Prove that, in a right triangle with a 15° angle, the altitude to the hypotenuse is one fourth of the hypotenuse (without using trigonometric functions).

Given: A Right triangle ABC in which ∠B=90°,∠C=15°.A perpendicular from B to Hypotenuse AC is Drawn such that ∠ADB=∠BDC=90°

To Prove: BD=1/4×AC

Proof: In Δ ABC

∠A+∠B+∠C=180°[Angle sum property of triangle]

∠A+90°+15°=180°

∠A=180°-105°

∠A=75°

Similarly In ΔABD ,∠ABD=15° →∠A+∠ABD+∠ADB=180°

In Δ ABC

Sin 15°=(1)

In Right ΔADB,

Cos 15°=(2)

Sin 30°=2 Sin 15°× Cos 15°  →Sin2A=2 SinA CosA

   →  1/2  = 2

Cancelling AB from numerator and denominator

→AC= 4 BD which is the result.

2,600 divided by 80 = 32.5

step-by-step explanation:

anser is a

step-by-step explanation:

negitive slope