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If angle abd is 110 and angle cbd is 60 then what is angle abc?
Wiki User
∙ 11y ago
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Wiki User
∙ 11y ago
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Is ABC and angle cbd a adjacent?
yes they are
In a given figure abc is an equilateral triangle and bcde is a square prove that angle Abe equals 15 degree?
Assuming the configuration is as follows. With the triangle inside the square D B A E C Angle CBD = 90 & Angle ABC = 60, therefore angle ABD = 30 (90-60) We also know that angle DBE = 45 (half square) So DBE-ABD=ABE which is 45 - 30 = 15 Its a bit confusing without proper diagrams but it works.
If ABCD is a rectangle and angle CBD equals 47 degrees what is the value of x?
The answer will depend on what x is!
If line segment ac is perpendicular to line segment cb and the measurement of angle cbd is 125 degrees then the measurement of angle a equals?
Here is the problem with this question. If AC is perpendicular to CB, then they must share a common endpoint, namely C. And we know that angle ACB must be a right angle. Now in order to make CBD 125 degrees, we need to draw it so it is greater than a right angle. So draw the line AC and at the point C, drop a perpendicular downwards and this is CB. Now D must be to the left and below B in order for the angle to be 125. Here is the catch.. We were never given any info about angle A, we only have point A. So one assumption would be to connect D and A and make segment AD. Then we have a quadralateral with two angle know, namely 125 and which is CBD and 90 which is ACB. The sum of the angles is 360 so the other two must be 360-(125+90)=145. This would be the sum of angles CAD and BDA. The length of segment BD would determine angle CAD and we do not know that So we don't have enough information. then the measurement of angle A equals:
The 5 other names for triangle BCD?
Triangles BDC, CBD, CDB, DBC and DCB.
Is ABC and angle cbd a adjacent?
yes they are
In a given figure abc is an equilateral triangle and bcde is a square prove that angle Abe equals 15 degree?
Assuming the configuration is as follows. With the triangle inside the square D B A E C Angle CBD = 90 & Angle ABC = 60, therefore angle ABD = 30 (90-60) We also know that angle DBE = 45 (half square) So DBE-ABD=ABE which is 45 - 30 = 15 Its a bit confusing without proper diagrams but it works.
In line segment Ac is perpendicular to line segment CB and the measurement of angle CBD equals 125 degrees the the measurement of angle A equals?
The problem is meaningless without a diagram but I am guessing that ABC make a triangle and D is on the extension of AB beyond B. In that case we use the exterior angle theorem to get CBD = C + A, so 125 = 90 + A and A = 35.
How do you prove that the diagonals in a rhombus divide the rhombus into four congruent triangles?
The proof is fairly long but relatively straightforward. You may find it easier to follow if you have a diagram: unfortunately, the support for graphics on this browser are hopelessly inadequate.Suppose you have a rhombus ABCD so that AB = BC = CD = DA. Also AB DC and AD BC.Suppose the diagonals of the rhombus meet at P.Now AB DC and BD is an intercept. Then angle ABD = angle BDC.Also, in triangle ABD, AB = AD. therefore angle ABD = angle ADC.while in triangle BCD, BC = CD so that angle DBC = angle BDC.Similarly, it can be shown that angle BAC = angle CAD = angle DCA = angle ACB.Now consider triangles ABP and CBP. angle ABP (ABD) = angle CBP ( CBD or DBC),sides AB = BCand angle BAP (BAC) = angle BCP (BCA = ACB).Therefore, by SAS, the two triangles are congruent.In the same way, triangles BCP and CPD can be shown to congruent as can triangles CPD and DPA. That is, all four triangles are congruent.
In an isosceles triangle does the median to the base bisect the vertex angle?
In the diagram, ABC is an isoscels triangle with the congruent sides and , and is the median drawn to the base . We know that ∠A ≅ ∠C, because the base angles of an isosceles triangle are congruent; we also know that ≅ , by definition of an isosceles triangle. A median of a triangle is a line segment drawn from a vertex to the midpoint of the opposite side. That means ≅ . This proves that ΔABD ≅ ΔCBD. Since corresponding parts of congruent triangles are congruent, that means ∠ABD≅ ∠CBD. Since the median is the common side of these adjacent angles, in fact bisects the vertex angle of the isosceles triangle.
If line segment ac is perpendicular to line segment cb and the measurment of angle cbd125 degrees then the mearsurment of a equals?
The problem is meaningless without a diagram but I am guessing that ABC make a triangle and D is on the extension of AB beyond B. In that case we use the exterior angle theorem to get CBD = C + A, so 125 = 90 + A and A = 35.
If ABCD is a rectangle and angle CBD equals 47 degrees what is the value of x?
The answer will depend on what x is!
How do you prove the diagonals of a rhombus divide the rhombus into four congruent triangles?
Take a rhombus ABCD. A rhombus as 4 equal sides, thus AB = BC = CD = DA Draw in 1 diagonal AC. This splits the rhombus into 2 triangles. ABC and CDA with side AB = CD, BC = DA and AC common to both triangles. Thus ABC and CDA are congruent by Side-Side-Side. Triangles ABC and CDA are isosceles triangles since they have two equal sides (AB = BC and CD = DA) thus angles DAC = DCA = BAC = BCA. Specifically DAC = BAC. But DAC + BAC = DAB, thus DAC = BAC = ½ DAB; similarly DCA = BCA = ½ BCD = ½ DAB Drawing in the other diagonal BD, the same arguments show triangles ABD and CDB are congruent and angles ADB = CDB = ABD = CBD = ½ ABC Let the point where the diagonals meet be E. We now have 4 triangles ABE, BCE, CDE and DAE with equivalent angles and sides: Angles DAE* = BAE = BCE = DCE (= ½ DAB) Angles ABE = CBE = CDE = ADE (= ½ ABD) Sides AB = BC = CD = DA Thus the 4 triangles are congruent by Angle-Angle-Side. *Angle DAE = DAC since E lines along AC; similarly for all the other angles involving point E, ie angle BCE = BCA, ADB = ADE, etc
Abcd is a convex quadrilateral in which angle bac equals 50 angle cad equals 60 angle cbd equals 30 if e is intersection point of ac and bd find angle aeb?
95 degrees (hint: construct the circumcircle of BCD
How far from Melbourne cbd is Frankston?
From Melbourne CBD to Frankston is 53km.
When was CBD-FM created?
CBD-FM was created on 1964-10-15.
How long does it take to travel from Melbourne to geelong?
About 1 hour from cbd to cbd.
