in a triangle ABC it is given that angle C=90 and tan =1÷√3 find the value of sinAcosB+cosAsinA
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Take AB as the Hypotenuse of the Triangle.
The Hypotenuse: AB^2 = AC^2 + BC^2 = 1^2 + √3^2 = 4..so AB = √4 = 2. So '2' is the Hypotenuse.
Tan=1/√3..so the Opposite or Altitude of the Triangle is equal to 1 and the Adjacent or Base of the Triangle is equal to √3.
Sin A=Opposite/Hypotenuse=√3/2 (60 Degrees).
Cos B=Adjacent/Hypotenuse=√3/2 (30 Degrees).
Cos A=Adjacent/Hypotenuse=1/2 (60 Degrees).
Sin A=Opposite/Hypotenuse=√3/2 (60 Degrees).
Sin A x Cos B + Cos A x SIn A = √3/2 x √3/2 + 1/2 x √3/2 = 3/4 + √3/4. (√3=1.73)
So, 3+1.73/4 = 1.18 (Approx.) is the Answer.
P.S. Correct me if I'm wrong.
The Hypotenuse: AB^2 = AC^2 + BC^2 = 1^2 + √3^2 = 4..so AB = √4 = 2. So '2' is the Hypotenuse.
Tan=1/√3..so the Opposite or Altitude of the Triangle is equal to 1 and the Adjacent or Base of the Triangle is equal to √3.
Sin A=Opposite/Hypotenuse=√3/2 (60 Degrees).
Cos B=Adjacent/Hypotenuse=√3/2 (30 Degrees).
Cos A=Adjacent/Hypotenuse=1/2 (60 Degrees).
Sin A=Opposite/Hypotenuse=√3/2 (60 Degrees).
Sin A x Cos B + Cos A x SIn A = √3/2 x √3/2 + 1/2 x √3/2 = 3/4 + √3/4. (√3=1.73)
So, 3+1.73/4 = 1.18 (Approx.) is the Answer.
P.S. Correct me if I'm wrong.
RoshanSingh1:
tan of which angle
given in question
??
Answered by
4
Answer:
since tan is cos /sin it has to be angle opposite to the hypoteneuse so it ia the 90 degree angle
Step-by-step explanation:
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