Question
Two supplementary angles are in the ratio $3: 7$, find the angles.
✓
Answer
Let the two supplementary angles be $3 x$ and $7 x$.
$
\therefore 3 x+7 x=180^{\circ}
$
$\left[\because\right.$ the sum of two supplementary angles $=180^{\circ}$ ]
$
\begin{array}{l}
\Rightarrow 10 x=180^{\circ} \\
\Rightarrow x=\frac{180^{\circ}}{10} \Rightarrow x=18^{\circ}
\end{array}
$
So, the angles are $3 \times 18=54^{\circ}$ and $7 \times 18=126^{\circ}$
$
\therefore 3 x+7 x=180^{\circ}
$
$\left[\because\right.$ the sum of two supplementary angles $=180^{\circ}$ ]
$
\begin{array}{l}
\Rightarrow 10 x=180^{\circ} \\
\Rightarrow x=\frac{180^{\circ}}{10} \Rightarrow x=18^{\circ}
\end{array}
$
So, the angles are $3 \times 18=54^{\circ}$ and $7 \times 18=126^{\circ}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Solve the following equations. Check your result in case.$3 + 2x = 1 - x$
→Find the radius of a circle, if its area is
$4 \pi cm^2$
→$4 \pi cm^2$
Write down the denominator of the following rational number:
$\frac{-15}{-82}$
→$\frac{-15}{-82}$
Divide:
$148$ by $0.074$
→$148$ by $0.074$
The enrolment in a school during six consecutive years was as follows
$ 1555,1670,1750,2013,2540,2820 . $
Find the mean enrolment of the school for this period.
→$ 1555,1670,1750,2013,2540,2820 . $
Find the mean enrolment of the school for this period.
Multiply the following fractions.
$3 \frac{4}{7} \times \frac{3}{5}$
→$3 \frac{4}{7} \times \frac{3}{5}$
Find the value of x in each of the following figures.
Two numbers are in the ratio $5 : 7.$ If the sum of these numbers is $84,$ find the numbers.
→Find the products:
$\left(2 a^2 b^3\right) \times\left(-3 a^3 b\right)$
→$\left(2 a^2 b^3\right) \times\left(-3 a^3 b\right)$
Write a positive integer and a negative integer whose sum is a negative integer.