$$ \\\\ tan{{60}^{0}}=\\dfrac{800}{x};tan{{30}^{0}}=\\dfrac{800}{y} \\\\ \\text{x }=\\dfrac{800}{\\sqrt{3}};y=800\\sqrt{3} \\\\ $$ Boat travelled in 30 sec $$=y-x$$ $$ \\\\ 800\\sqrt{3}-\\dfrac{800}{\\sqrt{3}}=923.7m$$ ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Angles of elevation and depression Quiz 1","text":" A flag pole is erected at the exact middle of a square shaped stadium. If the angle of elevation of the top of the pole from a corner of the stadium is $$30{}^\\text{o}$$, find the angle of elevation of the top of the pole from the midway of any side (boundary) of the stadium.","comment":{"@type":"Comment","text":" For both the triangles formed equate the tangent values to get the value of $\\theta $."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $${{\\tan }^{-1}}\\left( \\dfrac{1}{\\sqrt{2}} \\right)$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $${{45}^{0}}$$ ","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $${{\\tan }^{-1}}\\sqrt{2}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $${{\\tan }^{-1}}\\left( \\dfrac{\\sqrt{2}}{\\sqrt{3}} \\right)$$","position":0,"answerExplanation":{"@type":"Comment","text":" $$ \\\\ $$ Let $$BC=x$$, then $$BD=x\\sqrt{2} \\\\ \\text{In } \\Delta ADB,\\angle ADB={{30}^{0}} \\\\ \\tan {{30}^{0}}=\\dfrac{AB}{BD} \\\\ \\dfrac{AB}{x}=\\dfrac{\\sqrt{2}}{\\sqrt{3}} \\\\ \\text{Now, }\\tan \\theta =\\dfrac{AB}{x},\\theta ={{\\tan }^{-1}}\\left( \\dfrac{\\sqrt{2}}{\\sqrt{3}} \\right) $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Angles of elevation and depression Quiz 1","text":" Two persons started riding their bikes simultaneously from a common starting point A, and travelled in two different straight ways which subtend an angle of $$60{}^\\text{o}$$ at the starting point A. If the speeds of the first and the second persons are 60 kmph and 80 kmph respectively, what will be the shortest distance between their positions after 2 hours from their simultaneous start?","comment":{"@type":"Comment","text":" By the law of cosines, $B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}-2\\left( AB \\right)\\left( AC \\right)\\text{ c}osA$, we will get the value of BC."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$140km$$ ","position":0},{"@type":"Answer","encodingFormat":"text/html","text":"$$20\\sqrt{26}km$$ ","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$36\\sqrt{15}km$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$40\\sqrt{13}km$$","position":3,"answerExplanation":{"@type":"Comment","text":" Distance travelled by the first person in 2 hours $$=120km$$ $$ \\\\ $$ Distance travelled by the second person in 2 hours $$=160km$$ $$ \\\\ $$ Let the positions of the first and the second person after 2 hours are B and C respectively. $$ \\\\ $$
$$ \\\\ $$ As per the law of cosines; $$ \\\\ B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}-2\\left( AB \\right)\\left( AC \\right)\\text{ c}osA \\\\ \\text{i.e. } B{{C}^{2}}={{120}^{2}}+{{160}^{2}}-2\\times 120\\times 160\\times \\text{ c}os60{}^\\text{o} \\\\ B{{C}^{2}}=20800 \\\\ BC=40\\sqrt{13}km \\\\ $$ i.e. The shortest distance between their positions after 2 hours from their simultaneous start is $$40\\sqrt{13}km$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Angles of elevation and depression Quiz 1","text":" A structure stands on the bank of a river. A man looks out from a corner of the building's roof at the foot of an electric post on the opposite bank. What is the width of the river if the angle of depression of the foot of the light post at your eye is 60° and the height of the building is 14 metres?","comment":{"@type":"Comment","text":" Equate the tangent value with respect to angle perpendicular to its base."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 20.784 meters","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 16.84 meters","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" 10.982 meters","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 8.082meters","position":0,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ $$ From right-angled triangle ABC, $$\\dfrac{AB}{BC} = \\tan 60^\\circ$$ $$ \\\\ \\Rightarrow \\dfrac{14}{h}=\\sqrt{3} \\\\ \\Rightarrow 14=\\sqrt{3}h \\\\ \\Rightarrow h=\\dfrac{14\\sqrt{3}}{3} = 8.082mtrs\\\\ $$ Therefore, width of the river is 8.082meters","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Angles of elevation and depression Quiz 1","text":" From his college rooftop, Athul saw a bird flying at $$30{}^\\text{o}$$ angle of elevation near a well, Athul found that the reflection of the bird in the well water can be seen at $$60{}^\\text{o}$$ angle of depression. If the height of his college is 200 m, then how high was the bird flying from the ground?","comment":{"@type":"Comment","text":" From the trigonometric ratios of tangents for the two triangles ABD and ADC, we will get the relation of height, equating both we will get the required height."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 260 m","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 1600 m","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" 840 m","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 400 m","position":0,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ \\text{In }\\Delta ABD, BD=h-200 \\\\ tan30{}^\\text{o}=\\dfrac{BD}{AD} \\\\ \\dfrac{1}{\\sqrt{3}}=\\dfrac{h-200}{AD} \\\\ AD=\\sqrt{3}\\left( h-200 \\right) \\\\ \\text{In }\\Delta ADC,CD=h+200\\left( BE=CE \\right) \\\\ tan 60{}^\\text{o}=\\dfrac{CD}{AD} \\\\ \\sqrt{3}=\\dfrac{h+200}{AD} \\\\ i.e.\\sqrt{3}\\left( h-200 \\right)~~=\\dfrac{h+200}{\\sqrt{3}} \\\\ 3\\left( h-200 \\right)=\\left( h+200 \\right) \\\\ 3h-600=h+200 \\\\ h=400m $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}