Question

class 10 mathematics trigonometry​

Answer 1

Answer:

A=60 B=45 C=75

Step-by-step explanation:

sin(A+B-C)=$\frac{1}{2}$

we know that sin 30=$\frac{1}{2}$ ,so A+B-C=30 let this is taken as equation 1

cot(A-B+C)=0

cot 90=0 ,so A-B+C=90 let this be equation 2

if we solve equation 1 and 2 we get

2B-2C=-60 ;B-C=-30

if we sustitute this in equation1 we get A-30=30 ;A=60

cos(B+C-A)=$\frac{1}{2}$

we kow that cos 60=$\frac{1}{2}$;so B+C-A=60 let this be equation 3

solve 1 and 3 we get 2B=90;B=45

now if we substitute A and B in any equation and solve it we get C=75

Answer 2

Step-by-step explanation:

since

A, B,C<90°

=> A+B, B+C. AND C+A < 180°

=> A+B-C , A-B+C AND --A+B+C < 90°

USING THIS, WE HAVE

SIN (A+B-C) = 1/2= SIN30°

=> A+B-C= 30°................(1)

COT(A-B+C) = 0= COT 90°

=> A-B+C= 90°..............(2)

COS( -A+B+C)= 1/2= COS 60°

=> --A+B+C= 60°..............(3)

(1)+(2)

2A= 120° => A= 60°

(1)+(3)

2B= 90° => B= 45°

(2)+(3)

2C= 150°=> C= 75°

HENCE,

angle A= 60°

angle B = 45°

angle C = 75°