// Numbas version: finer_feedback_settings {"name": "2 triangle problem with ramp", "extensions": [], "custom_part_types": [], "resources": [["question-resources/ramp_problem_jGRPkmC.png", "/srv/numbas/media/question-resources/ramp_problem_jGRPkmC.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "2 triangle problem with ramp", "tags": [], "metadata": {"description": "
Students are given 2 right-angle triangles - two ramps of differing steepness up a step, and are asked to find one of a selection of randomly chosen lengths. The height of the step is given - it is randomised. Students are also given either the angle of incline of the steeper ramp or its length, both of which are randomised. They are also given the angle of incline of the shallower ramp, which is also randomised.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "A {height} metre high rise is represented by BT in the diagram. There is an existing ramp, XT, from the lower level to the upper level, but it has been deemed too steep, and a new ramp, YT, is to be built. The angle between the ground and the new ramp is to be {BYT}°.
\n\nnot to scale
", "advice": "$\\triangle BTX$ and $\\triangle BTY$ are right-angle triangles. So we can use the trigonometric ratios to determine the lengths of the sides.
\nTo find the difference between $YT$ and $XT$ we need to first find $YT$ and $XT$.
\nWe can find $YT$ using the sine ratio: $\\sin(\\angle BYT) = \\frac{opposite}{hypotenuse}$
\n$\\sin(\\var{BYT}°) = \\frac{\\var{height}}{YT}$
\nSo $YT = \\var{YT}$ m
\nWe can also find $XT$ using the sine ratio:
\n$sin(\\var{BXT}°) = \\frac{\\var{height}}{XT}$
\nSo $XT = \\var{XT}$ m
\nThe difference in lengths is $YT - XT = \\var{YT}-\\var{XT}=\\var{YT-XT}$ m
\nTo find length $XY$ we first need to find $XB$ and $YB$.
\nWe can find $XB$ using the tan ratio: $\\tan(\\angle BXT) = \\frac{opposite}{adjacent}$
\n$\\tan(\\var{BXT}°) = \\frac{\\var{height}}{XB}$
\nSo $XB=\\var{BX}$ m
\nWe can find $XB$ using Pythagoras' Theorem: $a^2 + b^2=c^2$
\n$\\var{XT}^2=\\var{height}^2+XB^2$
\nSo $XB=\\var{BX}$ m
\nWe can also find $YB$ using the tan ratio: $\\tan(\\angle BYT) = \\frac{opposite}{adjacent}$
\n$\\tan(\\var{BYT}°) = \\frac{\\var{height}}{YB}$
\nSo $YB=\\var{BY}$ m
\nTherefore, $XY = YB - XB = \\var{BY}-\\var{BX}=\\var{XY}$ m
\nYour answer may differ slightly (by up to 0.1) due to rounding as the computer solves the problem using a different method.
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\nRound your answer to 1 decimal place.
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