Deriving the Triangle Area Formula - Precalculus

1. A ramp rises 3 m over a 4 m run. What is the ramp length (hypotenuse)?

   $5\text{ m}$. Use Pythagorean theorem: length $= \sqrt{3^2+4^2} = \sqrt{25} = 5$ m.

2. What is the complementary-angle relationship between sine and cosine in a right triangle?

   $\sin(\theta)=\cos(90^\circ-\theta)$. Sine and cosine of complementary angles are equal.

3. Find $\tan(\theta)$ if opposite $=9$ and adjacent $=12$.

   $\tan(\theta)=\frac{3}{4}$. Simplify $\frac{9}{12}$ to lowest terms: $\frac{3}{4}$.

4. State the reciprocal identity for $\csc(\theta)$ in terms of $\sin(\theta)$.

   $\csc(\theta)=\frac{1}{\sin(\theta)}$. Cosecant is the reciprocal of sine.

5. Identify the missing leg: If $c=13$ and $a=5$, what is the other leg $b$?

   $b=12$. Use Pythagorean theorem: $b=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$.

6. Find the horizontal distance $d$ if the height is $10$ and the angle of elevation is $45^\circ$.

   $d=10$. With $\tan(45°)=1$, ground distance equals height.

7. State the reciprocal identity for $\sec(\theta)$ in terms of $\cos(\theta)$.

   $\sec(\theta)=\frac{1}{\cos(\theta)}$. Secant is the reciprocal of cosine.

8. Find $\cos(\theta)$ if adjacent $=24$ and hypotenuse $=25$.

   $\cos(\theta)=\frac{24}{25}$. Direct substitution into the cosine ratio formula.

9. What is the definition of $cos(theta)$ in a right triangle?

   $\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$. Ratio of the side adjacent to angle $\theta$ over the hypotenuse.

10. Identify the missing side: If $a=9$ and $b=12$, what is the hypotenuse $c$?

    $c=15$. Use Pythagorean theorem: $c=\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15$.

11. Identify the angle of depression in a right-triangle model of looking down at an object.

    Angle between horizontal and the line of sight downward. Measured from horizontal down to the line of sight.

12. Find the adjacent side: If hypotenuse $=20$ and $\cos(\theta)=\frac{4}{5}$, what is adjacent?

    $16$. Multiply: adjacent $= \cos(\theta) \times$ hypotenuse $= \frac{4}{5} \times 20 = 16$.

13. Find the missing leg $b$ if the hypotenuse is $13$ and the other leg is $5$.

    $b=12$. Use $b=\sqrt{c^2-a^2}=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$.

14. State the reciprocal identity for $\cot(\theta)$ in terms of $\tan(\theta)$.

    $\cot(\theta)=\frac{1}{\tan(\theta)}$. Cotangent is the reciprocal of tangent.

15. What is the definition of $sin(theta)$ in a right triangle?

    $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$. Ratio of the side opposite to angle $\theta$ over the hypotenuse.

16. State the complementary angle identity for sine and cosine in right triangles.

    $\sin(\theta)=\cos(90^\circ-\theta)$. Sine of an angle equals cosine of its complement in a right triangle.

17. What is $\cos(\theta)$ in a right triangle in terms of adjacent and hypotenuse?

    $\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$. Ratio of the side next to angle $\theta$ over the longest side.

18. A 13 ft ladder reaches a 5 ft height on a wall. What is the distance from wall to ladder base?

    $12\text{ ft}$. Use Pythagorean theorem: base $= \sqrt{13^2-5^2} = \sqrt{144} = 12$ ft.

19. Identify the angle of elevation in a right-triangle model of looking up at an object.

    Angle between horizontal and the line of sight upward. Measured from horizontal up to the line of sight.

20. What is the definition of $tan(theta)$ in a right triangle?

    $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$. Ratio of the side opposite to angle $\theta$ over the adjacent side.

21. State the Pythagorean Theorem for a right triangle with legs $a,b$ and hypotenuse $c$.

    $a^2+b^2=c^2$. Relates the squares of the legs to the square of the hypotenuse in any right triangle.

22. What is $\tan(\theta)$ in a right triangle in terms of opposite and adjacent?

    $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$. Ratio of opposite over adjacent sides, or $\frac{\sin(\theta)}{\cos(\theta)}$.

23. State the complementary angle identity for tangent in right triangles.

    $\tan(\theta)=\frac{1}{\tan(90^\circ-\theta)}$. Tangent of an angle equals the reciprocal of tangent of its complement.

24. Find the opposite side: If adjacent $=15$ and $\tan(\theta)=\frac{2}{3}$, what is opposite?

    $10$. Multiply: opposite $= \tan(\theta) \times$ adjacent $= \frac{2}{3} \times 15 = 10$.

25. Find $\sin(\theta)$ if opposite $=7$ and hypotenuse $=25$.

    $\sin(\theta)=\frac{7}{25}$. Direct substitution into the sine ratio formula.

26. Find the hypotenuse $c$ if the legs are $6$ and $8$.

    $c=10$. Apply Pythagorean theorem: $c=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$.

27. Find the angle $theta$ if $\text{opp}=3$ and $\text{adj}=3$ in a right triangle.

    $\theta=45^\circ$. Since $\tan(\theta)=\frac{3}{3}=1$, and $\tan(45°)=1$.

28. Find the height $h$ if the ground distance is $10$ and the angle of elevation is $45^\circ$.

    $h=10$. With $\tan(45°)=1$, height equals ground distance.

29. Find the adjacent side if $h=10$ and $\theta=60^\circ$ with $\cos(\theta)=\frac{\text{adj}}{h}$.

    $\text{adj}=5$. Since $\cos(60°)=0.5$, adjacent $=10 \times 0.5=5$.

30. State the complementary-angle relationship between tangent and cotangent in a right triangle.

    $\tan(\theta)=\cot(90^\circ-\theta)$. Tangent and cotangent of complementary angles are equal.

31. Find the opposite side: If hypotenuse $=10$ and $\sin(\theta)=\frac{3}{5}$, what is opposite?

    $6$. Multiply: opposite $= \sin(\theta) \times$ hypotenuse $= \frac{3}{5} \times 10 = 6$.

32. What is $\sin(\theta)$ in a right triangle in terms of opposite and hypotenuse?

    $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$. Ratio of the side opposite to angle $\theta$ over the longest side.

33. Find the opposite side if $h=10$ and $\theta=30^\circ$ with $\sin(\theta)=\frac{\text{opp}}{h}$.

    $\text{opp}=5$. Since $\sin(30°)=0.5$, opposite $=10 \times 0.5=5$.

34. A right triangle has legs $6$ and $8$. What is $\cos(\theta)$ for the angle adjacent to the $8$ side?

    $\cos(\theta)=\frac{4}{5}$. Hypotenuse $= \sqrt{6^2+8^2} = 10$, so $\cos(\theta) = \frac{8}{10} = \frac{4}{5}$.

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