Question : In $\triangle$ABC, $\angle$A = $90^{\circ}$, BP and CQ are two medians. Then the value of $\frac{BP^2 + CQ^2}{BC^2}$ is:
Option 1: $\frac{4}{5}$
Option 2: $\frac{5}{4}$
Option 3: $\frac{3}{4}$
Option 4: $\frac{3}{5}$
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Correct Answer: $\frac{5}{4}$
Solution : Since BP and CQ are two medians. So, Q and P are midpoints of AB and AC respectively. AQ = BQ = $\frac{1}{2}$AB AP = CP = $\frac{1}{2}$AC In $\triangle$AQC, $\angle$A = $90^{\circ}$ By Pythagoras theorem, ⇒ CQ2 = AC2 + QA2 ⇒ 4CQ2 = 4AC2 + 4QA2 ⇒ 4CQ2 = 4AC2 + (2QA)2 ⇒ 4CQ2 = 4AC2 + AB2------------(i) In $\triangle$ BPA, ⇒ BP2 = BA2 + AP2 ⇒ 4BP2 = 4BA2 + 4AP2 ⇒ 4BP2 = 4BA2 + (2AP)2 ⇒ 4BP2 = 4BA2 + AC2------------(ii) Adding (i) and (ii), ⇒ 4CQ2 + 4BP2 = 4AC2 + AB2 + 4BA2 + AC2 ⇒ 4(CQ2 + BP2) = 5AC2 + 5BA2 = 5(AC2 + AB2) = 5BC2 ⇒ $\frac{BP^2 + CQ^2}{BC^2}$ = $\frac{5}{4}$ Hence, the correct answer is $\frac{5}{4}$.
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Question : The side $BC$ of a triangle $ABC$ is extended to $D$. If $\angle ACD = 120^{\circ}$ and $\angle ABC = \frac{1}{2} \angle CAB$, then the value of $\angle ABC$ is:
Option 1: $80^{\circ}$
Option 2: $40^{\circ}$
Option 3: $60^{\circ}$
Option 4: $20^{\circ}$
Question : In a $\triangle$ ABC, BC is extended to D and $\angle$ ACD = $120^{\circ}$. $\angle$ B = $\frac{1}{2}\angle$ A. Then $\angle$ A is:
Option 1: $60^{\circ}$
Option 2: $75^{\circ}$
Option 3: $80^{\circ}$
Option 4: $90^{\circ}$
Question : In a $\triangle ABC$, if $2\angle A=3\angle B=6\angle C$, then the value of $\angle B$ is:
Option 2: $30^{\circ}$
Option 3: $45^{\circ}$
Question : $\triangle{ABC}$ is a right angled triangle. $\angle \mathrm{C}=90°$, AB = 25 cm and BC = 20 cm. What is the value of $\mathrm{sec}\; A$?
Option 1: $\frac{5}{3}$
Option 2: $\frac{4}{5}$
Option 3: $\frac{4}{3}$
Option 4: $\frac{5}{4}$
Question : In $\triangle \mathrm{ABC}, \angle \mathrm{A}=5 \mathrm{x}-60^{\circ}, \angle \mathrm{B}=2 \mathrm{x}+40^{\circ}, \angle \mathrm{C}=3 \mathrm{x}-80^{\circ}$. Find $\angle \mathrm{A}$.
Option 1: $75^{\circ}$
Option 2: $90^{\circ}$
Option 4: $60^{\circ}$
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