// Numbas version: exam_results_page_options {"name": "Geometry: right-angled triangle", "extensions": [], "custom_part_types": [], "resources": [["question-resources/triangle_6nKlln9.png", "/srv/numbas/media/question-resources/triangle_6nKlln9.png"], ["question-resources/triangle_2.png", "/srv/numbas/media/question-resources/triangle_2.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"preamble": {"css": "", "js": ""}, "extensions": [], "rulesets": {}, "functions": {}, "tags": [], "variable_groups": [], "advice": "

Each angle on a triangle is connected to two sides and is facing another.

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The longest side of the triangle is always the hypotenuse.

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The other side that makes the angle is called the adjacent.

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The final side not connected in any way to the angle is called the opposite.

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For example, using the image below, you can see which side is denoted by each term from the highlighted angle's perspective.

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One of the ways you can approach this style of question is by using SOHCAHTOA.

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This can be written more visually as

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\\[\\text{S}^\\text{O}_\\text{H}\\space\\text{C}^\\text{A}_\\text{H}\\space\\text{T}^\\text{O}_\\text{A}\\]

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It represents each trigonometric function and what they are equivalent to.

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Written out in full, we would have:

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SIN: opposite / hypotenuse

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COS: adjacent / hypotenuse

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TAN: opposite / adjacent

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For example, $\\sin$ is represented by the first S.

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If we were given an angle, say of $30^\\circ$,

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$\\sin(30^\\circ)=\\frac{\\text{opposite}}{\\text{hypotenuse}}$

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Evaluating $\\sin(30^\\circ)=\\frac{1}{2}$, we now know that $\\frac{\\text{opposite}}{\\text{hypotenuse}}=\\frac{1}{2}$

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If we were given one of these sides, we would then be able to work out the other one by multiplying accordingly.

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Similarly if we were given two sides, and told to work out a specific angle, we could.

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Referring to the image above, suppose we want to find the highlighted angle and we are given that the hypotenuse is equal to $5$ units, and the adjacent is $4$ units.

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We would determine from SOHCAHTOA that we need to use cos since we have the values for A and H.

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So, $\\cos(x)=\\frac{4}{5}$

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Hence, $x=\\cos^{-1}(\\frac{4}{5})=36.87^\\circ$

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Calculations are presented and students are asked to choose which angle/length the calculation corresponds to.

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Match the calculations with the appropriate length or angle.

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$\\sqrt{a^2+b^2}$

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$\\sqrt{a^2-b^2}$

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$b^2+c^2$

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$\\sqrt{c^2+b^2}$

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$\\frac{a}{c}$

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$\\frac{a}{b}$

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$\\frac{c}{a}$

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$\\frac{b}{a}$

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$\\frac{b}{c}$

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$\\frac{c}{b}$

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$a \\sin(x)$

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$a \\cos(x)$

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$a \\tan(x)$

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$c \\cos(x)$

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$\\frac{c}{\\sin(x)}$

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$\\frac{b}{\\sin(x)}$

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$\\frac{b}{\\cos(x)}$

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$b \\tan(y)$

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$c \\tan(y)$

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$\\frac{c}{\\tan(y)}$

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$a$

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$b$

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$c$

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$\\sin(x)$

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$\\cos(x)$

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$\\tan(x)$

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$\\sin(y)$

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$\\cos(y)$

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$\\tan(y)$

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None of these

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(Note the diagram is not drawn to scale)

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