Deriving the Triangle Area Formula - Geometry
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A $13$ ft ladder is $5$ ft from a wall. Find how high up the wall it reaches.
A $13$ ft ladder is $5$ ft from a wall. Find how high up the wall it reaches.
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$12\text{ ft}$. Using $h = \sqrt{13^2 - 5^2} = \sqrt{144} = 12$ ft.
$12\text{ ft}$. Using $h = \sqrt{13^2 - 5^2} = \sqrt{144} = 12$ ft.
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A right triangle has hypotenuse $10$ and acute angle $30^\circ$. Find the opposite leg.
A right triangle has hypotenuse $10$ and acute angle $30^\circ$. Find the opposite leg.
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$5$. Using opposite = hypotenuse × sin(30°) = $10 × \frac{1}{2} = 5$.
$5$. Using opposite = hypotenuse × sin(30°) = $10 × \frac{1}{2} = 5$.
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State the Pythagorean identity that connects $cot(\theta)$ and $csc(\theta)$.
State the Pythagorean identity that connects $cot(\theta)$ and $csc(\theta)$.
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$1+\cot^2(\theta)=\csc^2(\theta)$. One plus cotangent squared equals cosecant squared.
$1+\cot^2(\theta)=\csc^2(\theta)$. One plus cotangent squared equals cosecant squared.
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A right triangle has adjacent leg $8$ and angle $45^\circ$. Find the opposite leg.
A right triangle has adjacent leg $8$ and angle $45^\circ$. Find the opposite leg.
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$8$. Since tan(45°) = 1, opposite equals adjacent.
$8$. Since tan(45°) = 1, opposite equals adjacent.
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A ramp has length $10$ m and rises $6$ m. Find the horizontal run.
A ramp has length $10$ m and rises $6$ m. Find the horizontal run.
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$8\text{ m}$. Using $run = \sqrt{10^2 - 6^2} = \sqrt{64} = 8$ m.
$8\text{ m}$. Using $run = \sqrt{10^2 - 6^2} = \sqrt{64} = 8$ m.
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Identify the correct setup to find the missing leg $x$ if $c=17$ and $a=8$ in a right triangle.
Identify the correct setup to find the missing leg $x$ if $c=17$ and $a=8$ in a right triangle.
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$x=\sqrt{17^2-8^2}$. Using Pythagorean theorem: $x = \sqrt{c^2 - a^2}$.
$x=\sqrt{17^2-8^2}$. Using Pythagorean theorem: $x = \sqrt{c^2 - a^2}$.
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Find and correct the error: $\tan(\theta)=\frac{\text{adjacent}}{\text{opposite}}$.
Find and correct the error: $\tan(\theta)=\frac{\text{adjacent}}{\text{opposite}}$.
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Correct: $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$. Tangent uses opposite over adjacent, not adjacent over opposite.
Correct: $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$. Tangent uses opposite over adjacent, not adjacent over opposite.
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Find and correct the error: $\sin(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$.
Find and correct the error: $\sin(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$.
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Correct: $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$. Sine uses opposite over hypotenuse, not adjacent.
Correct: $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$. Sine uses opposite over hypotenuse, not adjacent.
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A right triangle has legs $9$ and $12$. What is $\tan(\theta)$ if $\theta$ is opposite the $9$ side?
A right triangle has legs $9$ and $12$. What is $\tan(\theta)$ if $\theta$ is opposite the $9$ side?
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$\tan(\theta)=\frac{3}{4}$. Tangent = opposite ÷ adjacent = $\frac{9}{12} = \frac{3}{4}$.
$\tan(\theta)=\frac{3}{4}$. Tangent = opposite ÷ adjacent = $\frac{9}{12} = \frac{3}{4}$.
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A right triangle has legs $9$ and $12$. What is $\cos(\theta)$ if $\theta$ is adjacent to the $9$ side?
A right triangle has legs $9$ and $12$. What is $\cos(\theta)$ if $\theta$ is adjacent to the $9$ side?
