Question

In the adjoining figure , AB = 6 , BC = 8 , " \angle" align="absmiddle" class="latex-formula"> ABC = 90° , BD ⊥ AC and $\angle$ ABD = $\theta$ , find the value of sin $\theta$.

Answer 1

Answer:

We can make three relation with the reference angle.

relation between perpendicular and hypotenuse

In the above figure, the ratio of AB (perpendicular) to AC (base) with reference angle is called sine θ.

∴ sin θ =

AB

AC

=

perpendicular

hypotenuse

=

p

h

relation between base and hypotenuse

In the above figure, the ratio of BC (base) to AC (base) with reference angle is called cosine θ.

∴ cos θ =

BC

AC

=

base

hypotenuse

=

b

h

relation between perpendicular and base

In the above figure, the ratio of AB (perpendicular) to BC (base) with reference angle is called tangent θ.

∴ tan θ =

AB

BC

=

perpendicular

base

=

p

b

Answer 2

Here we are given with a triangle with smaller triangles formed due to the altitude on AC. Given:

• AB = 6
• BC = 8
• <ABC = 90°
• BD ⊥ AC
• <ABD = $\theta$

We have to find the value for sin $\theta$

So, Let's start solving....

In ∆ADB and ∆ABC,

• <A = <A (common)
• <ABC = <ADB (90°)

So, ∆ADB ~ ∆ABC (By AA similarity)

The corresponding sides will be:

$\sf{ \dfrac{AD}{AB} = \dfrac{AB}{AC} }$

We know the value of AB and to find AC, we can use Pythagoras theoram that is:

AC = √6² + 8²

AC = 10

Coming back to the relation,

$\sf{ \dfrac{AD}{6} = \dfrac{6}{10} }$

$\sf{AD = \dfrac{6 \times 6}{10} = 3.6}$

In ∆ADB, we have to find sin $\theta$ which is given by perpendicular/base:

$\sf{\sin( \theta) = \dfrac{AD}{AB} }$

Plugging the values of AD and AB,

$\sf{\sin( \theta) = \dfrac{3.6}{6} }$

Simplifying,

$\sf{ \sin( \theta) = \dfrac{3}{5} = \boxed{ \red{0.6}}}$

And this is our final answer.....

Carry On Learning !