In isosceles triangle NTS, NT ≌ TS. If m∠N = 42° and NS = 30 cm, find the length of the altitude of △NTS. (Round answer to the nearest tenth.)

13.5 cm

Step-by-step explanation:

The altitude of the base of the isosceles triangle bisects the vertex angle, <T, and also bisects the base, NS.

Therefore, NS is divide into two, 15 cm each.

This means we now have two right triangles from the isosceles triangle with the following:

Reference angle = 42°

Opposite = altitude = h

Adjacent = 15 cm

To find h, apply trigonometric function, TOA:

Tan 42 = Opp/Adj

Tan 42 = h/15

15 * Tan 42 = h

h = 13.5060607 ≈ 13.5 cm (nearest tenth)

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~recklesschicken