∆KLM is an isosceles triangle because it has two equal sides, KL & KM. Therefore, the angles opposite to each of the two equal sides, which are referred to as the base angles are congruent to each other.

m<KML = m<KLM = 58°

m<MKL = 180 - (58 + 58) (Sum of triangle)

m<MKL = 64°

m<JKM = 180 - m<MKL (linear pair theorem)

m<JKM = 180 - 64 (Substitution)

m<JKM = 116°

∆JKM is also an isosceles triangle with two equal sides. Therefore, it's based angles (x & <J) would also be equal to each other.

Thus:

x = ½(180 - m<JKM)

x = ½(180 - 116) (Substitution)

x = 32°

3 0

3 years ago

Prove the identity (sina-cosatanb)/(cosa+sinatanb) = tan (a+b)

OLga [1]

Sina - (cosa)(tanb)/cosa + (sina)(tanb)
sina ≡ (tana)(cosa)

(tana)(cosa) - (cosa)(tanb)/cosa + (tana)(cosa)(tanb)
= cosa(tana - tanb)/cosa(1 + tanatanb)
(cosas cancel out)
= (tana - tanb)/(1 + tanatanb) ≡ tan(a-b)

7 0

3 years ago