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The altitude AD of a triangle ABC is 24 cm long. If BD=16 cm, DC=36 cm, prove that the...
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Middle School Teacher
I'm impressed! However, I'm guessing you don't want to have anything to do with tangents! Fortunately, there's no need for trigonometry.
By definition, the altitude divides the triangle into two right triangles: ADB and ADC.
You know the lengths of the legs of these two right triangles, so using the Pythagorean theorem you can figure out the lengths of the hypotenuses.
for ADB, AD^2 + DB^2 = AB(hypotenuse)^2
AD = 24 and DB = 16, so
24^2 + 16^2= AB^2
576 + 256 = 832
NOTE: For this problem, you don't need to find the square root of 832 -- read on!!
Now look at the other right triangle, ADC. We know that
AD^2 + DC^2 = AC (hypotenuse) ^2
AD = 24 and DC = 36
24^2 + 36^2 = AC^2
576 + 1296 = 1872
Again, don't bother finding the square root of 1872.
This problem asks whether angle A is a right angle. If it is, then triangle ABC will be a right triangle. AB and AC would be the legs, and BC would be the hypotenuse. So, if it is a right triangle,
AB^2 + AC^2 = BC^2
We know from our earlier work what AB^2 and AC^2 are!! So,
832 + 1872 = BC^2 = 2704
So, if ABC is a right triangle the hypotenuse^2 will be 2704. Well, we know what the hypotenuse is!
It is BD + DC = 16 + 36 = 52.
Lo and behold, 52^2 = 2704.
Posted by cburr on February 16, 2009 at 11:33 AM (Answer #1)
In triangle ABD,
∠ABD= tan^-1 (16/24)
In triangle ADC
tan ∠DAC= 36/24
∠BAC= 56.31 + 33.69
= 90° (shown)
Posted by revolution on September 7, 2011 at 12:25 AM (Answer #2)
High School Teacher
For triangle ABD, tan (A) = 24/16; tan (B) = 16/24. So let's look those tangets up-- what angle's tangent is 1.5, and what angle's tangent is 0.666? Let's hope that the two angles we find add up to 90 degrees, because the left over angle (D) is 90 degrees only if 180 minus the sum of angles A and B is 90 degrees.
Yep, 14 and 76 = 90, so good. And since BC is a line (180 degrees), the angle on the other side of the altitude is also 90 degrees.
Posted by kmieciakp on February 16, 2009 at 5:46 AM (Answer #3)
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