Change 2.1 to a mixed number
2 answers:
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<u>Answer:</u>
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.
<u>The solution is: m∠HCL = 30°</u>
<u>Step-by-step explanation:</u>
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°
<u>∴ ∠HCL = ∠ACL - ∠ACH = 45° - 15° = 30°</u>
