Question

what are the exact values of a and b? diagram: right triangle ( abc ) with right angle at ( c ), ( angle a = 30^circ ), hypotenuse ( ab = 7 ), ( bc = a ), ( ac = b ). multiple - choice options: ( a=\frac{7}{2}, b = \frac{7sqrt{3}}{2} ); ( a=\frac{7sqrt{3}}{2}, b=\frac{7}{2} ); ( a=\frac{7}{2}, b = 7sqrt{3} ); ( a=\frac{7}{2}, b=\frac{7sqrt{3}}{2} ) (exact option formatting may vary slightly)

Explanation:

Step1: Identify triangle type

This is a 30 - 60 - 90 right triangle (angle at A is \(30^{\circ}\), right angle at C). In a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite \(30^{\circ}\) (side \(a\)) is half the hypotenuse, and the side opposite \(60^{\circ}\) (side \(b\)) is \(\frac{\sqrt{3}}{2}\) times the hypotenuse. The hypotenuse \(AB = 7\).

Step2: Calculate side \(a\)

Side \(a\) is opposite \(30^{\circ}\), so \(a=\frac{1}{2}\times\) hypotenuse.
\(a = \frac{1}{2}\times7=\frac{7}{2}\)

Step3: Calculate side \(b\)

Side \(b\) is opposite \(60^{\circ}\), so \(b=\frac{\sqrt{3}}{2}\times\) hypotenuse.
\(b=\frac{\sqrt{3}}{2}\times7 = \frac{7\sqrt{3}}{2}\)

Wait, but looking at the options, maybe I mixed up \(a\) and \(b\). Wait, angle at A is \(30^{\circ}\), so angle at B is \(60^{\circ}\). Side \(a\) is opposite angle B (\(60^{\circ}\))? Wait, no, in the triangle, \(C\) is right angle, \(A\) is \(30^{\circ}\), so side opposite \(A\) is \(a\) (BC), side opposite \(B\) is \(b\) (AC), hypotenuse AB = 7.

So angle \(A = 30^{\circ}\), so side opposite \(A\) (BC = \(a\)) is \(\frac{1}{2}\) hypotenuse: \(a=\frac{7}{2}\)

Side opposite \(B\) (AC = \(b\)): angle \(B=60^{\circ}\), so \(b = \sin(60^{\circ})\times\) hypotenuse or \(\cos(30^{\circ})\times\) hypotenuse. \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(b = 7\times\frac{\sqrt{3}}{2}=\frac{7\sqrt{3}}{2}\)? Wait, no, wait the options: let's check the options again.

Wait the options:

First option: \(a=\frac{7}{2},b = \frac{7\sqrt{3}}{2}\)

Second option: \(a=\frac{7}{2},b = 7\sqrt{3}\)

Third option: \(a=\frac{7\sqrt{3}}{2},b=\frac{7}{2}\)

Fourth option: \(a=\frac{7}{2},b=\frac{7\sqrt{3}}{2}\) (wait maybe the first and fourth are same? Wait maybe I misread. Wait the triangle: right - angled at C, so AC is \(b\), BC is \(a\), AB is hypotenuse 7, angle at A is \(30^{\circ}\).

So \(\sin(30^{\circ})=\frac{a}{AB}\), so \(a = AB\times\sin(30^{\circ})=7\times\frac{1}{2}=\frac{7}{2}\)

\(\cos(30^{\circ})=\frac{b}{AB}\), so \(b = AB\times\cos(30^{\circ})=7\times\frac{\sqrt{3}}{2}=\frac{7\sqrt{3}}{2}\)

So the correct option is the one where \(a = \frac{7}{2}\) and \(b=\frac{7\sqrt{3}}{2}\) (probably the first or fourth option, assuming the first option is \(a=\frac{7}{2},b=\frac{7\sqrt{3}}{2}\))

Answer:

The exact values are \(a=\frac{7}{2}\) and \(b = \frac{7\sqrt{3}}{2}\) (corresponding to the option with \(a=\frac{7}{2},b=\frac{7\sqrt{3}}{2}\))