Class 8: Triangles – Exercise 30A – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Two angles of a triangle measure $58^{\circ} \ and\ 87^{\circ}$ . Find the measure of the third angle.

Answer:

Sum of all the angles of a triangle $= 180^{\circ}$

Let the third angle $= x$

Therefore $58 + 87 + x = 180^o$

Or $x= 35^{\circ}$

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Question 2: Three angles of a triangle are in the ratio of $2 : 5 : 8$ . Find the angles of the triangle.

Answer:

Let the angles be $2x, 5x, \ and\ 8x$

Therefore $2x+5x+8x = 180^o$

Or $x= 12^o$

Therefore the angles are $24^{\circ}, 60^{\circ}, \ and\ 96^{\circ}.$

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Question 3: One angle of the triangle is $76^o$ and the other two angles are equal. Find the measure of each angle of the triangle.

Answer:

Let the two equal angles be $x^{\circ}$

Therefore $76 + 2x = 180^o$

Or $x = 52^o$

Therefore the angles are $76^{\circ}, 52^{\circ} \ and\ 52^{\circ}.$

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Question 4: One angle of the triangle is $85^{\circ}$ and the other two angles are in the ratio $8 :11$ . Find the measures of each of the unknown angles.

Answer:

Let the two unknown angles be $8x \ and\ 11 x$

Therefore $85+8x+11x = 180^o$

Or $19x = 95$

Or $x = 5^o$

Therefore the two angles are $40^{\circ} \ and\ 55^{\circ}.$

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Question 5: Three angles of a triangle measures $(2x)^{\circ}, (4x-7)^{\circ} \ and\ (5x-11)^{\circ}$ . Find the value of $x$ and the measures of all the angles.

Answer:

$2x + (4x-7) + (5x-11) = 180^o$

Or $x = 18^o$

Hence the angles are $36^{\circ}, 65^{\circ}, \ and\ 79^{\circ}$

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Question 6: In $\Delta ABC, \angle A=x^{\circ}, \angle B=(2x-15)^{\circ}, \ and\ \angle C=(3x+21)^{\circ}$ . Find the value of $x$ and the measure of all the angles of the triangle.

Answer:

$x+(2x-15)+(3x+21)=180^o$

or $x=29^o$

Hence $\angle A=29^{\circ}, \angle B=43^{\circ}, \ and\ \angle C=108^{\circ}$

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Question 7: In $\Delta ABC, \angle A=(2x+15)^{\circ}, \angle B=x^{\circ}, \ and\ \angle C=(3x-15)^{\circ}$ . Find the value of x and the measure of all the angles of the triangle and show that the triangle is an isosceles triangle.

Answer:

$(2x+15)+x+(3x-15)=180^o \ or\ x=30^o$

Hence $\angle A=75^{\circ}, \angle B=30^{\circ}, \ and\ \angle C=75^{\circ}.$

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Question 8: In $\Delta ABC, \angle A=3\ \angle B=6 \ \angle C$ . Find the angles of the triangle.

Answer:

Let $\angle A=3\angle B=6 \angle C=6x$

$\Rightarrow \angle A=6x, \angle B=2x, and \angle C=x$

$\Rightarrow 6x+2x+x=180^o \ or\ x=20^o$

$\Rightarrow \angle A=120^{\circ}, \angle B=40^{\circ} \ and\ \angle C=20^{\circ}$

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Question 9: In the adjoining figure, it is given that $\angle ABC=40^{\circ}, \angle ACD=70^{\circ}, \angle ACB=x^{\circ} \ and\ \angle EAC=y$ . Find the value of $latex $x and y$ .

