Question

5. In the adjoining figure, ABCD is a rhombus in which angle BAC = 40°.
Find the measures of:(i) angle ACB (ii) angle ABC (iii) angle ADC (iv) angle ACD (v) angle CAD
IT'S VERY IMPORTANT NEED THE ANSWER URGENTLY!!!​

Answer 1

Answer:

∠ACB =40°(since AB=BC therefore, angle opposite to equal sides of a triangle are equal)

∠ABC =100°(∵sum of all the sides of a triangle is equal to 180°)

∠ADC =100°(vertically opposite angle)

∠DCA =40°(alternate interior angles)

∠DAC =40°(alternate interior angles)

Step-by-step explanation:

Answer 2

GivEn:-

• ABCD is a Rhombus

• ∠BAC = 40°

To find:-

1. ∠ACB
2. ∠ABC
3. ∠ADC
4. ∠ACD
5. ∠CAD

SoluTion:-

♡DIAGRAM:

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(8.6,3){\large{D}}\put(7.7,0.9){\large{A}}\put(11.1,0.9){\large{B}}\put(8,1){\line(1,0){3}}\qbezier(11,1)(11.5,2)(12,3)\put(9,3){\line(3,0){3}}\put(8,1){\line(2,1){4}}\qbezier(8,1)(8.5,2)(9,3)\put(12.1,3){\large{C}}\qbezier(8.6,1.3)(8.96,1.4)(8.9,1)\put(9,1.2){\sf 40^{\circ} }\end{picture}$

$\dag\;{\underline{\underline{\bf{\blue{According\;to\;question:-}}}}}$

━━━━━━━━━━━━━━━

i) ∠BAC = 40°

∴ AC = BC

∴ ∠ACB = ∠BAC = 40°

$\dashrightarrow$ ∠ACB = 40°

━━━━━━━━━━━━━━━

ii) Now, In ∆ABC :

$\dashrightarrow$ ∠ABC + ∠BAC + ∠BAC = 180°

$\dashrightarrow$ ∠ABC + 40° + 40° = 180°

$\dashrightarrow$ ∠ABC = 180° - 80°

$\dashrightarrow$ ∠ABC = 100°

━━━━━━━━━━━━━━━

iii) Now, ∠ADC = ∠ABC

[ ∴ In Rhombus opposite angles are equal]

$\dashrightarrow$ ∠ADC = 100°

━━━━━━━━━━━━━━━

iv) AD = AC

∠ACD = ∠CAD = $\sf \dfrac{ 180^\circ - 100^\circ}{2}$

$\dashrightarrow\sf \cancel{ \dfrac{80^\circ }{2}}$

$\dashrightarrow\sf 40^\circ$

$\dashrightarrow$ ∠ACD = 40°

━━━━━━━━━━━━━━━

v) ∠CAD = ∠ACB [ ∴alternate opposite angle]

$\dashrightarrow$ ∠CAD = 40°

━━━━━━━━━━━━━━━

$\dag\;{\underline{\underline{\bf{\purple{Hence\;SolvEd!!}}}}}$