pls answer will mark as brainliest
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Answer:
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Step-by-step explanation:
A+B+C=180
0
⟶(1)
sin(A+B−C)=
2
1
⟶(2)
cos(B+C−A)=
2
1
⟶(3)
Now from (1)
A+B=Π−C⟶(4)
∴ Using (4) we can rewrite (2) as
sin(Π−2C)=
2
1
⇒sin2C=
2
1
(∵sin(Π−θ)=sinθ)
∴ 2C=30
0
or 150
0
(∵ sin30
0
=1/2
⇒C=15
0
or 75
0
sin150
0
=1/2)
Again from (1)
B+C=Π−A
∴ cos(B+C−A)=
2
1
⇒cos(Π−2A)=
2
1
⇒−cos2A=
2
1
(∵ cos(Π−θ)=−cosθ)
⇒cos2A=
2
−1
∴ 2A=135
0
(∵ cos135
0
=
2
−1
)
⇒A=67.5
0
Case I : For C=15
0
Case II : For C=75
0
A=67.5
0
A=67.5
0
∴ B=180
0
−(15
0
+67.5
0
) B=180
0
−(67.5
0
+75
0
)
=180
0
−82.5=97.5
0
=37.5
0
Since ΔABC is acute angled, thereby n none of the angles can be more than 90
0
.
Hence ∠A=67.5
0
, ∠C=75
0
, ∠B=37.5
0