Question : If $\triangle \mathrm{BPQ} \cong \triangle \mathrm{ASR}$ and $\angle \mathrm{A}=\frac{1}{3} \angle \mathrm{R}=\angle \mathrm{S}$, then find $\angle \mathrm{Q}$. (All angles are in degrees).
Option 1: 108°
Option 2: 36°
Option 3: 72°
Option 4: 118°
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 108°
Solution : Given, $\angle \mathrm{A}=\frac{1}{3} \angle \mathrm{R}=\angle \mathrm{S}$ ⇒ $3\angle \mathrm{A}=\angle \mathrm{R}$ and $\angle \mathrm{A}=\angle \mathrm{S}$ In $\triangle \mathrm{ASR},$ $\angle A + \angle S +\angle R = 180°$ ⇒ $\angle \mathrm{A} + \angle \mathrm{A}+3\angle \mathrm{A}=180°$ ⇒ $5\angle \mathrm{A} = 180°$ ⇒ $\angle \mathrm{A} = \frac{180°}{5}=36°$ Now, $\triangle \mathrm{BPQ} \cong \triangle \mathrm{ASR}$ ⇒ $\angle \mathrm{Q}=\angle \mathrm{R}$ ⇒ $\angle \mathrm{Q}=3\angle \mathrm{A}$ ⇒ $\angle \mathrm{Q}=3\times 36° = 108°$ Hence, the correct answer is 108°.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If it is given that for two right-angled triangles $\triangle$ABC and $\triangle$DFE, $\angle$A = 25°, $\angle$E = 25°, $\angle$B = $\angle$F = 90°, and AC = ED, then which one of the following is TRUE?
Question : If $\triangle \mathrm{ABC} \cong \triangle \mathrm{PQR}, \mathrm{BC}=6 \mathrm{~cm}$, and $\angle \mathrm{A}=75^{\circ}$, then which one of the following is true?
Question : If $p+q+r=pqr=\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1$, find $\mathrm{p}^3+\mathrm{q}^3+\mathrm{r}^3$.
Question : In a $\triangle \mathrm{XYZ}, \mathrm{XO}$ is the median and $\mathrm{XO}=\frac{1}{2} \mathrm{YZ}$. If $\angle \mathrm{YXO}=30^{\circ}$, then what is the value of $\angle \mathrm{XYZ}$?
Question : If $\mathrm{p}=\frac{\sqrt{2}+1}{\sqrt{2}-1}$ and $\mathrm{q}=\frac{\sqrt{2}-1}{\sqrt{2}+1}$ then, find the value of $\frac{\mathrm{p}^2}{\mathrm{q}}+\frac{\mathrm{q}^2}{\mathrm{p}}$.
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile