Question

3. Find the value(s) of the unknown(s) in each of the
following kites.
Kindly give answer for both parts, thanks in anticipation ​

Answer 1

Q. (a)

✧ Given :-

• $\tt{\angle}ADC = 100°$

✧ To find :-

• $\rm{\bold{The unknown angle i.e.,{\angle} a° = ?}}}$

✧ Solution :-

• we know that CAD = DCA

according to the question,

=> CAD + DCA + CDA = 180° (angle sum property)

=> a° + a° + 100° = 180°

=> 2a° = 80° (on subtracting 100 both side)

$\rm\purple{\underline{\bold{<strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong>=</strong><strong>></strong><strong> </strong><strong>a°</strong><strong> </strong><strong>=</strong><strong> </strong><strong>4</strong><strong>0</strong><strong>°</strong><strong> </strong>( on\:\: dividing \:both \:side \:by \:2)}}}$

Q. (b) :-

✧ Given :-

• DCA = 26° = DAC (opposite angles are equal)

✧To find :-

• The unknown angle i.e., c° = ??

✧ Solution :-

• $\tt{we \:know \:that\: {\angle}CAD = {\angle}DCA = 26°}$

$\tt\green{\underline{according \:to\: the \:question,}}$

=> 26°(2) + c° = 180°

=> 52° + c° = 180°

$\rm\orange{\underline{\bold{<strong>=</strong><strong>></strong><strong> </strong><strong>c°</strong><strong> </strong><strong>=</strong><strong> </strong><strong>1</strong><strong>2</strong><strong>8</strong><strong>°</strong> }}}$

Answer 2

Q.(a)

✧ Given :-

• $\bf{\angle}ADC = 100°$

✧ To find :-

• $\bf{the \: unknown \:{\angle}a°=?}$

✧ Solution :-

• $\bf{we \:know \:that \;{\angle}DAC= {\angle} ACD = a°}$

According to the question,

=> $\bf{\angle}{DAC + {\angle}{ACD + CDA = 100° }}$

=> 2a° + 100° = 180°

=> 2a° = 180° - 100°

=> a° = 40°

✯Now, in ∆ ABC :-

✧ Given :-

• $\bf{\angle}{ACB = 61°}$

✧ To find :-

• $\bf {The \:unknown {\angle}b° = ??}$

✧ Solution :-

• $\bf{\angle{As \: we \:know \: that \: {\angle}BAC = {\angle}BCA = 61°}}$

According to the question,

=> $\bf{\angle}BAC + {\angle}BCA + {\angle} ABC = 180°$

=> 2(61°) + b° = 180°

=> 122 + b° = 180°

=> b° = 180° - 122° => b° = 38°

Q.(b)

✧ Given :-

• $\bf{\angle}ACD = 26°$
• $\bf{\angle}BAC = 40°$

✧To find :-

• $\bf{the \: unknown \: angle \:i.e., c° = ?}$

✧Solution :-

• $\bf{as we know here {\angle}DAC= {\angle}DCA = 26° </li></ul><p><strong>Acc</strong><strong>ording</strong><strong> to</strong><strong> the</strong><strong> question</strong><strong>,</strong><strong> </strong></p><p><strong>=</strong><strong>></strong><strong> </strong>[tex]\bf{\angle}DAC + {\angle}DCA + c° = 180°$

  => 2(26)° + c° = 180°

  => 52° + c° = 180°

  => c° = 128°