Question

solve for \a\.
30°
70°
10√3
concept check

Explanation:

Step1: Find the height (h) of the large triangle

In the 30 - 60 - 90 triangle (since the right angle and 30° angle imply the third angle is 60°), the hypotenuse - related side: let the base of the large triangle (with 30° angle) have adjacent side (let's say the horizontal part) and the height \( h \). For a 30 - 60 - 90 triangle, if we consider the side opposite 30° is \( x \), hypotenuse is \( 2x \), and the side opposite 60° is \( x\sqrt{3} \). Wait, actually, the total base is \( 10\sqrt{3} \), but the large triangle has angle 30° at the left, right angle at the bottom - right. So, \( \tan(30^{\circ})=\frac{h}{10\sqrt{3}} \)? Wait, no. Wait, the large triangle: angle at left is 30°, right angle at bottom - right. So the height \( h \) (vertical side) and the base (horizontal side) is \( 10\sqrt{3} \)? Wait, no, the total horizontal length is \( 10\sqrt{3} \), and the large triangle has angle 30°, so \( \tan(30^{\circ})=\frac{h}{b_1} \), and the small triangle (with angle 70°) has \( \tan(70^{\circ})=\frac{h}{b_2} \), and \( b_1 + b_2=10\sqrt{3} \). Wait, maybe better: in the large right - triangle (30° angle), \( \tan(30^{\circ})=\frac{h}{10\sqrt{3}} \)? No, wait, actually, the large triangle: the angle at the left is 30°, the right angle is at the bottom - right, so the height \( h \) (vertical leg) and the horizontal leg is \( 10\sqrt{3} \)? Wait, no, that can't be. Wait, maybe the large triangle has angle 30°, and the side adjacent to 30° is the total base \( 10\sqrt{3} \), and the height \( h \) is opposite to 30°? No, \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \). So \( \tan(30^{\circ})=\frac{h}{10\sqrt{3}} \)? Wait, \( \tan(30^{\circ})=\frac{1}{\sqrt{3}} \), so \( h = 10\sqrt{3}\times\tan(30^{\circ})=10\sqrt{3}\times\frac{1}{\sqrt{3}} = 10 \). So the height \( h = 10 \).

Step2: Use the height to find \( a \) in the small right - triangle (angle 70°)

In the small right - triangle with angle 70° and height \( h = 10 \) (opposite side to 70°? Wait, no: the small triangle has angle 70° at the bottom, right angle at the bottom - right. So the height \( h \) is the opposite side to the 70° angle? Wait, no: in the small right - triangle, angle at the bottom is 70°, right angle at the bottom - right, so the vertical side is \( h \) (opposite to 70°), and the horizontal side is adjacent to 70°, and \( a \) is the hypotenuse? Wait, no, the side labeled \( a \) is the hypotenuse of the small right - triangle? Wait, no, looking at the diagram, \( a \) is a side of the small triangle (with angle 70° and right angle). Wait, maybe \( a \) is the hypotenuse? No, the diagram shows \( a \) as a side from the vertex at the top to the point on the base (between the two triangles). Wait, actually, the small triangle is a right - triangle with angle 70° at the base, right angle at the bottom - right, and \( a \) is the hypotenuse? Wait, no, the height \( h = 10 \) (from step 1) is the opposite side to the 70° angle in the small triangle. So \( \sin(70^{\circ})=\frac{h}{a} \), so \( a=\frac{h}{\sin(70^{\circ})} \). Since \( h = 10 \), \( a=\frac{10}{\sin(70^{\circ})} \approx\frac{10}{0.9397}\approx10.64 \). Wait, but let's re - check step 1.

Wait, in the large right - triangle (angle 30°), \( \tan(30^{\circ})=\frac{h}{x} \), where \( x \) is the horizontal leg of the large triangle, and the horizontal leg of the small triangle is \( 10\sqrt{3}-x \). But actually, the large triangle: angle 30°, right angle, so \( \tan(30^{\circ})=\frac{h}{10\sqrt{3}} \)? No, that's wrong. Wait, the total base is \( 10\sqrt{3} \), which is the sum of the two horizontal segments (from the left vertex to the vertex between the two triangles, and from that vertex to the right - angle vertex). Let's call the left - most horizontal segment \( x \) and the right - most horizontal segment \( y \), so \( x + y=10\sqrt{3} \). In the large triangle (angle 30°), \( \tan(30^{\circ})=\frac{h}{x} \), so \( x = h\cot(30^{\circ})=h\sqrt{3} \). In the small triangle (angle 70°), \( \tan(70^{\circ})=\frac{h}{y} \), so \( y = h\cot(70^{\circ}) \). Then \( h\sqrt{3}+h\cot(70^{\circ}) = 10\sqrt{3} \). Factor out \( h \): \( h(\sqrt{3}+\cot(70^{\circ})) = 10\sqrt{3} \). \( \cot(70^{\circ})=\frac{1}{\tan(70^{\circ})}\approx\frac{1}{2.7475}\approx0.3640 \). \( \sqrt{3}\approx1.732 \). So \( h(1.732 + 0.3640)=10\times1.732 \). \( h(2.096)=17.32 \). \( h=\frac{17.32}{2.096}\approx8.26 \). Wait, that contradicts the previous step. I must have misinterpreted the diagram.

Wait, maybe the diagram is: a right - triangle with a segment from the right - angle vertex to a point on the hypotenuse of the large triangle (with 30° angle), creating a small right - triangle with angle 70° and side \( a \). Wait, the large triangle: angle at left is 30°, right angle at bottom - right. The segment from the right - angle vertex to a point on the hypotenuse of the large triangle, making a 70° angle with the horizontal. So the height (vertical leg) is \( h \), and in the large triangle, \( \tan(30^{\circ})=\frac{h}{10\sqrt{3}} \), so \( h = 10\sqrt{3}\tan(30^{\circ})=10\sqrt{3}\times\frac{1}{\sqrt{3}} = 10 \). Then in the small triangle (angle 70°), \( \sin(70^{\circ})=\frac{h}{a} \), so \( a=\frac{h}{\sin(70^{\circ})}=\frac{10}{\sin(70^{\circ})}\approx\frac{10}{0.9397}\approx10.64 \). Alternatively, \( \cos(70^{\circ})=\frac{\text{adjacent}}{a} \), but no, the small triangle is a right - triangle with angle 70°, right angle at the bottom - right, so the vertical side is \( h = 10 \) (opposite to 70°), and \( a \) is the hypotenuse? Wait, no, if the angle is 70° at the bottom, then the vertical side is opposite to 70°, so \( \sin(70^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{h}{a} \), so \( a=\frac{h}{\sin(70^{\circ})} \). Since \( h = 10 \), \( a=\frac{10}{\sin(70^{\circ})}\approx10.64 \).

Answer:

\( a\approx10.64 \) (or more precisely, \( a = \frac{10}{\sin(70^{\circ})}\approx10.64 \))