Solve acute angles? Need Help!?

Solve acute angles? Need Help!?

Ashish

Can someone explain it and give steps?

Diagram: http://www.flickr.com/photos/27678773@N06/48815133...

1. A Triangle always have three angles ABC in the Diagram whose sum = 180°.

2. A Right Triangle has one angle B = 90°3. Acute angles are angle that measure lesser than 90°

4. When the 180° which is the sum of the three angles ABC in theTriangle minus the B=90° leave the sum of the two angles = 90°

5. This 90° is share by two angles AC.

6. If for example one Angle say Angle C = 10° then other angle = 80°

So to your Question: → One of the acute angles of a right triangle has twice the measure of the other acute angle. What is the measure of the larger acute angle in the right triangle

Let C = angle to be doubled or twice the measure of angle A

A + C = 90° ← eq1

C = 2 x A ← eq2

Substitute eq2 to eq1

A + (2 x A) = 90

3A = 90

A = 30° ← the smaller Angle

C = 2 x A

C = 2 x 30°

C = 60° ← the bigger angle

I hope this explain everything

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God Bless

Lim♥E

An acute angle is less than 90 deg. Inside angles of a triangle = 180 deg.

180 minus the 90 deg right angle leaves 90 deg.

The ratio of 2/1 of the unknown angles means that if you devide 90 by 3 it will give you

one of the angles and multiply by 2 for the other. 90/3 = 30....30 x 2 = 60. You have

a 30 - 60 degree right triangle. 30 + 60 + 90 = 180 deg.

Let x be the smaller angle, then 2x is the measure of the other angle. Since you are in a right triangle, you have:

x+2x=90 or 3x=90 so x=30.

So the smaller angle is 30 and the bigger is 2*30=60