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In the triangle ABC, AD is perpendicular to BC. Angle CAD=y and angle DBA=x. AC=17 and BD=25 units Find the value of 5cosx -17siny+tanx
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well, 25tanx = 17cosy, so see what you can do with that.
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In triangle ABC angle B > angle C if AM is the bisector of angle BAC and AN perpendicular BC. prove that angle
MAN =1/2(angle B -
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To prove that angle MAN is equal to half of the difference between angle B and angle C, we will use
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AB = 8 cm, AC = 6 cm, AD = 7 cm, CD = 2.82 and CAB=50°. Find the length BC, angle ABC, angle CAD , area of triangle ACD
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Start off with the cosine law to find BC BC^2= 8^2 + 6^2 - 2(8)(6)cos50° .. .. BC = 6.188 Now use
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AB = 8 cm, AC = 6 cm, AD = 7 cm, CD = 2.82 and CAB =50°. Find the length BC, angle ABC, angle CAD , area of triangle ACD.
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Can't do it. If you draw A,B,C then D can be anywhere on a circle of radius 2.82 with center at C.
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In a triangle ABC, angle B> angle C. If AM is the bisector of angle A and An is perpendicular to BC. Prove that angle
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angle ACB=90° angle ABC=20° and angle CAD =angle DAB what is the measure of angle ADC
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no idea. Where is D?
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In a triangle ABC,right angled at B, BD is drawn perpendicular to AC. prove that:
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(i) ABD is the complement of A. C is also the complement of A Therefore ABD = C (ii) CBD is the
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In a triangle ABC,right angled at B, BD is drawn perpendicular to AC. prove that:
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To prove the given statements, we will use the properties of right-angled triangles and the
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AB = 8 cm, AC = 6 cm, AD = 7 cm, CD = 2.82 and CAB=50°.
Calculate to 2 decimal places (a) the length BC (b) The size of angle
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To solve this problem, we can use the Law of Cosines and the Law of Sines. Let's go step by step to
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AB = 8 cm, AC = 6 cm, AD = 7 cm, CD = 2.82 and CAB=50°.
(a) Find the length BC, (BC)^2 = (AB)^2 +(AC)^2 -2(AB)(AC)cosA (BC)^2 =
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sinB/6 = sin50/6.188 2.82^2 = g^2+7^2-2*6*7 cos(CAD) area of ACD = 6*(7*sin(CAD))/2
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AB = 8 cm, AC = 6 cm, AD = 7 cm, CD = 2.82 and CAB=50° (a) the length BC (b) The size of angle ABC; (c) The size of angle CAD
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To find the answers to these questions, we can use various properties and formulas related to
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