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Question Number 115642 Answers: 0 Comments: 0

A vertical post of height h m rises from a plane which slopes down towards the South at an angle α to the horizontal. Prove that the length of its shadow when the sun is S𝛉W at an elevation β is ((h(√((1+tan^2 α cos^2 θ) )))/(tanβ + tanα cos θ))m

$${A}\:{vertical}\:{post}\:{of}\:{height}\:{h}\:{m}\:{rises}\:{from}\:{a}\:{plane}\:{which}\: \\ $$$${slopes}\:{down}\:{towards}\:{the}\:{South}\:{at}\:{an}\:{angle} \\ $$$$\alpha\:{to}\:{the}\:{horizontal}.\:{Prove}\:{that}\:{the}\:{length} \\ $$$${of}\:{its}\:{shadow}\:{when}\:{the}\:{sun}\:{is}\:\boldsymbol{{S}\theta{W}}\:\: \\ $$$${at}\:{an}\:{elevation}\:\beta\:{is} \\ $$$$ \\ $$$$\frac{{h}\sqrt{\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\:{cos}^{\mathrm{2}} \theta\right)\:}}{{tan}\beta\:+\:{tan}\alpha\:\mathrm{cos}\:\theta}{m} \\ $$

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if the sum of three consecutive num ber in a geometric progression(G.P) is 19 and their multiple is 216.find the number

$${if}\:{the}\:{sum}\:{of}\:{three}\:{consecutive}\:{num} \\ $$$${ber}\:{in}\:{a}\:{geometric}\:{progression}\left({G}.{P}\right) \\ $$$${is}\:\mathrm{19}\:{and}\:{their}\:{multiple}\:{is}\:\mathrm{216}.{find} \\ $$$${the}\:{number} \\ $$

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Let Ω denote the circumcircle of ABC. The tangent to Ω at A meets BC at X. Let the angle bisectors of ∠AXB meet AC and AB at E and F respectively. D is the foot of the angle bisector from ∠BAC on BC. Let AD intersect EF at K and Ω again at L(other than A). Prove that AEDF is a rhombus and further prove that the circle defined by triangle KLX passes through the midpoint of line segment BC.

$$\mathrm{Let}\:\Omega\:\mathrm{denote}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{of}\:\mathrm{ABC}. \\ $$$$\mathrm{The}\:\mathrm{tangent}\:\mathrm{to}\:\Omega\:\mathrm{at}\:\mathrm{A}\:\mathrm{meets}\:\mathrm{BC}\:\mathrm{at}\:\mathrm{X}. \\ $$$$\mathrm{Let}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{bisectors}\:\mathrm{of}\:\angle\mathrm{AXB}\:\mathrm{meet} \\ $$$$\mathrm{AC}\:\mathrm{and}\:\mathrm{AB}\:\mathrm{at}\:\mathrm{E}\:\mathrm{and}\:\mathrm{F} \\ $$$$\mathrm{respectively}.\:\mathrm{D}\:\mathrm{is}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angle} \\ $$$$\mathrm{bisector}\:\mathrm{from}\:\angle\mathrm{BAC}\:\mathrm{on}\:\mathrm{BC}.\:\mathrm{Let}\:\mathrm{AD} \\ $$$$\mathrm{intersect}\:\mathrm{EF}\:\mathrm{at}\:\mathrm{K}\:\mathrm{and}\:\Omega\:\mathrm{again}\:\mathrm{at} \\ $$$$\mathrm{L}\left(\mathrm{other}\:\mathrm{than}\:\mathrm{A}\right).\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{AEDF}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{rhombus}\:\mathrm{and}\:\mathrm{further}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{circle}\:\mathrm{defined}\:\mathrm{by}\:\mathrm{triangle}\:\mathrm{KLX}\:\mathrm{passes} \\ $$$$\mathrm{through}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{line}\:\mathrm{segment} \\ $$$$\mathrm{BC}. \\ $$

Question Number 112209 Answers: 0 Comments: 6

A triangle ABC has the following properties BC=1, AB=BC and that the angle bisector from vertex B is also a median. Find all possible triangle(s) with its/their side−lengths and angles.

