1)
given: m∠kal = 100º
m∠l = 25°, oa = 21
find: ak, al, kl
2)
given: δabc, m∠bac = 77°, ad = 18
ad⊥cb, ∠ c = 18°,
find: ac, ab, bc

Step-by-step explanation:

The diagram of the triangles are shown in the attached photo.

1) Looking at ∆AOL, to determine AL, we would apply the sine rule

a/SinA = b/SinB = c/SinC

21/Sin25 = AL/Sin 105

21Sin105 = ALSin25

21 × 0.9659 = 0.4226AL

AL = 20.2839/0.4226

AL = 50

Looking at ∆KAL,

AL/Sin55 = KL/Sin100

50/0.8192 = KL/0.9848

50 × 0.9848 = KL × 0.8192

KL = 49.24/0.8192

KL = 60

AK/Sin25 = AL/Sin 55

AKSin55 = ALSin25

AK × 0.8192 = 0.4226 × 50

AK = 21.13/0.8192

AK = 25.8

2) looking at ∆AOC,

Sin 18 = AD/AC = 18/AC

AC = 18/Sin18 = 18/0.3090

AC = 58.25

Sin 85 = AD/AB = 18/AB

AB = 18/Sin85 = 18/0.9962

AB = 18.1

To determine BC, we would apply Sine rule.

BC/Sin77 = 58.25/Sin85

BCSin85 = 58.25Sin77

BC = 58.25Sin77/Sin85

BC = 58.25 × 0.9744/0.9962

BC = 56.98

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