Question : In triangle ABC, $\angle$ABC = 15°. D is a point on BC such that AD = BD. What is the measure of $\angle$ADC (in degrees)?
Option 1: 15
Option 2: 30
Option 3: 45
Option 4: 60
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Correct Answer: 30
Solution : Given: In triangle ABC, $\angle$ABC = 15°. D is a point on BC such that AD = BD. Given that AD = BD. Then, $\angle$ABD = $\angle$BAD = 15° Now, in $\triangle$ABD ⇒ $\angle$ABD + $\angle$BAD + $\angle$ADB = 180° ⇒ $\angle$ADB = 180° – 15° – 15° ⇒ $\angle$ADB = 150° We know that, $\angle$ADB + $\angle$ADC = 180° ⇒ $\angle$ADC = 180° – 150° ⇒ $\angle$ADC = 30° Hence, the correct answer is 30.
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Question : In $\triangle $ABC, AD$\perp$ BC and AD2 = BD × DC. The measure of $\angle$ BAC is:
Option 1: 60°
Option 2: 75°
Option 3: 90°
Option 4: 45°
Question : In $\triangle ABC$, $\angle B=60°$, $\angle C=40°$. AD is the bisector of $\angle A$ and AE is drawn perpendicular on BC from A. Then the measure of $\angle EAD$ is:
Option 1: $40^{\circ}$
Option 2: $30^{\circ}$
Option 3: $10^{\circ}$
Option 4: $80^{\circ}$
Question : $ABC$ is an isosceles triangle with $AB = AC$, The side $BA$ is produced to $D$ such that $AB = AD$. If $\angle ABC = 30^{\circ}$, then $\angle BCD$ is equal to:
Option 1: $45^{\circ}$
Option 2: $90^{\circ}$
Option 3: $30^{\circ}$
Option 4: $60^{\circ}$
Question : In $\triangle$ABC, $\angle$A = $\angle$B = 60°, AC = $\sqrt{13}$ cm, the lines AD and BD intersect at D with $\angle$D = 90°. If DB = 2 cm, then the length of AD is:
Option 1: 3 cm
Option 2: 3.5 cm
Option 3: 4 cm
Option 4: 4.7 cm
Question : $\angle A$ of $\triangle ABC$ is a right angle. $AD$ is perpendicular on $BC$. If $BC= 14$ cm and $BD= 5$ cm, then measure of $AD$ is:
Option 1: $2\sqrt5$ cm
Option 2: $\sqrt5$ cm
Option 3: $3\sqrt5$ cm
Option 4: $3.5\sqrt5$ cm
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