In ΔABC, m∠A = 90° and AN is an altitude. If AB = 20 in and AC = 15 in, find BC, BN, NC, AN.

Question

03-05-2017
Mathematics

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In ΔABC, m∠A = 90° and AN is an altitude. If AB = 20 in and AC = 15 in, find BC, BN, NC, AN.

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sharmaenderp sharmaenderp · Answer 1 · 2017-05-12T16:15:46+02:00

Since ΔABC is a right triangle, we can use the Pythagorean theorem to find the missing side, BC as shown below.

[tex] \overline{BC}= \sqrt{20^{2} + 15^{2}} = 25 [/tex]

Now, since AN is an altitude formed along BC, we can form two smaller triangles that are both similar with ΔABC.
(An image is attached along the file to show the illustrations of the triangles with the larger one.)

Thus, we have ΔNBA~ΔABC and ΔNAC~ΔABC. So, for ΔNBA and ΔABC, we have

[tex]\frac{BN}{BA} = \frac{BA}{BC} [/tex]
[tex] \frac{BN}{20} = \frac{20}{25} [/tex]
[tex] \overline{BN} = \frac{20(20)}{25} = 16 [/tex]

and applying the same process with AN, we have

[tex] \frac{AN}{BA} = \frac{AC}{BC} [/tex]
[tex] \frac{AN}{20} = \frac{15}{25} [/tex]
[tex] \overline{AN} = \frac{20(15)}{25} = 12 [/tex]

We can also use the same method to find the missing sides in ΔANC. But, we can immediately find the value of NC as shown.

[tex] \overline{NC} = \overline{BC} - \overline{BN} [/tex]
[tex] \overline{NC} = 25 - 16 = 9 [/tex]

Since ΔANC is a right triangle and two given lengths, we can find the third side, AN, through Pythagorean theorem.

[tex] \overline{AN} = \sqrt{15^{2} - 12^{2}} = 9 [/tex]

Thus, we have found all the missing sides. The picture attached also shows the missing sides' lengths in red.

Answer: BC = 25, BN = 16, NC = 9, and AN = 12

jayilych4real jayilych4real · Answer 2 · 2022-04-25T19:11:48+02:00

The values of each side of the triangle are:

BC = 25,
BN = 16,
NC = 9,
AN = 12

Calculations and Parameters:

Because ΔABC is a right triangle, the Pythagorean theorem will be used to find the missing side, BC

BC= [tex]\sqrt{20^2 + 15^2}[/tex]= 25.

Based on the fact that AN is an altitude formed along BC, we can form two smaller triangles that are both similar with ΔABC

Therefore, we have:

ΔNBA~ΔABC
ΔNAC~ΔABC.

So, for ΔNBA and ΔABC, we have:

[tex]BN/BC= BA/BC\\BN/20= 20/25\\BN= 20(20)/25\\=16[/tex]

For AN,

[tex]AN/BA= AC/BC\\AN/20= 15/25\\AN= 20(15)/25\\=12[/tex]

We already know that

NC= 9

AN= 9

Therefore, all the sides are:

BC = 25,
BN = 16,
NC = 9,
AN = 12