<br />Find (i) sin C (ii) cos C and<br />(iii) tan C in the adjacent triangle.<br />2. In a triangle XYZ, Z Y is right angle,<br />13 cm<br />5 cm<br />XZ=17cm and YZ = 15 cm, then find<br />SA<br />.<br />(i) sin X (ii) cos Z (iii) tan X<br />In a triangle PQR with right angle at Q, the value of Pis x, PQ = 7 cm and QR=<br />24 cm, then find sin x and cos x.
Answers
Answer:
Given :-
If sin (A + B) = 1 and tan (A - B) = 1/√3
To Find :-
value of:
i) tan A+ tan B
ii) sec A- cosec B
Solution :-
We know that
sin 90 = 1
tan 30 = 1/√3
So,
sin(A + B) = 90
tan(A - B) = 30
A + B + A - B = 90 + 30
(A + A) + (B - B) = 120
2A = 120
A = 120/2
A = 60
By putting value of A in 2
A - B = 30
60 - B = 30
-B = 30 - 60
-B = -30
B = 30
Finding values
tan 60 + tan 30
We know that
tan 60 = √3
tan 30 = 1/√3
√3 + 1/√3
√3 × √3 + 1/√3
3 + 1/√3
4/√3
sec A- cosec B
sec 60 - cosec 30
We know that
sec 60 = 2
cosec 30 = 2
2 - 2 = 0
..Given :-
If sin (A + B) = 1 and tan (A - B) = 1/√3
To Find :-
value of:
i) tan A+ tan B
ii) sec A- cosec B
Solution :-
We know that
sin 90 = 1
tan 30 = 1/√3
So,
sin(A + B) = 90
tan(A - B) = 30
A + B + A - B = 90 + 30
(A + A) + (B - B) = 120
2A = 120
A = 120/2
A = 60
By putting value of A in 2
A - B = 30
60 - B = 30
-B = 30 - 60
-B = -30
B = 30
Finding values
tan 60 + tan 30
We know that
tan 60 = √3
tan 30 = 1/√3
√3 + 1/√3
√3 × √3 + 1/√3
3 + 1/√3
4/√3
sec A- cosec B
sec 60 - cosec 30
We know that
sec 60 = 2
cosec 30 = 2
2 - 2 = 0
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Step-by-step explanation:
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Sin c= P/H
perpendicular = √(13)^2-(5)^2
= √144= 12
= 12/13
Cosc = √1-sin^2c
1-144/169= 5/13
TanC= sinc/cosc = 12/13/(5/13)
= 12/5
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