In the given figure- Delta ABD cong Delta ACD- If angle BDC - 110-circ- and angle DAC - 30-circ- then the measure of angle DBA is -70-circ-40-circ-30-circ-25-circ-

Question

Toppr Admin · Accepted Answer

-x394-ABD-x2245-x394-ACD -given-x2234-x2220-DAC-x2220-DAB-30-x2218- -By CPCT-Also- -x2220-ADC-x2220-ADB-x -By CPCT-Now- -x2220-ADC-x2220-ADB-x2220-BDC-360-x21D2-x-x-110-360 -xA0- -xA0-Angles at a point-x21D2-x-125 -xA0- -xA0-x21D2-x2220-ADC-x2220-ADB-125-x2218-

In the given figure, $Δ A B D ≅ Δ A C D$ . If $∠ B D C = 110^{\circ}$ and $∠ D A C = 30^{\circ}$ , then the measure of $∠ D B A$ is :

$70^{\circ}$
$40^{\circ}$
$30^{\circ}$
$25^{\circ}$

$Δ A B D ≅ Δ A C D$ (given)
$∴ ∠ D A C = ∠ D A B = 30^{\circ}$ (By CPCT)
Also, $∠ A D C = ∠ A D B = x$ (By CPCT)
Now, $∠ A D C + ∠ A D B + ∠ B D C = 360$
$\Rightarrow x + x + 110 = 360$ ...Angles at a point
$\Rightarrow x = 125$
$\Rightarrow ∠ A D C = ∠ A D B = 125^{\circ}$

In the given figure, ΔABD≅ΔACD. If ∠BDC=110∘ and ∠DAC=30∘, then the measure of ∠DBA is :70∘40∘30∘25∘

ΔABD≅ΔACD (given)∴∠DAC=∠DAB=30∘ (By CPCT)Also, ∠ADC=∠ADB=x (By CPCT)Now, ∠ADC+∠ADB+∠BDC=360⇒x+x+110=360 ...Angles at a point⇒x=125 ⇒∠ADC=∠ADB=125∘

In the given figure, $Δ A B D ≅ Δ A C D$ . If $∠ B D C = 110^{\circ}$ and $∠ D A C = 30^{\circ}$ , then the measure of $∠ D B A$ is :

$70^{\circ}$
$40^{\circ}$
$30^{\circ}$
$25^{\circ}$

$Δ A B D ≅ Δ A C D$ (given)
$∴ ∠ D A C = ∠ D A B = 30^{\circ}$ (By CPCT)
Also, $∠ A D C = ∠ A D B = x$ (By CPCT)
Now, $∠ A D C + ∠ A D B + ∠ B D C = 360$
$\Rightarrow x + x + 110 = 360$ ...Angles at a point
$\Rightarrow x = 125$
$\Rightarrow ∠ A D C = ∠ A D B = 125^{\circ}$