MCQ
In Figure. $\angle\text{BAC}=90^{\circ},$ $\text{AD}\bot\text{BC}$ and$\angle\text{ BAD}=50^{\circ},$ then$\angle\text{ ACD}$ is:
- ✓$50^\circ$
- B$40^\circ$
- C$70^\circ$
- D$60^\circ$
Given, $\angle\text{BAC}=90^{\circ},$ $\text{AD}\bot\text{BC}$ and $\angle\text{ BAD}=50^{\circ}$
In $\triangle\text{ABD},$
$\angle\text{ABD}+\angle\text{DAB}+\angle\text{ADB}=180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow \ \angle\text{ABD}+50^{\circ}+90^{\circ}=180^{\circ}$
$\Rightarrow \ \angle\text{ABD}+140^{\circ}=180^{\circ}$
$\Rightarrow \ \angle\text{ABD}=180^{\circ}-140^{\circ}$
$\Rightarrow \ \angle\text{ABD}=40^{\circ}$
Now In$\triangle\text{ABD},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow \ 90^{\circ}+40^{\circ}+\angle\text{C}=180^{\circ}$
$\Rightarrow \ \angle\text{C}=180^{\circ}-130^{\circ}$
$\Rightarrow \ \angle\text{C}=50^{\circ}$
$\therefore\ \angle\text{ACD}=50^{\circ}$
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