$$\mathrm{A}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{properties}\:\mathrm{BC}=\mathrm{1},\:\mathrm{AB}=\mathrm{BC}\:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{bisector}\:\mathrm{from}\:\mathrm{vertex}\:\mathrm{B}\:\mathrm{is} \\ $$$$\mathrm{also}\:\mathrm{a}\:\mathrm{median}.\:\mathrm{Find}\:\mathrm{all}\:\mathrm{possible} \\ $$$$\mathrm{triangle}\left(\mathrm{s}\right)\:\mathrm{with}\:\mathrm{its}/\mathrm{their} \\ $$$$\mathrm{side}−\mathrm{lengths}\:\mathrm{and}\:\mathrm{angles}. \\ $$

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Question Number 112318 Answers: 2 Comments: 2

A triangle ABC has the following properties BC=1, AB=AC and that the angle bisector from vertex B is also a median. Find all possible triangle(s) with its/their side−lengths and angles.

$$\mathrm{A}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{properties}\:\mathrm{BC}=\mathrm{1},\:\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{AC}}\:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{bisector}\:\mathrm{from}\:\mathrm{vertex}\:\mathrm{B}\:\mathrm{is} \\ $$$$\mathrm{also}\:\mathrm{a}\:\mathrm{median}.\:\mathrm{Find}\:\mathrm{all}\:\mathrm{possible} \\ $$$$\mathrm{triangle}\left(\mathrm{s}\right)\:\mathrm{with}\:\mathrm{its}/\mathrm{their} \\ $$$$\mathrm{side}−\mathrm{lengths}\:\mathrm{and}\:\mathrm{angles}. \\ $$

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Using the cosine rule(c^2 =a^2 +b^2 −2abcosC), prove the triangle inequality: if a,b and c are sides of a triangle ABC, then a+b≥c and explain when equality holds. Further prove that sin α + sin β ≥ sin(α+β) for 0° ≤α,β≤180°

$$\mathrm{Using}\:\mathrm{the}\:\mathrm{cosine} \\ $$$$\mathrm{rule}\left(\mathrm{c}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{2abcosC}\right),\:\mathrm{prove}\:\mathrm{the} \\ $$$$\mathrm{triangle}\:\mathrm{inequality}:\:\mathrm{if}\:\mathrm{a},\mathrm{b}\:\mathrm{and}\:\mathrm{c}\:\mathrm{are} \\ $$$$\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{ABC},\:\mathrm{then}\:\mathrm{a}+\mathrm{b}\geqslant\mathrm{c} \\ $$$$\mathrm{and}\:\mathrm{explain}\:\mathrm{when}\:\mathrm{equality}\:\mathrm{holds}. \\ $$$$\mathrm{Further}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{sin}\:\alpha\:+\:\mathrm{sin}\:\beta\:\geqslant \\ $$$$\mathrm{sin}\left(\alpha+\beta\right)\:\mathrm{for}\:\mathrm{0}°\:\leqslant\alpha,\beta\leqslant\mathrm{180}° \\ $$

Question Number 112060 Answers: 1 Comments: 0

In a trapezium, ABCD, with AB parallel to CD. If M is the midpoint of line segment AD and P is a point on line BC such that MP is perpendicular to BC. Show that, we need only the lengths of line segments MP and BC to calculate the area ABCD.

$$\mathrm{In}\:\mathrm{a}\:\mathrm{trapezium},\:\mathrm{ABCD},\:\mathrm{with}\:\mathrm{AB} \\ $$$$\mathrm{parallel}\:\mathrm{to}\:\mathrm{CD}.\:\mathrm{If}\:\mathrm{M}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of} \\ $$$$\mathrm{line}\:\mathrm{segment}\:\mathrm{AD}\:\mathrm{and}\:\mathrm{P}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on} \\ $$$$\mathrm{line}\:\mathrm{BC}\:\mathrm{such}\:\mathrm{that}\:\mathrm{MP}\:\mathrm{is}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\mathrm{BC}.\:\mathrm{Show}\:\mathrm{that},\:\mathrm{we}\:\mathrm{need}\:\mathrm{only}\:\mathrm{the} \\ $$$$\mathrm{lengths}\:\mathrm{of}\:\mathrm{line}\:\mathrm{segments}\:\mathrm{MP}\:\mathrm{and}\:\mathrm{BC} \\ $$$$\mathrm{to}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{ABCD}. \\ $$

