Question

In the given figure, triangle ABC is right angled at B and BD is perpendicular to AC. Prove that sin theta is equal to 1 by underroot 10.

Answer 1

Answer:

Step-by-step explanation:

answer is attached ...........

Answer 2

Answer:

$\sin\theta =\frac{1}{\sqrt{10}}$

Step-by-step explanation:

Given information: In triangle ABC is right angled at B and BD is perpendicular to AC.

To prove: $\sin\theta =\frac{1}{\sqrt{10}}$

Proof:

According to Pythagoras theorem:

$hypotenuse^2=base^2+perpendicular^2$

Use Pythagoras theorem in triangle ABD,

$AB^2=AD^2+BD^2$

$AB^2=(\frac{1}{3}b)^2+(b)^2$

$AB^2=\frac{1}{9}b^2+b^2$

$AB^2=\frac{10}{9}b^2$

Taking square root both sides.

$AB=\frac{\sqrt{10}}{3}b$

In triangle triangles ABC and ADB,

$\angle BAC\cong \angle DAB$          (Reflexive property)

$\angle ABC\cong \angle ADB$          (Right angle)

By AA property,

$\triangle ABC\sim \triangle ADB$

The corresponding sides of similar triangles are proportional.

$\frac{AB}{AC}=\frac{AD}{AB}$

Substitute the value of AD and AB.

$\frac{AB}{AC}=\frac{\frac{1}{3}b}{\frac{\sqrt{10}}{3}b}$

$\frac{AB}{AC}=\frac{1}{\sqrt{10}}$             .... (1)

In a right angled triangle

$\sin \theta =\frac{adjacent}{hypotenuse}$

In triangle ABC,

$\sin \theta =\frac{AB}{AC}$

Using equation (1).

$\sin \theta =\frac{1}{\sqrt{10}}$

Hence proved.