Download App

MATHEMATICS [2 Mark Question Answer]

Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle] · ICSE STD 9 (ICSE - English Medium). 11 questions.

Questions

[2 Mark Question Answer]

🎯

Test yourself on this topic

11 questions · timed · auto-graded

Question 12 Marks
Find angle $'A\ '$ if:
Answer
From the figure we have,
$ \tan A =\frac{\text { Perpendicular }}{\text { Base }}$
$\tan A =\frac{10 \sqrt{3}}{10}$
$\tan A =\sqrt{3}$
$\tan A =\tan 60^{\circ}$
$\therefore A =60^{\circ}$
View full question & answer→
Question 22 Marks
Find angle$ 'A\ '$ if :
Answer
From the figure we have
$\sin A=\frac{\frac{10}{\sqrt{2}}}{10} $
$ \sin A=\frac{1}{\sqrt{2}} $
$\sin A=\sin 45^{\circ} $
$A=45^{\circ}$
View full question & answer→
Question 32 Marks
Find angle $ 'A\ '$ if :
Answer
From the figure we have
$\cos A=\frac{10}{20}$
$ \cos A=\frac{1}{2}$
$ \cos A=\cos 60^{\circ} $
$ A=60^{\circ}$
View full question & answer→
Question 42 Marks
Find $'x\ ',$ if :
Answer
From the figure we have
$\sin 45^{\circ}=\frac{20}{x}$
$\frac{1}{\sqrt{2}}=\frac{20}{x}$
$x=20 \sqrt{2}$
View full question & answer→
Question 52 Marks
Find$: \text{AC}$
Answer
From right triangle $\text{ABC}$
$\sin B=\frac{AB}{A C} $
$ \sin 30^{\circ}=\frac{A B}{A C} $
$ \frac{1}{2}=\frac{12}{A C} $
$\text{AC}=24\ cm$
View full question & answer→
Question 62 Marks
Find$:\text{AD}$
Answer
From the right triangle $\text{ABD}$
$\cos A =\frac{Ac }{ AB } $
$\cos 60^{\circ}=\frac{ AD }{ AB }$
$ \frac{1}{2}=\frac{ AD }{12} $
$\text{AD} =\frac{12}{2} $
$=6\ cm$
View full question & answer→
Question 72 Marks
Find$: \text{BC}$
Answer
From right triangle $\text{ABC}$
$\tan 30^{\circ}=\frac{ AB }{ BC } $
$ \frac{1}{\sqrt{3}}=\frac{12}{ BC } $
$ \text{BC} =12 \sqrt{3}\ cm .$
View full question & answer→
Question 82 Marks
Find $'x\ ',$ if :
Answer
From the figure we have
$\tan 30^{\circ}=\frac{20}{x} $
$ \frac{1}{\sqrt{3}}=\frac{20}{x} $
$ x=20 \sqrt{3}$
View full question & answer→
Question 92 Marks
Find $ 'x\ ',$ if :
Answer
From the figure we have
$\sin 60^{\circ}=\frac{20}{x} $
$\frac{\sqrt{3}}{2}=\frac{20}{x} $
$ x=\frac{40}{\sqrt{3}}$
View full question & answer→
Question 102 Marks
In the given figure, $\text{AB}$ and $\text{EC}$ are parallel to each other. Sides $\text{AD}$ and $\text{BC}$ are $2 \ cm$ each and are perpendicular to $\text{AB}.$

Given that $\angle AED = 60^\circ$ and $\angle ACD = 45^\circ.$ Calculate $:\text{AE}.$
Answer
From the right triangle $\text{ADE}$ we have
$\sin 60^{\circ}=\frac{ AD }{ AE }$
$\frac{\sqrt{3}}{2}=\frac{2}{ AE }$
$\text{AE} =\frac{4}{\sqrt{3}} .$
View full question & answer→
Question 112 Marks
In the given figure, $\text{AB}$ and $\text{EC}$ are parallel to each other. Sides $\text{AD}$ and $\text{BC}$ are $2\  cm$ each and are perpendicular to $\text{AB}.$

Given that $\angle AED = 60^\circ$ and $\angle ACD = 45^\circ.$ Calculate $: \text{AC}$
Answer
Again
$\sin 45^{\circ}=\frac{ AD }{ AC }$
$\frac{1}{\sqrt{2}}=\frac{2}{ AC }$
$\text{AC} =2 \sqrt{2}.$
View full question & answer→
Generate Paper FreeAll Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle] topics
[2 Mark Question Answer] - MATHEMATICS STD 9 Questions - Vidyadip