Question

Two angles of a triangle are in the ratio $4:5.$ If the sum of these angles is equal to the third angle. Find all angles of the triangle.

• A. ${40}^{\circ },$ ${50}^{\circ }$ and ${90}^{\circ }$
• B. ${30}^{\circ },$ ${60}^{\circ }$ and ${90}^{\circ }$
• C. ${50}^{\circ },$ ${50}^{\circ }$ and ${100}^{\circ }$
• D. ${40}^{\circ },$ ${70}^{\circ }$ and ${110}^{\circ }$

Answer 1

The correct option is A ${40}^{\circ },$ ${50}^{\circ }$ and ${90}^{\circ }$

Let the two angles of the triangle be $4x$ and $5x$.

Then, as per the question, we know
third angle $=4x+5x=9x$
We know that sum of all the angles of a triangle is ${180}^{\circ }$.
$\therefore 4x+5x+9x={180}^{\circ }\Rightarrow 18x={180}^{\circ }\Rightarrow x={10}^{\circ }$ ....[On dividing both sides by 18]
So, first angle $=$ ${4x}^{\circ }$ $=$ $4\times {10}^{\circ }={40}^{\circ }$,
second angle $=$ ${5x}^{\circ }$ $=$ $5\times {10}^{\circ }={50}^{\circ }$ and
and third angle $=$ ${9x}^{\circ }$ $=$ $9\times {10}^{\circ }={90}^{\circ }$