1 \implies \cos^2(\theta) = \frac{576}{625}
Step 2: Since θ\theta is acute, cos⁡(θ)\cos(\theta) is positive.
cos⁡(θ)=2425\cos(\theta) = \frac{24}{25}
2425\boxed{\dfrac{24}{25}}

Example 5: Angle of Elevation Word Problem

A person stands 50 feet from the base of a building and looks up at the top of the building at an angle of elevation of 60∘60^\circ. What is the height of the building, in feet? Step 1: The horizontal distance (50 feet) is the side adjacent to the angle. The height of the building is the side opposite the angle. Use tangent.
tan⁡(60∘)=h50\tan(60^\circ) = \frac{h}{50}
Step 2: Since tan⁡(60∘)=3\tan(60^\circ) = \sqrt{3}:
3=h50  ⟹  h=503\sqrt{3} = \frac{h}{50} \implies h = 50\sqrt{3}
The building is 503\boxed{50\sqrt{3}} feet tall.

Example 6: Finding tan⁡(θ)\tan(\theta) from sin⁡(θ)\sin(\theta)

If sin⁡(θ)=35\sin(\theta) = \dfrac{3}{5} and θ\theta is acute, what is tan⁡(θ)\tan(\theta)? Step 1: Find cos⁡(θ)\cos(\theta) using the Pythagorean identity.
cos⁡2(θ)=1−sin⁡2(θ)=1−925=1625  ⟹  cos⁡(θ)=45\cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25} \implies \cos(\theta) = \frac{4}{5}
Step 2: Use tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \dfrac{\sin(\theta)}{\cos(\theta)}.
tan⁡(θ)=3/54/5=34\tan(\theta) = \frac{3/5}{4/5} = \frac{3}{4}
34\boxed{\dfrac{3}{4}}

Key Takeaways

  • Always identify ``opposite'' and ``adjacent'' relative to the specific angle in the problem, not relative to the triangle in general. Drawing a quick mental picture of which sides are which prevents mix-ups.
  • The complementary angle relationship sin⁡(x)=cos⁡(90−x)\sin(x) = \cos(90 - x) is one of the most commonly tested trig facts on the SAT. If you see a question with sin⁡\sin and cos⁡\cos of two different angles, check whether those angles add to 90∘90^\circ.
  • The Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 turns one known trig ratio into another. You can also think of it as reconstructing the right triangle: given sin⁡(θ)=ac\sin(\theta) = \frac{a}{c}, the missing side is b=c2−a2b = \sqrt{c^2 - a^2}.
  • For angle of elevation and depression problems, the horizontal distance is always adjacent, the vertical distance is always opposite, and the line of sight is the hypotenuse.