Question : $ABC$ is a right-angled triangle, right-angled at B and $\angle A=60°$ and $AB=20$ cm, then the ratio of sides $BC$ and $CA$ is:
Option 1: $\sqrt{3}:1$
Option 2:
$1:\sqrt{3}$
Option 3:
$\sqrt{3}:\sqrt{2}$
Option 4:
$\sqrt{3}:2$
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Correct Answer:
Solution : Given: $ABC$ is a right-angled triangle, right-angled at $B$. $\angle A=60°$ and $AB=20$ cm So, $\angle C=180°-90°-60°= 30°$ ⇒ $\frac{BC}{CA}=\cos C$ ⇒ $\frac{BC}{CA}=\cos 30°$ ⇒ $\frac{BC}{CA}=\frac{\sqrt{3}}{2}$ ⇒ $BC:CA=\sqrt{3}:2$ Hence, the correct answer is $\sqrt{3}:2$.
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Question : The sides of similar triangle $\triangle ABC$ and $\triangle DEF$ are in the ratio of $\frac{\sqrt{3}}{\sqrt{5}}$. If the area of $\triangle ABC$ is $90 \text{ cm}^2$, then the area of $\triangle DFF\left(\right.$ in $\left.\text{cm}^2\right)$ is:
Option 1: 150
Option 2: 152
Option 3: 154
Option 4: 156
Question : ABC is an isosceles right-angled triangle with $\angle$B = 90°. On the sides AC and AB, two equilateral triangles ACD and ABE have been constructed. The ratio of the area of $\triangle$ABE and $\triangle$ACD is:
Option 1: $1 : 3$
Option 2: $2 : 3$
Option 3: $1 : 2$
Option 4: $1 : \sqrt{2}$
Question : $\triangle\mathrm{ABC}$ is a right angled triangle. $\angle \mathrm{A}=90°$, $AB = 4$ cm, and $BC = 5$ cm. What is the value of $\cos B + \cot C$?
Option 1: $\frac{17}{20}$
Option 2: $\frac{29}{20}$
Option 3: $\frac{23}{20}$
Option 4: $\frac{31}{20}$
Question : If $\sin(A+B)=\sin A\cos B+\cos A \sin B$, then the value of $\sin75°$ is:
Option 1: $\frac{\sqrt{3}+1}{\sqrt{2}}$
$\frac{\sqrt{2}+1}{2\sqrt{2}}$
$\frac{\sqrt{3}+1}{2\sqrt{2}}$
$\frac{\sqrt{3}+1}{2}$
Question : Let $ABC$ and $PQR$ be two congruent right-angled triangles such that $\angle A=\angle P=90^{\circ}$. If $BC=13\ \text{cm}$ and $PR=12\ \text{cm}$, then find the length of $AB$.
Option 1: 25 cm
Option 2: 20 cm
Option 3: 10 cm
Option 4: 5 cm
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