Question

if the sides of the triangle are 18cm,18√3cm and 36cm, then show that it is 30°-60°-90° triangle.​

Answer 1

Step-by-step explanation:

use p.t to prove its a right angled triangle

then use sinQ = perpendicular /base

Answer 2

It is proved that if the sides of the triangle are 18 cm, 18$\sqrt{3}$ cm and 36 cm, then the angles of the triangle are 30°, 60°, and 90°, using the concepts such as the Pythagoras theorem, the trigonometric identity for tan θ, and the sum of the angles in a triangle.

Given:

The sides of the triangle = 18 cm, 18$\sqrt{3}$ cm, and 36 cm.

To Find:

We have to show that if the sides of the triangle are 18 cm,1 8$\sqrt{3}$ cm and 36 cm, then the angles of the triangle are 30°, 60°, and 90°.

Solution:

Given that, the sides of the triangle = 18 cm, 18$\sqrt{3}$ cm, and 36 cm.

The hypotenuse of the triangle = 36 cm.

The base of the triangle = 18 cm.

The altitude of the triangle = 18$\sqrt{3}$ cm.

The Pythagoras theorem for right triangles is given by,

$(Hypotenuse)^{2} = (Base)^2 + (Altitude)^2$.

Applying this theorem to the given triangle, we get,

$(36)^{2} = (18)^2 + (18\sqrt{3} )^2$

⇒ $1296 = 324 + 972$

⇒ $1296 = 1296.$

Since $L.H.S = R.H.S$, one angle of the given triangle (i.e., the angle against the hypotenuse) is 90°.

Now, consider the following trigonometric equation for right triangles.

tan θproved that if the sides of the triangle are 18 cm,1 8 cm and 36 cm, then the angles of the triangle are 30°, 60°, and tan θ = $\frac{Opposite side}{Adjacent side}$.

tan θ = $\frac{18}{18\sqrt{3} } = \frac{1}{\sqrt{3} } .$

∴, θ = $tan^{-1} \frac{1}{\sqrt{3} }$ = 30°.

That is, another angle of the given triangle is 30°.

The sum of the three angles inside a triangle = 180°.

∴, The third angle of the given triangle = $180 - (90 + 30)$ = 60°.

That is, the third angle of the given triangle is 60°.

Hence, proved that if the sides of the triangle are 18 cm, 18$\sqrt{3}$ cm and 36 cm, then the angles of the triangle are 30°, 60°, and 90°.

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