Question

In the given figure ABCD is a rhombus and Δ DCE is an equilateral triangle. Also, ∠BAD=78°. Calculate ∠CBE, ∠ DBE and ∠ DEB

Answer 1

$∠BAD=180−78=102 {}^{0}$

$In \: triangle ABC,  \: ∠BAC=∠BCA \: = \: 51 {}^{0}$

[ABC is an isosceles triangle]

In triangle EDC, angle EDC ⬇️

$60+78=138 {}^{0}$

$So,  ∠DCE = ∠DEC \: = \: 21 {}^{0}$

$We \: know \: angle \: BCA \: is 102 {}^{0}$

Out of which angle BCA is 51 deg and angle DCE is 21 deg.

Hence the remaining angle, angle ACE is:

$102 \: − \: (51+21) \: = \: 30 {}^{0}$

$∠DCE \: = \: 21{}^{0}∠ACE \: = \: 30 {}^{0}$

Answer 2

Step-by-step explanation:

∠BAD=180−78=102

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In \: triangle ABC, \: ∠BAC=∠BCA \: = \: 51 {}^{0}IntriangleABC, ∠BAC=∠BCA=51

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[ABC is an isosceles triangle]

In triangle EDC, angle EDC ⬇️

60+78=138 {}^{0}60+78=138

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So, ∠DCE = ∠DEC \: = \: 21 {}^{0}So, ∠DCE=∠DEC=21

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We \: know \: angle \: BCA \: is 102 {}^{0}WeknowangleBCAis 102

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Out of which angle BCA is 51 deg and angle DCE is 21 deg.

Hence the remaining angle, angle ACE is:

102 \: − \: (51+21) \: = \: 30 {}^{0}102−(51+21)=30

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∠DCE \: = \: 21{}^{0}∠ACE \: = \: 30 {}^{0}∠DCE=21

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∠ACE=30

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