can anyone slice plzz
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as tanα = 4/5
Then opp(O) = 4k units and adj(A) = 5k units.
(as tanα = opp/adj)
Then in Δ By Pythagoras theorem,
H² =O² + A²
H² = (4k)² + (5k)²
H² = 16k² + 25k²
H = √39k² ⇒ √39 k units.
Now,
sinα = opp/hypo ⇒ 4k/√39k ⇒ 4/√39
cosα = adj/hypo ⇒ 5k/√39k ⇒ 5/√39
Then substituting these values,
= 5 sinα - 3 cosα/5 sinα + 2 cosα
= 5(4/√39) - 3(5/√39)/5(4/√39) + 2(5/√39)
= 20 -15/√39 ÷ 20 + 10/√39
(taking reciprocals)
= 5/√39 × √39/30
(cancelling √39 on numerator and denominator)
= 5/30 ⇒ 1/6
Answer = 1/6
Then opp(O) = 4k units and adj(A) = 5k units.
(as tanα = opp/adj)
Then in Δ By Pythagoras theorem,
H² =O² + A²
H² = (4k)² + (5k)²
H² = 16k² + 25k²
H = √39k² ⇒ √39 k units.
Now,
sinα = opp/hypo ⇒ 4k/√39k ⇒ 4/√39
cosα = adj/hypo ⇒ 5k/√39k ⇒ 5/√39
Then substituting these values,
= 5 sinα - 3 cosα/5 sinα + 2 cosα
= 5(4/√39) - 3(5/√39)/5(4/√39) + 2(5/√39)
= 20 -15/√39 ÷ 20 + 10/√39
(taking reciprocals)
= 5/√39 × √39/30
(cancelling √39 on numerator and denominator)
= 5/30 ⇒ 1/6
Answer = 1/6
☺ Hope this Helps ☺
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