Question
$ABCD$ is a rhombus. If $\angle BAC = 38^\circ$ , find : $(i)\ \angle ACB\ (ii)\ \angle DAC\ (iii)\ \angle ADC.$
✓
Answer
$ABCD$ is Rhombus $($Given$)AB = BC$
$\angle BAC = \angle ACB (\angle s$ opp. to equal sides$)$
$But \angle BAC = 38^\circ ($Given$)$
$\angle ACB = 38^\circ$
In $\triangle ABC,$
$\angle ABC + \angle BAC + \angle ACB = 180^\circ$
$\angle ABC + 38^\circ + 38^\circ = 180^\circ$
$\angle ABC = 180^\circ – 76^\circ = 104^\circ$
But $\angle ABC = \angle ADC ($opp. $ \angle s$ of rhombus$)$
$\angle ADC = 104^\circ$
$\angle DAC = \angle DCA ( AD = CD)$
$\angle DAC = `1/2` [180^\circ - 104^\circ ]$
$\angle DAC = `1/2 x\times` 76^\circ = 38^\circ$
Hence $(i)\ \angle ACB = 38^\circ\ (ii)\ \angle DAC = 38^\circ\ (iii)\ \angle ADC = 104^\circ$ Ans.
$\angle BAC = \angle ACB (\angle s$ opp. to equal sides$)$
$But \angle BAC = 38^\circ ($Given$)$
$\angle ACB = 38^\circ$
In $\triangle ABC,$
$\angle ABC + \angle BAC + \angle ACB = 180^\circ$
$\angle ABC + 38^\circ + 38^\circ = 180^\circ$
$\angle ABC = 180^\circ – 76^\circ = 104^\circ$
But $\angle ABC = \angle ADC ($opp. $ \angle s$ of rhombus$)$
$\angle ADC = 104^\circ$
$\angle DAC = \angle DCA ( AD = CD)$
$\angle DAC = `1/2` [180^\circ - 104^\circ ]$
$\angle DAC = `1/2 x\times` 76^\circ = 38^\circ$
Hence $(i)\ \angle ACB = 38^\circ\ (ii)\ \angle DAC = 38^\circ\ (iii)\ \angle ADC = 104^\circ$ Ans.
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