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Consider the triangle ABC with C=90∘. Compute the following trigonometric values when a=1 and b=5.

what does

sinA=

cosA=

sinB=

cosB=

### 6 Answers

- sepiaLv 72 months ago
Consider the triangle ABC with C = 90∘.

Compute the following trigonometric values when a = 1 and b = 5. sin A =cos A =sin B =cos B =

- Engr. RonaldLv 72 months ago
we 1st need to solve for the hypotenuse by using pythagorean theorem.

c = √(a^2 + b^2)

c = √(1^2 + 5^2)

c = √(26)

so

sin(A) = 1/√(26) or √(26)/26

cos(A) = 5/√(26) or 5√(26)/26

sin(B) = 5/√(26) or 5√(26)/26

cos(B) = 1/√(26) or √(26)/26

- PhilipLv 62 months ago
In triangle ABC, (A,B,C) is opposite (a,b,c) = (small arm,larger arm,hypotenuse)

Clearly, c^2 = a^2+b^2 = 1^2+5^2 = (rt26)^2, where rt = ''square root of''. Then

c = rt26.;

By law of sines, a/sinA = b/sinB = c/sinC;

c/sinC = rt26/sin90 = rt26/1 = rt26;

a/sinA = rt26, ie., sinA = a/rt26 = 1/rt26 = (1/26)rt26.;

b/sinB = rt26, ie., sinB = b/rt26 = 5/rt26 = (5/26)rt26.;

cos^2(A) = 1 - sin^2(A) = 1 - (1/26) = (25/26). Then cos(A) = (5/26)rt26.;

cos^2(B) = 1 - sin^2(B) = 1 - (25/26) = (1/26). Then cos(B) = (1/26)rt26.;

- micatkieLv 72 months ago
Refer to the figure below.

c² = a² + b² (Pythagorean theorem)

c = √(1² + 5²) = √26

sinA = a/c = 1/√26 = (√26)/26

cosA = b/c = 5/√26 = 5(√26)/26

sinB = b/c = 5/√26 = 5(√26)/26

cosB = a/c = 1/√26 = (√26)/26

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- ted sLv 72 months ago
the reference triangle for angle A is { 5 , 1 , √26 } ...thus sin A = 1 / √26...you certainly can finish