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$\cos(\theta)=\frac{3}{5}$. Hypotenuse = $\sqrt{9^2 + 12^2} = 15$, so cos = $\frac{9}{15} = \frac{3}{5}$.
$\cos(\theta)=\frac{3}{5}$. Hypotenuse = $\sqrt{9^2 + 12^2} = 15$, so cos = $\frac{9}{15} = \frac{3}{5}$.
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A right triangle has legs $9$ and $12$. What is $\sin(\theta)$ if $\theta$ is opposite the $9$ side?
A right triangle has legs $9$ and $12$. What is $\sin(\theta)$ if $\theta$ is opposite the $9$ side?
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$\sin(\theta)=\frac{3}{5}$. Hypotenuse = $\sqrt{9^2 + 12^2} = 15$, so sin = $\frac{9}{15} = \frac{3}{5}$.
$\sin(\theta)=\frac{3}{5}$. Hypotenuse = $\sqrt{9^2 + 12^2} = 15$, so sin = $\frac{9}{15} = \frac{3}{5}$.
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A right triangle has opposite leg $10$ and angle $30^\circ$. Find the hypotenuse.
A right triangle has opposite leg $10$ and angle $30^\circ$. Find the hypotenuse.
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$20$. Using $\text{hypotenuse} = \text{opposite} \div \sin(30^\circ) = 10 \div \frac{1}{2} = 20$.
$20$. Using $\text{hypotenuse} = \text{opposite} \div \sin(30^\circ) = 10 \div \frac{1}{2} = 20$.
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State the Pythagorean Theorem for a right triangle with legs $a,b$ and hypotenuse $c$.
State the Pythagorean Theorem for a right triangle with legs $a,b$ and hypotenuse $c$.
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$a^2+b^2=c^2$. Fundamental relationship between legs and hypotenuse in right triangles.
$a^2+b^2=c^2$. Fundamental relationship between legs and hypotenuse in right triangles.
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State the definition of $sin(\theta)$ in a right triangle.
State the definition of $sin(\theta)$ in a right triangle.
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$\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$. Ratio of the side opposite the angle to the hypotenuse.
$\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$. Ratio of the side opposite the angle to the hypotenuse.
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State the definition of $cos(\theta)$ in a right triangle.
State the definition of $cos(\theta)$ in a right triangle.
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$\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$. Ratio of the side adjacent to the angle to the hypotenuse.
$\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$. Ratio of the side adjacent to the angle to the hypotenuse.
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State the definition of $tan(\theta)$ in a right triangle.
State the definition of $tan(\theta)$ in a right triangle.
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$\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$. Ratio of the side opposite the angle to the adjacent side.
$\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$. Ratio of the side opposite the angle to the adjacent side.
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State the definition of $csc(\theta)$ in a right triangle.
State the definition of $csc(\theta)$ in a right triangle.
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$\csc(\theta)=\frac{\text{hypotenuse}}{\text{opposite}}$. Reciprocal of sine function.
$\csc(\theta)=\frac{\text{hypotenuse}}{\text{opposite}}$. Reciprocal of sine function.
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State the definition of $sec(\theta)$ in a right triangle.
State the definition of $sec(\theta)$ in a right triangle.
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$\sec(\theta)=\frac{\text{hypotenuse}}{\text{adjacent}}$. Reciprocal of cosine function.
$\sec(\theta)=\frac{\text{hypotenuse}}{\text{adjacent}}$. Reciprocal of cosine function.
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State the definition of $cot(\theta)$ in a right triangle.
State the definition of $cot(\theta)$ in a right triangle.
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$\cot(\theta)=\frac{\text{adjacent}}{\text{opposite}}$. Reciprocal of tangent function.
$\cot(\theta)=\frac{\text{adjacent}}{\text{opposite}}$. Reciprocal of tangent function.
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What identity relates $tan(\theta)$ to $sin(\theta)$ and $cos(\theta)$?
What identity relates $tan(\theta)$ to $sin(\theta)$ and $cos(\theta)$?
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$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$. Tangent equals sine divided by cosine.
