Answer: 12 unit.
Step-by-step explanation:
Given : In triangle ABC,
m∠NMO=90°, MN=MO, BK⊥AC, NO∥AC, M∈AC, BK=10, AC=30,
We have to find : NO
Since, NO∥AC,
By the alternative interior angle theorem,
[tex]\angle BNO\cong \angle BAC[/tex]
[tex]\angle BON\cong \angle BCA[/tex]
Also,
[tex]\angle NBO\cong \angle ABC[/tex]
Thus, by AAA similarity postulate,
[tex]\triangle NBO\cong \triangle ABC[/tex]
Let S ∈ NO such that BS ⊥ NO,
By the property of similar triangles,
[tex]\frac{BS}{BK}=\frac{NO}{AC}[/tex]
[tex]\implies \frac{BK-SK}{BK}=\frac{NO}{AC}[/tex] -------- (1),
Now, m∠NMO=90° and MN=MO,
Let J ∈ NO, such that MJ⊥NO
⇒ Triangle NMO is a isosceles triangle,
⇒ ∠MNJ = 45°,
[tex]\implies tan 45^{\circ} = \frac{MJ}{NJ}[/tex]
[tex]\implies MJ = NJ = SK[/tex]
[tex]\implies NO = 2 SK[/tex] -------(2)
From equation (1),
[tex]\implies \frac{BK-SK}{BK}=\frac{2 SK}{AC}[/tex]
Since, BK=10, AC=30
[tex]\implies \frac{10-SK}{10}=\frac{2 SK}{30}[/tex]
[tex]\implies 300 - 30 SK = 20 SK \implies 50 SK = 300\implies SK = 6[/tex]
From equation (2),
NO = 2 × 6 = 12 unit.
