Test: Trigonometric Functions - Commerce MCQ

Thus, C/2 = 45º ⇒ C = 90º

∠C = 180 − 45 − 75 = 60

sin 75 = sin (45 + 30) = sin45.cos30 + sin30.cos45
sin 75 = 1/√2 x √3/2 + 1/2 1/√2

Let the angles be x, 2x, and 7x respectively.
⇒ x + 2x + 7x = 180° [angle sum property of triangle]
⇒ 10x = 180°
⇒ x = 18°
Hence, the angles are: 18°,36°,126°

⇒ Greatest Side = K sin 126°
⇒ Smallest side = K sin 18°
So, required ratio =K sin 126°/ K sin 18°
= sin(90°+36°)/sin18°= cos36°/sin18°    [sin(90°+x)=cosx]
= (√5+1)/(√5−1)

a = 13, b = 7, c = 8

 By cosine formula,

⇒ A = 120º = 2π/3

The sine formula is 
a/sinA = b/sinB 
⇒ a sin B = b sin A
(a) b sin A = a 
⇒ a sin B = a 
⇒ B = π/2 
Since, ∠A  < π/2 therefore, the triangle is possible.
(b) b sin A < a 
⇒ a sin B < a 
⇒ sinB < 1 
⇒ ∠B exists 
Now, b > a 
⇒ B > A since A < π/2 
∴ The triangle is possible.

(b+c) cos A + (c + a) cos B + (a + b) cos C
⇒ b cos A + c cos A + c cos B + a cos B + a cos C + b cos C
⇒ (b cos C + c cos B) + (c cos A + a cos C) + (a cos B + b cos A) ...(1)
Using projection formula,
a = (b cos C + c cos B) ...(2)
b = (c cos A + a cos C) ...(3)
c = (a cos B + b cos A) ...(4)

On adding the projection formula, we get the initial expression
i.e. (2) + (3) + (4) = (1)
∴ (b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c

Let the angles be a,a+d,a+2d.
Then,
a+a+d+a+2d=180º
a+d=60º
So, ∠B=60º,
By cosine formula,

⇒ b² = a² + c² − ac

Area of a triangle = 80 cm², a + b + c = 8 cm
s = perimeter/2=(a + b + c)/2 = 8/2 = 4 cm

Radius of inscribed circle (r) = (area of the triange)/s = 80/4 = 20 cm

c² − 7 = 3c ⇒ c² − 3c − 7 = 0

Replace c by x : x² - 3x - 7 = 0

cos A + cos B = 4 sin²(C/2​)
⇒ 2 cos (A+B)/2​ cos (A−B)/2 ​= 4 sin²(C/2​)

∵ A + B + C = π ⇒ A + B = π − C

⇒ cos (π−C)/2 ​cos (A−B)/2 ​= 2 sin²(C/2​)
⇒ sin C/2 ​cos (A−B)/2 ​= 2 sin²(C/2​)
⇒ cos (A−B)/2 = 2 sin (C/2​)
⇒ cos C/2 ​cos (A−B)/2 = 2 sin (C/2​) cos (C/2)​
⇒ cos (π−(A+B)​)/2 cos (A−B)/2 = sin C
⇒ 2 sin (A+B)/2 ​cos (A−B)/2​ = sin C
⇒ sin A + sin B = 2 sin C

∵ a/sinA​ = b/sinB​ = c/sinC​ = k
⇒sinA = ak, sin B = bk , sin C = ck

⇒ ak + bk = 2(ck)
⇒ a+b=2c

Therefore the sides of triangle a,b,c are in A.P.