Question

Construct a triangle with sides $AB=4cm,BC=5cm$ and $AC=6cm$ and then find $\angle ACB$ using cosine's rule

• A. $co{s}^{-1}\frac{3}{4}$
• B. $co{s}^{-1}\frac{5}{8}$
• C. $co{s}^{-1}\frac{2}{4}$
• D. $co{s}^{-1}\frac{1}{4}$

Answer 1

The correct option is A $co{s}^{-1}\frac{3}{4}$

1. Mark a point B that will be one vertex of the new triangle.
2. Set the compasses' width to $5$ cm, the length of the segment BC.
3. With the compasses' point on B, make an arc near the future vertex C of the triangle.
4. Mark a point C on this arc. This will become the next vertex of the new triangle.
5. Set the compasses' width to the length of the line segment AB = $4$ cm .
6. Place the compasses' point on B and make an arc in the vicinity of where the third vertex of the triangle (A) will be.
7. Set the compasses' width to the length of the line segment AC = $6$ cm .
8. Place the compasses' point on C and cut the arc drawn by keeping the compass at B. The point where these two arcs meet is the vertex A of the triangle.
9. Thus a triangle ABC with given sides is formed.
For the given triangle a= $5$,b=$6$,c=$4$

Acc. to cosine rule

$cosC=\frac{a^2+b^2-c^2}{2ab}$

So $cosC=\frac{5^2+6^2-4^2}{2\times 5\times 6}$

$cosC=\frac{25+36-16}{60}$

$cosC=\frac{45}{60}$

$cosC=\frac{3}{4}$

$\therefore C=co{s}^{-1}\frac{3}{4}⟹\angle ACB=co{s}^{-1}\frac{3}{4}$