Answer:

$x+70=180^o \ hence\ x=110^{\circ}$

Note: Sum of angles of adjacent angles on a straight line $=180^o$

$\Rightarrow 40+110+(180-y)=180^o$

Note:Sum of all angles of a triangle $=180^o$

$\Rightarrow y=150^{\circ}$

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Question 10: In the adjoining figure, it is given that $\angle A=50^{\circ} \ and\ \angle B=70^{\circ}$ and $BO \ and\ CO$ are bisectors of $\angle B \ and\ \angle C$ respectively. Find the measure of $\angle BOC.$

Answer:

Let $\angle OCA= \angle OCB=x^{\circ}$

Since $BO \ bisects \angle A=ABC, \angle ABO= \angle OBC=35^{\circ}.$

$\Rightarrow 50^o+70^o+2x=180^o$

$\Rightarrow x=30^{\circ}=\angle OCB$

$\Rightarrow 35^o+30^o+ \angle BOC=180^o$

$\Rightarrow \angle BOC=115^{\circ}$

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Question 11: Using the information given in the adjoining figure, find the value of $x \ and\ y.$ P12

Answer:

$36^o+y+x=180^o$

$(24^o+36^o)+2x+64^o=180^o$

$\Rightarrow x=28^{\circ}$

$\Rightarrow 36^o+y+28^o=180^o \ or\ y=116^{\circ}$

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Question 12: In the adjoining figure, $DCBE$ is a straight line, $\angle ABD=135^{\circ} \ and\ \angle ACE=125^{\circ}$ . Find $\angle BAC.$

Answer:

$\angle ABC=180^o-135^o=45^{\circ}$

$\angle ACB=180^o-125^o=55^{\circ}$

Therefore $\angle BAC=180^o- \angle ABC- \angle ACB$

$=180^o-45^o-55^o=80^o$

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Question 13: Find, giving reasons, the unknown angles marked by $x, \ y$ in the following figures.

Answer:

i) $58^o+x+(180^o-123^o)=180^o$

$\Rightarrow x=65^{\circ}$

ii) $42^o+y=180^o$

$\Rightarrow y=138^{\circ}$

iii) $x+72^o+42^o=180^o$

$\Rightarrow x=68^{\circ}$

iv) $50^o+x+x=180^o$

$\Rightarrow x=65^{\circ}$

v) $65^o+y=180^o$

$\Rightarrow y=115^{\circ}$

vi) $30^o+y+110^o=180^o$

$\Rightarrow y=40^{\circ}$

Given $AC=AD$ which means $\angle ACD= \angle ADC=70^{\circ}$

$x+70^o+70^o=180^o$

$\Rightarrow x=40^{\circ}$

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Question 14: In the adjoining figure, if $\angle A=(x+5)^{\circ}, \angle B=(2x+3)^{\circ} \ and\ \angle BCD=(5x-14)^{\circ}$ , find the value of $x$ and hence, find the measure of $i)\ \angle BCD \ ii)\ \angle ACB$

Answer:

$\angle A+ \angle B+ \angle C=180^{\circ}$

$\Rightarrow (x+5)+(2x+3)+{180^o-(5x-14)}=180^o$

$\Rightarrow x=11 \ Therefore\ \angle A=16^{\circ}, \angle B=25^{\circ}$

$\Rightarrow i) \angle BCD=41^{\circ} \ \ and \ \ ii) \angle ACB=180^o-41^o=139^{\circ}$

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Question 15: In $\Delta ABC$ , side AC has been produced to D. If $\angle BCD=125^{\circ} \ and\ \angle A : \angle B=2 :3$ , find the measure of $\angle A \ and\ \angle B.$

Answer:

Let $\angle A=2x \ and\ \angle B=3x.$

$\Rightarrow 2x+3x+(180^o-125^o)=180^o$

$\Rightarrow x=25^o$

Therefore $\angle A=50^{\circ} \ and\ \angle B=75^{\circ}$

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Question 16: In the adjoining figure, it is given that: $BC \parallel DE,\ \angle BAC=35^o \ and\ \angle BCE=102^o$ . Find the measure of $i) \angle ABC ii)\ \angle ADE \ and\ iii) \angle CED$

Answer:

i) $\angle ABC+35^o+(180^o-102^o)=180^o$

$\Rightarrow \angle ABC=67^{\circ}$

$\angle DEA+102^o=180^o$ Note:Co-interior Angles

$\Rightarrow \angle DEA=78^{\circ}$

ii) $\angle ADE+78^o+35^o=180^o$

$\Rightarrow \angle ADE=67^{\circ}$

iii) $\angle CED+67^o+35^o=180^o$

$\Rightarrow \angle CED=78^{\circ}$

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Question 17: In the adjoining figure, it is given that: $AB=AC, \angle BAC=36^{\circ}, \angle ADB=45^{\circ} and \angle AEC=40^{\circ}$ . Find $i) \angle ABC, \hspace{1.0cm} ii) \angle ACB \hspace{1.0cm} iii) \angle DAB \ and \hspace{1.0cm} iv) \angle EAC$

