Question : In $\triangle$ABC, $\angle$C = 90° and CD is perpendicular to AB at D. If $\frac{\text{AD}}{\text{BD}}=\sqrt{k}$, then $\frac{\text{AC}}{\text{BC}}$=?
Option 1: $\sqrt{k}$
Option 2: $\frac{1}{\sqrt{k}}$
Option 3: $\sqrt[4]{k}$
Option 4: $k$
Correct Answer: $\sqrt[4]{k}$
Solution : In $\triangle$ABC is a right-angled triangle. The CD is perpendicular to AB ⇒ $\frac{AD}{BD}$ = $(\frac{AC}{BC})^{2}$ Also, ⇒ $\sqrt{k}$ = $(\frac{AC}{BC})^2$ ⇒ $\frac{AC}{BC}$ = $\sqrt{k}^{\frac{1}{2}}$ = $\sqrt[4]{k}$ Hence, the correct answer is $\sqrt[4]{k}$.
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Question : In a triangle$\frac{AB}{AC}=\frac{BD}{DC}$, $\angle$B = 70° and $\angle$C = 50°, then $\angle$BAD =?
Option 1: 60°
Option 2: 20°
Option 3: 30°
Option 4: 50°
Question : In a triangle, ABC, BC is produced to D so that CD = AC. If $\angle BAD=111^{\circ}$ and $\angle ACB=80^{\circ}$, then the measure of $\angle ABC$ is:
Option 1: 31°
Option 2: 33°
Option 3: 35°
Option 4: 29°
Question : In $\triangle ABC, \angle B=90^{\circ}$ and AB : BC = 1 : 2. The value of $\cos A+\tan C$ is:
Option 1: $\frac{5+\sqrt{5}}{2 \sqrt{5}}$
Option 2: $\frac{1+\sqrt{5}}{2 \sqrt{5}}$
Option 3: $\frac{2 \sqrt{5}}{2+\sqrt{5}}$
Option 4: $\frac{2+\sqrt{5}}{2 \sqrt{5}}$
Question : In $\triangle$ ABC, $\angle$ C = 90$^{\circ}$. M and N are the midpoints of sides AB and AC, respectively. CM and BN intersect each other at D and $\angle$ BDC = 90$^{\circ}$. If BC = 8 cm, then the length of BN is:
Option 1: $6 \sqrt{3} {~cm}$
Option 2: $6 \sqrt{6} {~cm}$
Option 3: $4 \sqrt{6} {~cm}$
Option 4: $8 \sqrt{3} {~cm}$
Question : In $\triangle A B C $, AB and AC are produced to points D and E, respectively. If the bisectors of $\angle C B D$ and $\angle B C E$ meet at the point O, and $\angle B O C=57^{\circ}$, then $\angle A$ is equal to:
Option 1: 93°
Option 2: 66°
Option 3: 114°
Option 4: 57°
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