Question

In triangle ABC AB = AC and side D is pointed on AC like this that BD = BC if ab = 12.5 cm and BD = 5 cm then tell measure of DC​

Answer 1

2

In triangle ABC AB = AC and side D is pointed on AC like this that BD = BC if ab = 12.5 cm and BD = 5 cm

1)First make a figure of the given details

2)The given triangle is a Isosceles triangle, apply sin rule which is:

a/sin A=b/sin B=c/sin C

3)using a/sin A=b/sin B, find sin B by substituting A=180-2B

4)As B=C so sin C=sin B,then apply sin rule on triangle BDC

5)then you'll get the length of DC

Answer 2

DC = 2 cm.

Step-by-step explanation:

See the attached diagram.

Here, AB = AC = 12.5 cm and BD = BC = 5 cm.

Now, draw perpendicular on BC from vertex A and it will bisect BC at E.

Again, draw perpendicular on DC from vertex B and it will bisect DC at F.

Now, from the right triangle Δ AEC, AC = 12.5 cm and  CE = 1/2 BC = 2.5 cm.

So, $\cos C = \frac{CE}{AC} = \frac{2.5}{12.5} = 0.2$

⇒ $C = \cos^{-1}(0.2) = 78.463^{\circ}$

Again, from the right triangle Δ BCF, BC = 5 cm and ∠ C = 78.463°.

So, $\cos C = \frac{CF}{BC} = \frac{CF}{5}$

⇒ CF = 5 cos 78.463° = 1 cm

Hence, DC = 2 × CF = 2 cm. (Answer)