In a right angled triangle the one acute angle is double the other.Show that hypotenuse is double the smallest?

The sum of the angles of a triangle is 180 degrees. When you take away 90 degrees you are left with 90 degrees.

So the other angles are 30 degrees and 60 degrees.

Since sin(30 degrees) = (1/2), and sin(x) = the length of the side opposite it divided by the length of the hypotenuse, then that side is (1/2) the size of the hypotenuse. That means that the hypotenuse is twice as long the side opposite the 30 degree angle.

.

In the triangle you have 90, 2a and a

They have to add up to 180, so 90 + 3a = 180

and 3a = 90 and a=90/30 so a = 30

You can use your knowledge of trig ratios.

The smallest side is opposite the smallest angle, therefore we look at the sin 30.

sin 30 = 1/2 therefore smallest side/ hypotenuse = 1/2

therefore hypotenuse is double the smallest side

the only possibility is in that case will be one angle is 30 and another 60. if you talk in trigonometry it says thar sin@= P/H =1/2

so H=2P

You should learn to do it yourself, this looks like homework. What will you do in the exam? Take a phone browser to put the exam questions on Yahoo Answers?

Given: In Δ ABC , <B = 900 and <ACB = 2 <CAB

Prove that AC = 2BC

Construction: Produce CB to D such that BC = BD Join to AD

Proof : In Δ ABD, and ABC

BD = BC ; AB = AB and <B = <B = 900

By SAS congruency , D ABD ≅ D ABC

By CPCT, AD = AC

<DAB = <BAC = X0

So, < DAC = 2X0

=> <ACB = <ACD

Now in Triangle Δ ADC, <DAC = <ACD= 2X0

So, AD = DC

=> AC = DC = 2BC Proved