Question
In $\triangle\text{ABC},$ if $\angle\text{A}+\angle\text{B}=108^\circ$ and $\angle\text{B}+\angle\text{C}=130^\circ,$ find $\angle\text{A},\angle\text{B}$ and $\angle\text{C}.$
✓
Answer
$\angle\text{A}+\angle\text{B}=108^\circ$ [Given] But as $\angle\text{A},\angle\text{B}$ and $\angle\text{C}$ are the angles of a triangle,
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow108^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-108^\circ=72^\circ$
Also, $\angle\text{B}+\angle\text{C}=130^\circ$ [Given]
$\Rightarrow\angle\text{B}+72^\circ=130^\circ$
$\Rightarrow\angle\text{B}+72^\circ=130^\circ$
$\Rightarrow\angle\text{B}=130^\circ-72^\circ=58^\circ$
Now as, $\angle\text{A}+58^\circ=108^\circ$
$\Rightarrow\angle\text{A}+58^\circ=108^\circ$
$\Rightarrow\angle\text{A}=108^\circ-58^\circ=50^\circ$
$\therefore\angle\text{A}=50^\circ,\angle\text{B}=58^\circ$ and $\angle\text{C}=72^\circ$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow108^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-108^\circ=72^\circ$
Also, $\angle\text{B}+\angle\text{C}=130^\circ$ [Given]
$\Rightarrow\angle\text{B}+72^\circ=130^\circ$
$\Rightarrow\angle\text{B}+72^\circ=130^\circ$
$\Rightarrow\angle\text{B}=130^\circ-72^\circ=58^\circ$
Now as, $\angle\text{A}+58^\circ=108^\circ$
$\Rightarrow\angle\text{A}+58^\circ=108^\circ$
$\Rightarrow\angle\text{A}=108^\circ-58^\circ=50^\circ$
$\therefore\angle\text{A}=50^\circ,\angle\text{B}=58^\circ$ and $\angle\text{C}=72^\circ$
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
Find six rational numbers between $2$ and $3.$
→A coin is tossed $500$ times and we get: heads: $285$ times and tails:$ 215$ times. When a coin is tossed at random, what is the probability of getting
$i.$ A head?
$ii.$ A tail?
→$i.$ A head?
$ii.$ A tail?
The bisectors of $\angle\text{B}$ and $\angle\text{C}$ of an isosceles triangle with $AB = AC$ intersect each other at a point $O. BO$ is produced to meet $AC$ at a point $M.$ Prove that $\angle\text{MOC}=\angle\text{ABC}.$
→$\text{ABC}$ is an isosceles triangle in which $\text{AB = AC. BE}$ and $\text{CF}$ are its two medians. Show that $\text{BE = CF.}$
→If the area of an equilateral triangle is $36\sqrt{3}\text{cm}^2,$ find its perimeter.
→In the given figure, $AB\ ||\ CD$ and $EF$ is a transversal, cutting them at $G$ and $H$ respectively. If $\angle\text{EGB}=35^\circ$ and $\text{QP}\perp\text{EF},$ find the measure of $\angle\text{PQH}.$
→Prove that: $\Big(\frac{64}{125}\Big)^{-\frac{2}{3}}+\frac{1}{\Big(\frac{256}{625}\Big)^{\frac{1}{4}}}+\frac{\sqrt{25}}{\sqrt[3]{64}}=\frac{65}{16}$
→Find two irrational numbers between $0.5$ and $0.55.$
→Find five rational numbers between $\frac{3}{5}{\text{ and }}\frac{4}{5}$
→Locate $\sqrt{2}$ on the number line.
→