Question

1) The angles of a triangle are in the ratio 2:3:4. Find the angles of a triangle.
2) The sum of three consecutive multiples of 11 is 333. Find these multiples.​

Answer 1

${ \boxed{\boxed{\begin{array}{cc} \underline{ \bf \: solution - 1} \\ \\ \bf \: given \\ \\ \rm \to \:The \: angles \: of \: a \: triangle \: are \\ \rm in \: the \: ratio \: 2:3:4 \\ \\ \sf \to \: we \: have \: to \: find \: : \\ \\ \odot \sf \: the \: angles \: of \: that \: triangle \\ \\ \\ \bf \: let \: common \: ratio \: of \: angle \: is \: = x \\ \\ \rm \: So \: let \: us \: consider \: the \: angles \: \\ \rm \: are \: 2x, \: 3x \: and \: 4x \\ \\ \sf \: we \: know \: that \\ \\ \rm \to \: Angle \: sum \: of \: triangle \: is \: = {180}^{ \circ} \\ \\ \sf \: \therefore \: according \: to \: the \: question \\ \\ \rm \: 2x + 3x + 4x = {180}^{ \circ} \\ \\ \rm \implies \: 9x = {180}^{ \circ} \\ \\ \rm \implies \: x = \frac{ {180}^{ \circ} }{9} \\ \\ \rm \implies \: x = {20}^{ \circ} \\ \\ \boxed{ \begin{array}{cc} \rm \: \to \: {1}^{{st}} \: angle = 2x = 2 \times 20 = {40}^{ \circ} \\ \\ \rm \to \: {2}^{{nd}} \: angle = 3x = 3 \times 20 = {60}^{ \circ} \\ \\ \rm \to \: {3}^{{rd}} \: angle = 4x = 4 \times 20 = {80}^{ \circ} \end{array}} \\ \\ \end{array}}}}$

Solution -2:

${ \boxed{\boxed{\begin{array}{cc} \sf \: Let \: the \: three \: \: consecutive \: multiples \\ \rm \: of \: 11  \: are \: 11x,  \: 11x+11  \: and  \: 11x+22 \\ \\ \rm \: and \: their \: sum \: is \: = 333 \: \\ \\ \therefore \: \rm \: 11x+11x+11+11x+22=333 \\ \\ \rm \implies \: 33x+33=333 \\ \\ \rm \implies \: 33x=333−33 \\ \\ \rm \implies \: 33x=300 \\ \\ \rm \implies \: x = \frac{300}{33} \\ \\ \rm \implies \: x = \frac{100}{11} \\ \\ \small{\boxed{ \begin{array}{cc} \rm \to \: First \: multiple  \: =11x=11× \frac{100}{11} =100 \\ \\ \rm \to \: Second \: multiple  \: =11x+11=11 \times \frac{100}{11} +11=111 \\ \\ \rm \to \: Third \: multiple  \:=11x+22=11 \times \frac{100}{11} +22=122 \end{array}}} \\ \\ \end{array}}}}$

Answer 2

Question 1/-

$\large \bold { \underline{Given :}}$

$\sf \: {Ratio \: of \: three \: angles \: of \: a \: triangle \: is \: 3 : 4 : 2.}$

$\large \bold { \underline{To \: Find:}}$

$\sf {All \: angles \: of \: the \: triangle. }$

$\large \bold { \underline{Solution: \: }}$

$\sf {Put \: k \: in \: the \: ratio. }$

$\sf{Now, angles \: are - } \\ \\ </p><p> \sf{First \: angle = 2 \: k } \\ </p><p> \sf{Second angle = 3k } \\ </p><p> \sf{Third angle = 4k } \\ \\ </p><p></p><p> \sf{As \: we \: know \: that - \: } \\ \\ </p><p></p><p> \sf{Sum \: of \: all \: angles \: of \: a \: triangle \: is \: 180°. } \\ </p><p> \sf{i.e. } \\ \\ \star \: \underline{\boxed{\sf\pink{ \angle1 + \angle2 + \angle3 = 180 \degree}}}$

$\it{\bold{\underline{According \: to \: the \: question \: : }}} </p><p> \\ \\ \sf:\implies \: 2k + 3k + 4k = 180 \\ \\ \sf:\implies \: 5k + 4k = 180 \\ \\ \sf:\implies \: 9k = 180 \\ \\ \sf:\implies \: k = \cancel\frac{180}{9} \\ \\ \sf:\implies \: \underline{\boxed{\mathfrak\purple{k = 20}}} \: \bigstar \\ \\ \\ \\ \sf \therefore{ \underline{Value \: of \: k \: is \: \bold{20}.}}$

$\sf{Now, \: put \: the \: value \: of \: k \: in \: the \: ratio. }</p><p> \\ \\ </p><p> \sf{First \: angle = 2k = 2(20) = 40°} \\ </p><p> \sf{Second \: angle = 3k = 3(20) = 60° } \\ </p><p> \sf{Third \: angle = 4k = 4(20) = 80° } \\ \\ </p><p> \sf \therefore{ \underline{Hence, \: the \: all \: angles \: of \: the \: triangle \: are \: \bold{40 \degree} \: \bold{60 \degree} \: and \: \bold{80 \degree} \: respectively.}} \\ \\ </p><p> </p><p> \sf{ \: \: \: \: \: \: \: The numbers are 108, 111, 114}</p><p></p><p>$

Question2/-

$\large \bold { \underline{Given :}}$

The sum of three consecutive multiples of 11 is 333.

$\large \bold { \underline{To \: Find:}}$

The numbers

$\large \bold { \underline{Solution: \: }}$

Let the numbers be x, (x+3), (x+6)

[As they are multiples of 3]

Now,

According to the question,

Their sum is 333

⇒ x + (x+3) + (x+6) = 333

⇒ x + x + x + 3 + 6 = 333

⇒ 3x + 9 = 333

⇒ 3x = 333 - 9

[by taking 9 to RHS]

⇒ 3x = 324

⇒ x = 324 ÷ 3

[By taking 3 to RHS]

⇒ x = 108

Now,

The numbers are

→ x = 108

→ (x + 3) = 108 + 3 = 111

→ (x + 4) = 108 + 6 = 114

----

Verification:

Their sum is 333

So,

108 + 111 + 114

= 219 + 114

= 333

The numbers

Let the numbers be x, (x+3), (x+6)

[As they are multiples of 3]

Now,

According to the question,

Their sum is 333

⇒ x + (x+3) + (x+6) = 333

⇒ x + x + x + 3 + 6 = 333

⇒ 3x + 9 = 333

⇒ 3x = 333 - 9

[by taking 9 to RHS]

⇒ 3x = 324

⇒ x = 324 ÷ 3

[By taking 3 to RHS]

⇒ x = 108

Now,

The numbers are

→ x = 108

→ (x + 3) = 108 + 3 = 111

→ (x + 4) = 108 + 6 = 114

Verification:

Their sum is 333

So,

108 + 111 + 114

= 219 + 114

= 333