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If ∣z∣ = 3 , what is the maximum and minimum value of ∣z−1+i(√3) ∣ ?

$${If}\:\mid{z}\mid\:=\:\mathrm{3}\:,\:{what}\:{is}\:{the}\:{maximum} \\ $$$${and}\:{minimum}\:{value}\:{of}\:\mid{z}−\mathrm{1}+{i}\sqrt{\mathrm{3}}\:\mid\:? \\ $$

Question Number 111539 Answers: 1 Comments: 0

In a trapezoid ABCD, sides AB and CD are parallel and side BC=CD=(√5). If DC^ B =120° and BA^ D=60°. Find the area of ABCD.

$$\mathrm{In}\:\mathrm{a}\:\mathrm{trapezoid}\:\mathrm{ABCD},\:\mathrm{sides}\:\mathrm{AB}\:\mathrm{and} \\ $$$$\mathrm{CD}\:\mathrm{are}\:\mathrm{parallel}\:\mathrm{and}\:\mathrm{side}\:\mathrm{BC}=\mathrm{CD}=\sqrt{\mathrm{5}}. \\ $$$$\mathrm{If}\:\mathrm{D}\hat {\mathrm{C}B}\:=\mathrm{120}°\:\mathrm{and}\:\mathrm{B}\hat {\mathrm{A}D}=\mathrm{60}°.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{ABCD}. \\ $$

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In a quadrilateral ABCD, ∠B is a right angle, diagonal AC is perpendicular to CD,BC=21cm,CD=14cm and AD=31cm. Find the area of ABCD.

$$\mathrm{In}\:\mathrm{a}\:\mathrm{quadrilateral}\:\mathrm{ABCD},\:\angle\mathrm{B}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{right}\:\mathrm{angle},\:\mathrm{diagonal}\:\mathrm{AC}\:\mathrm{is} \\ $$$$\mathrm{perpendicular}\:\mathrm{to} \\ $$$$\mathrm{CD},\mathrm{BC}=\mathrm{21cm},\mathrm{CD}=\mathrm{14cm}\:\mathrm{and} \\ $$$$\mathrm{AD}=\mathrm{31cm}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{ABCD}. \\ $$

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Triangle ABC has AB=2∙AC. Let D and E be on AB and BC respectively such that ∠BAE =∠ACD. Let F be the intersections of segments AE and CD, and suppose that △CFE is equilateral. What is ∠ACB?

$$\mathrm{Triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{AB}=\mathrm{2}\centerdot\mathrm{AC}.\:\mathrm{Let} \\ $$$$\mathrm{D}\:\mathrm{and}\:\mathrm{E}\:\mathrm{be}\:\mathrm{on}\:\mathrm{AB}\:\mathrm{and}\:\mathrm{BC} \\ $$$$\mathrm{respectively}\:\mathrm{such}\:\mathrm{that}\:\angle\mathrm{BAE} \\ $$$$=\angle\mathrm{ACD}.\:\mathrm{Let}\:\mathrm{F}\:\mathrm{be}\:\mathrm{the}\:\mathrm{intersections}\:\mathrm{of} \\ $$$$\mathrm{segments}\:\mathrm{AE}\:\mathrm{and}\:\mathrm{CD},\:\mathrm{and}\:\mathrm{suppose} \\ $$$$\mathrm{that}\:\bigtriangleup\mathrm{CFE}\:\mathrm{is}\:\mathrm{equilateral}.\:\mathrm{What}\:\mathrm{is} \\ $$$$\angle\mathrm{ACB}? \\ $$

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