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:
$\sqrt{3}:2$
New: SSC CHSL Tier 2 answer key released | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
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 $\sqrt{3}:2$.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
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:
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:
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 $\cos B + \cot C$?
Question : If $\sin(A+B)=\sin A\cos B+\cos A \sin B$, then the value of $\sin75°$ is:
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 $AB$.
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile