// 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}°. 

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\"A

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not 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.

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To find the difference between $YT$ and $XT$ we need to first find $YT$ and $XT$. 

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We can find $YT$ using the sine ratio: $\\sin(\\angle BYT) = \\frac{opposite}{hypotenuse}$

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$\\sin(\\var{BYT}°) = \\frac{\\var{height}}{YT}$

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So $YT = \\var{YT}$ m

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We can also find $XT$ using the sine ratio:

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$sin(\\var{BXT}°) = \\frac{\\var{height}}{XT}$

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So $XT = \\var{XT}$ m

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The difference in lengths is $YT - XT = \\var{YT}-\\var{XT}=\\var{YT-XT}$ m

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To find length $XY$ we first need to find $XB$ and $YB$.

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We can find $XB$ using the tan ratio: $\\tan(\\angle BXT) = \\frac{opposite}{adjacent}$

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$\\tan(\\var{BXT}°) = \\frac{\\var{height}}{XB}$

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So $XB=\\var{BX}$ m

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We can find $XB$ using Pythagoras' Theorem: $a^2 + b^2=c^2$

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$\\var{XT}^2=\\var{height}^2+XB^2$

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So $XB=\\var{BX}$ m

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We can also find $YB$ using the tan ratio: $\\tan(\\angle BYT) = \\frac{opposite}{adjacent}$

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$\\tan(\\var{BYT}°) = \\frac{\\var{height}}{YB}$

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So $YB=\\var{BY}$ m

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Therefore, $XY = YB - XB = \\var{BY}-\\var{BX}=\\var{XY}$ m

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Your 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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If the {infolist[info]}, {questionlist[question]}?

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Round your answer to 1 decimal place.

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