Question : An isosceles triangle ABC is right-angled at B. D is a point inside the triangle ABC. P and Q are the feet of the perpendicular drawn from D on the sides AB and AC respectively of ABC. If $AP = a$ cm , $AQ = b$ cm and $\angle$BAD = 15°, $\sin$ 75°= ?
Option 1: $\frac{2b}{\sqrt 3 a }$
Option 2: $\frac{a}{2b}$
Option 3: $\frac{\sqrt 3a}{2b}$
Option 4: $\frac{2a}{\sqrt 3b}$
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Correct Answer: $\frac{\sqrt 3a}{2b}$
Solution : Since $\triangle$ABC is a right-angled isosceles triangle, $\angle$A = $\frac{180 - 90}{2}$ = 45° $\angle$DAQ = $\angle$A–$\angle$BAD = 45° – 15° = 30° $\sin$ $\angle$ADQ = $\sin$ 60° = $\frac{AQ}{AD}$ $\frac{b}{AD}$ = $\frac{\sqrt3}{2}$ AD = $\frac{2b}{\sqrt3}$ From $\triangle$APD, $\sin$ 75° = $\frac{AP}{AD}$ = $\frac{a}{\frac{2b}{\sqrt3}}$ = $\frac{\sqrt3a}{2b}$ Hence, the correct answer is $\frac{\sqrt3a}{2b}$.
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Question : In a triangle ABC, AB = 6$\sqrt{3}$ cm, AC = 12 cm and BC = 6 cm. Then the measure of $\angle B$ is equal to:
Option 1: 90°
Option 2: 45°
Option 3: 70°
Option 4: 60°
Question : Suppose $\triangle ABC$ be a right-angled triangle where $\angle A=90°$ and $AD\perp BC$. If the area of $\triangle ABC =40$ cm$^{2}$ and $\triangle ACD =10$ cm$^{2}$ and $\overline{AC}=9$ cm, then the length of $BC$ is:
Option 1: 12 cm
Option 2: 18 cm
Option 3: 4 cm
Option 4: 6 cm
Question : $\triangle ABC$ is an equilateral triangle with a side of 12 cm and AD is the median. Find the length of GD if G is the centroid of $\triangle ABC$.
Option 1: $6 \sqrt{3}$ cm
Option 2: $3 \sqrt{3} $ cm
Option 3: $4 \sqrt{3} $ cm
Option 4: $2 \sqrt{3}$ cm
Question : In $\triangle{XYZ}$, right-angled at $Y$, if $\sin X = \frac{1}{2}$, find the value of $\cos X \cos Z + \sin X \sin Z$.
Option 1: $\frac{\sqrt{3}}{2}$
Option 2: $\frac{\sqrt{3}}{4}$
Option 3: $\frac{2}{\sqrt{3}}$
Option 4: $\sqrt{3}$
Question : In triangle ABC, $\angle$ B = 90°, and $\angle$C = 45°. If AC = $2 \sqrt{2}$ cm then the length of BC is:
Option 1: 3 cm
Option 2: 2 cm
Option 3: 1 cm
Option 4: 4 cm
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