Question

In ∆ABC acute angles are in the ratio 5:1, i.e.

Answer 1

Answer:

The quen is as following:

ABC is a right triangle at C,

Acute angles are in the ratio 5:1, i.e. ∠BAC : ∠ABC = 5:1

If CH is an altitude to AB and CL is an angle bisector of ∠ACB, find m∠HCL.

The solution is:  m∠HCL = 30°

Step-by-step explanation:

See the attached figure.

∵The triangle is right at C   ∴∠C = 90°

∴∠A + ∠B = 90° ⇒(1)

∵ Acute angles are in the ratio 5:1, i.e. ∠BAC : ∠ABC = 5:1

∴∠A = 5 times ∠B

Substitute at (1)

∴ 5 ∠B + ∠B = 90° ⇒⇒⇒ ∴∠B = 15°  and ∠A = 75°

∵CL is an angle bisector of ∠ACB

∴ ∠ACL = 90°/2 = 45°

∵ CH is an altitude to AB ⇒ ∠CHA = 90°

At the triangle AHC:

∠ACH = 180° - (∠CHA + ∠CAH) = 180° - (90° + 75°) = 15°

∴ ∠HCL = ∠ACL - ∠ACH = 45° - 15° = 30°