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The diagram shows a rhombus ABCD. The diagonal DB is produced to E such that BC=BE and C hat(DE)=46^(@). Calculate (i) B hat(AD), (ii) B hat(CE).

The diagram shows a rhombus ABCD A B C D . The diagonal DB D B is produced to E E such that BC=BE B C=B E and CDE^=46∘ C \hat{D E}=46^{\circ} . Calculate\newline(i) BAD^ B \hat{A D} ,\newline(ii) BCE^ B \hat{C E} .

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Q. The diagram shows a rhombus ABCD A B C D . The diagonal DB D B is produced to E E such that BC=BE B C=B E and CDE^=46∘ C \hat{D E}=46^{\circ} . Calculate\newline(i) BAD^ B \hat{A D} ,\newline(ii) BCE^ B \hat{C E} .
  1. Rhombus Properties: Since ABCDABCD is a rhombus, all sides are equal, so AD=BCAD = BC. Also, since BC=BEBC = BE, we have AD=BEAD = BE.
  2. Triangle BDE Angles: In triangle BDE, we have C^(DE)=46∘\hat{C}(DE) = 46^{\circ}, and since BE=BDBE = BD (as it's a rhombus), triangle BDE is isosceles with angles B^(DE)\hat{B}(DE) and B^(ED)\hat{B}(ED) being equal.
  3. Calculating Angle B^(DE)\hat{B}(DE): The sum of angles in a triangle is 180∘180^\circ, so we can calculate B^(DE)\hat{B}(DE) by subtracting C^(DE)\hat{C}(DE) from 180∘180^\circ and dividing by 22.\newlineB^(DE)=(180∘−46∘)2=134∘2=67∘\hat{B}(DE) = \frac{(180^\circ - 46^\circ)}{2} = \frac{134^\circ}{2} = 67^\circ.
  4. Exterior Angle of Triangle ABD: Since extBextextasciicircumextDE ext{B} ext{ extasciicircum} ext{DE} is an exterior angle to triangle ABD, and triangle ABD is also isosceles (AD=BDAD = BD), extBextextasciicircumextDE ext{B} ext{ extasciicircum} ext{DE} is equal to the sum of the two opposite interior angles, which are extBextextasciicircumextAD ext{B} ext{ extasciicircum} ext{AD} and extAextextasciicircumextBD ext{A} ext{ extasciicircum} ext{BD}.
  5. Final Angle Calculation: So, B^(AD)=B^(DE)−A^(BD)\hat{B}(AD) = \hat{B}(DE) - \hat{A}(BD). But since ABDABD is isosceles, A^(BD)=B^(AD)\hat{A}(BD) = \hat{B}(AD).
  6. Final Angle Calculation: So, B^(AD)=B^(DE)−A^(BD)\hat{B}(AD) = \hat{B}(DE) - \hat{A}(BD). But since ABDABD is isosceles, A^(BD)=B^(AD)\hat{A}(BD) = \hat{B}(AD).Therefore, B^(AD)=67∘2=33.5∘\hat{B}(AD) = \frac{67^{\circ}}{2} = 33.5^{\circ}. This is the measure of angle B^(AD)\hat{B}(AD).

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