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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.)

In isosceles triangle NTS NT TS If mN 42 and NS 30 cm find the length of the altitude of NTS Round answer to the nearest tenth class=

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akposevictor

Answer:

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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