Answer:

$AB=AC \Rightarrow \angle ABC= \angle ACB=x$

Therefore $36^o+2x=180^o \Rightarrow x=72^o$

i) & ii) Hence $\angle ABC= \angle ACB=72^{\circ}$

iv) $\angle CAE+(180^o-72^o)+40^o=180^o$

$\Rightarrow \angle CAE=32^{\circ}$

iii) $\angle DAB+(180^o-72^o)+4^o5=180^o$

$\Rightarrow \angle DAB=27^{\circ}$

$\angle DAE=27^o+36^o+32^o=95^{\circ}$

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Question 18: Using the information given in the adjoining figure, calculate the values of $x \ and\ y.$ P19

Answer:

In $\Delta ABC$

$34+(180^o-x)+62^o=180^o$

$\Rightarrow x=96^{\circ}$

In $\Delta DCE 24^o+96^o+(180^o-y)=180^o$

$\Rightarrow y=120^{\circ}$

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Question 19: Sides $AB \ and\ AC \ of \ \Delta ABC$ have been produced to D and E respectively. $BI \ and\ CI$ are bisectors of $\angle CBD \ and\ \angle BCE$ respectively. If $\angle BAC=70^{\circ} \ and \angle ACB 50^{\circ}$ , find the measure of $\angle BIC$ P20

Answer:

Given $\angle DBI= \angle IBC=x \ and \ \angle BCI= \angle ICE=y$

$\therefore 50^o+2y=180^o \Rightarrow y=65^o$

$\angle ABC+50^o+70^o=180^o$

$\Rightarrow \angle ABC=60^{\circ}$

$60^o+2x=180^o$

$\Rightarrow x=60^o$

$\therefore 60^o+ \angle BIC+65^o=180^o$

$\Rightarrow \angle BIC=55^{\circ}$

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Question 20: In the adjoining figure, it is being given that $BC \parallel DE, \angle CED=70^{\circ}, \angle CBA=84^{\circ} \ and\ \angle BAC=x^{\circ}$ . Find the value of $x$ .

Answer:

$\angle DEC= \angle BCA=70^{\circ}$ Note:Alternate angles

$x+84^o+70^o=180^o \Rightarrow x=26^{\circ}$

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Question 21: In the adjoining figure, it is being given that $BC \parallel DE, \angle EDC=25^{\circ}, \angle ECD=20^{\circ}, \angle ABC=70^{\circ}$ $\angle BAC=x^{\circ} and \angle DEA=y^{\circ}$ . Find the value of $x \ and\ y$ .

Answer:

In $\Delta DEC,$

$25^o+20^o+ \angle DEC=180^o$

$\Rightarrow \angle DEC=135^{\circ}$

$\therefore y+135^o=180^o$

$\Rightarrow y=45^{\circ}$

$BC \parallel DE \Rightarrow \angle CBD= \angle EDA=70^{\circ}$

In $\Delta ADE,$

$70^o+45^o+x=180^o$

$\Rightarrow x=65^{\circ}$

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Question 22: In the adjoining figure, it is given that $\angle ACB=42^{\circ}, \angle DEB= \angle CAB=90^{\circ}, \angle ABC=x^{\circ},$ $\angle BDE=y^{\circ} \ and\ \angle AFE=z^{\circ}.$ Find the value of $x, \ y \ and \ z.$

Answer:

In $\triangle FEC , 90^o+42^o+(180^o-z) = 180^o$

$\Rightarrow z=132^{\circ}$

In $\Delta ADF, (180^o-132^o)+90^o+y=180^o$

$\Rightarrow y=42^{\circ}$

In $\Delta DBE, 42^o+90^o+x=180^o$

$\Rightarrow x=48^{\circ}$

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Question 23: In the adjoining figure, $\displaystyle AE \parallel BC, \angle DAE=x^{\circ} , \angle ACB=(x-15)^{\circ}, \angle BAC=( \frac{x}{2} +y)^{\circ} \ and\ \angle ABC=(y+15)^{\circ}$