$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$. Tangent equals sine divided by cosine.
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What is the complementary-angle relationship for sine and cosine in right triangles?
What is the complementary-angle relationship for sine and cosine in right triangles?
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$\sin(\theta)=\cos(90^\circ-\theta)$. Sine and cosine of complementary angles are equal.
$\sin(\theta)=\cos(90^\circ-\theta)$. Sine and cosine of complementary angles are equal.
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What is the complementary-angle relationship for tangent and cotangent in right triangles?
What is the complementary-angle relationship for tangent and cotangent in right triangles?
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$\tan(\theta)=\cot(90^\circ-\theta)$. Tangent and cotangent of complementary angles are equal.
$\tan(\theta)=\cot(90^\circ-\theta)$. Tangent and cotangent of complementary angles are equal.
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State the Pythagorean identity that connects $sin(\theta)$ and $cos(\theta)$.
State the Pythagorean identity that connects $sin(\theta)$ and $cos(\theta)$.
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$\sin^2(\theta)+\cos^2(\theta)=1$. Sum of squares of sine and cosine equals 1.
$\sin^2(\theta)+\cos^2(\theta)=1$. Sum of squares of sine and cosine equals 1.
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State the Pythagorean identity that connects $tan(\theta)$ and $sec(\theta)$.
State the Pythagorean identity that connects $tan(\theta)$ and $sec(\theta)$.
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$1+\tan^2(\theta)=\sec^2(\theta)$. One plus tangent squared equals secant squared.
$1+\tan^2(\theta)=\sec^2(\theta)$. One plus tangent squared equals secant squared.
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A road rises $200$ m over a horizontal distance of $1000$ m. Find the angle of incline.
A road rises $200$ m over a horizontal distance of $1000$ m. Find the angle of incline.
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$\theta=\tan^{-1}!\left(\frac{1}{5}\right)$. Using arctangent of rise over run ratio.
$\theta=\tan^{-1}!\left(\frac{1}{5}\right)$. Using arctangent of rise over run ratio.
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A kite is $40$ m high and the string makes a $60^\circ$ angle with the ground. Find the string length.
A kite is $40$ m high and the string makes a $60^\circ$ angle with the ground. Find the string length.
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$\frac{40}{\sin(60^\circ)}\text{ m}$. String length = height ÷ sin(angle with ground).
$\frac{40}{\sin(60^\circ)}\text{ m}$. String length = height ÷ sin(angle with ground).
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A kite string is $50$ m long and makes a $30^\circ$ angle with the ground. Find the kite height.
A kite string is $50$ m long and makes a $30^\circ$ angle with the ground. Find the kite height.
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$50\sin(30^\circ)\text{ m}$. Height = string length × sin(angle with ground).
$50\sin(30^\circ)\text{ m}$. Height = string length × sin(angle with ground).
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From a point $30$ m from a building, the angle of elevation to the top is $40^\circ$. Find the height.
From a point $30$ m from a building, the angle of elevation to the top is $40^\circ$. Find the height.
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$30\tan(40^\circ)\text{ m}$. Height = distance × tan(elevation angle).
$30\tan(40^\circ)\text{ m}$. Height = distance × tan(elevation angle).
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A $12$ m tree casts a $5$ m shadow. Find the angle of elevation of the sun.
A $12$ m tree casts a $5$ m shadow. Find the angle of elevation of the sun.
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$\theta=\tan^{-1}!\left(\frac{12}{5}\right)$. Using arctangent of tree height over shadow length.
$\theta=\tan^{-1}!\left(\frac{12}{5}\right)$. Using arctangent of tree height over shadow length.
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An airplane climbs at $15^\circ$ for $200$ m along its path. Find the altitude gained.
An airplane climbs at $15^\circ$ for $200$ m along its path. Find the altitude gained.
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$200\sin(15^\circ)\text{ m}$. Altitude = path length × sin(angle of climb).
$200\sin(15^\circ)\text{ m}$. Altitude = path length × sin(angle of climb).
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