in triangle abc angle b =90 degree tan=3/4.find the value of sinacosc+cosasinc
Answers
Answer:
sin(A+B) = sinA.cosB + cosA.sinB
cos(A+B)= cosA. cosB - sinA.sinB
Given:
tanA=1/√3
A=30°
since B is right angle, C=60°
Now,
sinA.cosC + cosA.sinC
=sin(A+C)
=sin(30°+60°)
=sin90°
=1
cosA.cosC - sinA.sinC
=cos(A+C)
=cos(30°+60°)
=cos90°
=0
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Answer:
SinACosC + CosASinC = 0
CosACosC - SinASinC = 0
Solution is independent of tanA value
Step-by-step explanation:
In triangle ABC,right-angle at B,if tanA=1/√3,find the value of
sina cosc+cosa sinc
cosa cosc-sina sinc
Its a right angle triangle at B
so ∠A + ∠C = 90
∠A = ∠90 - ∠C
Sin∠A = Sin(∠90 - ∠C) = CosC
Cos∠A = Cos(∠90 - ∠C) = SinC
SinACosC + CosASinC
= SinASinA + CosACosA
=Sin²A + Cos²A
= 1
CosACosC - SinASinC
= SinCCosC - CosCSInC
= SinCCosC - SinCCosC
= 0
Step-by-step explanation:
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It took 5 min to write the answer