In δbca, cb = 11 cm, cg = 6 cm, ah = 9 cm. find the perimeter of δbca. triangle bca with inscribed circle d. segments bf and bh, cf and cg, and ag and gh are tangent to circle d. 32 cm 40 cm 35 cm 42 cm

The perimeter of Δ ABC is 40 cm ⇒ 2nd answer

Step-by-step explanation:

* Lets explain how to solve the problem

- Circle D is inscribed in triangle ABC

- The circle touches the side AB at H , side BC at F , side CA at G

- BF and BH are tangents to circle D from point B

∴ BF = BH ⇒ tangents drawn from a point outside the circle

- CF and CG are tangents to circle D from point C

∴ CF = CG ⇒ tangents drawn from a point outside the circle

- AG and AH are tangents to circle D from point A

∴ AG = AH ⇒ tangents drawn from a point outside the circle

∵ CG = 6 cm ⇒ given

∴ CF = 6 cm

∵ CB = 11 cm ⇒ given

∵ CB = CF + FB

∴ 11 = 6 + FB ⇒ subtract 6 from both sides

∴ FB = 5 cm

∵ FB = BH

∴ BH = 5 cm

∵ AH = 9 cm ⇒ given

∵ AH = AG

∴ AG = 9 cm

∵ AB = AH + HB

∴ AB = 9 + 5 = 14 cm

∵ AC = AG + GC

∴ AC = 9 + 6 = 15 cm

∵ BC = 11 cm ⇒ given

∵ The perimeter of Δ ABC = AB + BC + CA

∴ The perimeter of Δ ABC = 14 + 11 + 15 = 40 cm

* The perimeter of Δ ABC is 40 cm

(1) Recall that

sin(x - y) = sin(x) cos(y) - cos(x) sin(y)

sin²(x) + cos²(x) = 1

Given that α lies in the third quadrant, and β lies in the fourth quadrant, we expect to have

• sin(α) < 0 and cos(α) < 0

• sin(β) < 0 and cos(β) > 0

Solve for cos(α) and sin(β) :

cos(α) = -√(1 - sin²(α)) = -3/5

sin(β) = -√(1 - cos²(β)) = -5/13

Then

sin(α - β) = sin(α) cos(β) - cos(α) sin(β) =  (-4/5) (12/13) - (-3/5) (-5/13)

==>   sin(α - β) = -63/65

(2) In the second identity listed above, multiplying through both sides by 1/cos²(x) gives another identity,

sin²(x)/cos²(x) + cos²(x)/cos²(x) = 1/cos²(x)

==>   tan²(x) + 1 = sec²(x)

Rewrite the equation as

3 sec²(x) tan(x) = 4 tan(x)

3 (tan²(x) + 1) tan(x) = 4 tan(x)

3 tan³(x) + 3 tan(x) = 4 tan(x)

3 tan³(x) - tan(x) = 0

tan(x) (3 tan²(x) - 1) = 0

Solve for x :

tan(x) = 0   or   3 tan²(x) - 1 = 0

tan(x) = 0   or   tan²(x) = 1/3

tan(x) = 0   or   tan(x) = ±√(1/3)

x = arctan(0) + nπ   or   x = arctan(1/√3) + nπ   or   x = arctan(-1/√3) + nπ

x = nπ   or   x = π/6 + nπ   or   x = -π/6 + nπ

where n is any integer. In the interval [0, 2π), we get the solutions

x = 0, π/6, 5π/6, π, 7π/6, 11π/6

(3) You only need to rewrite the first term:

Then