Question

you might need: calculator
find the value of ( x ) in the triangle shown below.
( x = square^circ )
(there is a triangle with sides 3.5, 4, 3.5 and angles ( x^circ ), ( 55^circ ) as shown in the image)

Explanation:

Step1: Identify triangle type

The triangle has two sides of length 3.5, so it's isosceles. In an isosceles triangle, angles opposite equal sides are equal. Wait, no—wait, the sides: two sides are 3.5, so the angles opposite them? Wait, no, the side lengths: 3.5, 3.5, and 4. So the angles opposite the 3.5 sides: the angle opposite the side of length 4 is 55°? Wait, no, let's look at the diagram. The triangle has sides: one side 3.5, another 3.5, and the base 4? Wait, no, the angles: one angle is 55°, and the other two angles? Wait, no, the sides: the two sides of length 3.5 are adjacent to angle \( x \) and the angle opposite the side of length 4? Wait, no, let's re-examine. The triangle has sides: length 3.5 (left), length 3.5 (bottom), length 4 (right). The angle at the bottom right is 55°, and the angle at the bottom left is \( x \). Wait, in a triangle, if two sides are equal, the angles opposite them are equal. Wait, the sides of length 3.5: the angle opposite the side of length 4 is 55°? No, wait, the side of length 4 is opposite angle \( x \)? No, wait, the side of length 4 is between the two sides of length 3.5? Wait, no, let's use the Law of Sines or Law of Cosines. Wait, but maybe it's an isosceles triangle with two sides equal (3.5 and 3.5), so the angles opposite those sides are equal. Wait, the side of length 4 is opposite angle \( x \), and the side of length 4 is opposite angle \( x \), and the side of length 3.5 is opposite the 55° angle? Wait, no, let's label the triangle: let's call the vertices A, B, C. Let’s say vertex A is the top, B is bottom left (angle \( x \)), C is bottom right (angle 55°). Then side AB = 3.5, BC = 3.5, and AC = 4. Wait, no, AB = 3.5, AC = 4, BC = 3.5. So sides AB and BC are both 3.5, so triangle ABC is isosceles with AB = BC = 3.5. Therefore, the angles opposite those sides: angle at C (55°) is opposite side AB (3.5), and angle at A is opposite side BC (3.5), so angle at A would be equal to angle at C? Wait, no, that can't be. Wait, maybe I got the sides wrong. Alternatively, maybe the triangle has two sides of length 3.5, so the base angles are equal. Wait, no, let's use the Law of Sines. The Law of Sines states that \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\), where \(a\), \(b\), \(c\) are the lengths of the sides opposite angles \(A\), \(B\), \(C\) respectively. So let's denote: side \(a = 4\) (opposite angle \(A\) = \(x\)), side \(b = 3.5\) (opposite angle \(B\) = 55°), side \(c = 3.5\) (opposite angle \(C\) = let's say angle at the top). Wait, no, the diagram: the side of length 4 is opposite angle \(x\), and the two sides of length 3.5 are opposite angles 55° and the top angle? Wait, no, the angle at the bottom right is 55°, and the side opposite to it is the left side (length 3.5), and the side opposite to \(x\) is the right side (length 4). Wait, this is confusing. Wait, maybe the triangle is isosceles with two sides of length 3.5, so the angles opposite those sides are equal. Wait, the two sides of length 3.5: the angles opposite them are \(x\) and the angle opposite the side of length 4? No, that doesn't make sense. Wait, maybe I made a mistake. Let's start over. The triangle has sides: length 3.5 (left), length 3.5 (bottom), length 4 (right). The angle at the bottom right (between the bottom side (3.5) and the right side (4)) is 55°, and the angle at the bottom left (between the left side (3.5) and the bottom side (3.5)) is \(x\). Wait, no, the bottom side is 3.5, left side is 3.5, right side is 4. So the two sides of length 3.5 are adjacent to angle \(x\) (bottom left) and the top angle. Wait, no, in a triangle, the sum of angles is 180°. If two sides are equal, the angles opposite them are equal. So if two sides are 3.5, then the angles opposite those sides are equal. So the side of length 4 is opposite angle 55°, and the two sides of length 3.5 are opposite angles \(x\) and the other angle? Wait, no, that can't be. Wait, maybe the angle of 55° is opposite one of the 3.5 sides, and \(x\) is opposite the other 3.5 side, so they are equal? But that would mean \(x = 55°\), but that can't be, because the sum would be 55 + 55 + angle = 180, so angle = 70, but that doesn't fit. Wait, no, maybe I have the sides and angles mixed up. Let's use the Law of Sines. Let's denote: side \(a = 3.5\) (opposite angle \(A = x\)), side \(b = 3.5\) (opposite angle \(B\)), side \(c = 4\) (opposite angle \(C = 55°\)). Then by Law of Sines: \(\frac{a}{\sin A} = \frac{c}{\sin C}\). So \(\frac{3.5}{\sin x} = \frac{4}{\sin 55°}\). Wait, no, if side \(c = 4\) is opposite angle \(C = 55°\), and side \(a = 3.5\) is opposite angle \(A = x\), then \(\frac{3.5}{\sin x} = \frac{4}{\sin 55°}\). Solving for \(\sin x\): \(\sin x = \frac{3.5 \times \sin 55°}{4}\). Let's calculate that. \(\sin 55° \approx 0.8192\), so \(3.5 \times 0.8192 \approx 2.8672\), then divide by 4: \(2.8672 / 4 \approx 0.7168\). Then \(x \approx \arcsin(0.7168) \approx 45.8°\)? Wait, that doesn't seem right. Wait, maybe I mixed up the sides. Wait, maybe the angle of 55° is between the two sides of length 3.5 and 4, and the two sides of length 3.5 are adjacent to angle \(x\). Wait, no, the diagram: the triangle has two sides of length 3.5 (left and bottom) and one side of length 4 (right). The angle at the bottom right (between bottom side (3.5) and right side (4)) is 55°, and the angle at the bottom left (between left side (3.5) and bottom side (3.5)) is \(x\). Wait, in that case, the triangle has sides: left = 3.5, bottom = 3.5, right = 4. So the two sides of length 3.5 form a triangle where the bottom side is 3.5, left side is 3.5, so it's an isosceles triangle with base 4? No, the base would be 4, and the two equal sides are 3.5. Then the base angles (the angles at the bottom) would be equal? Wait, no, in an isosceles triangle, the base angles are equal, and the vertex angle is opposite the base. Wait, so if the base is 4, then the two equal sides are 3.5, so the base angles (the angles at the bottom) are equal? Wait, no, the base is the unequal side. Wait, no, in an isosceles triangle, the two equal sides are called the legs, and the unequal side is the base. The angles opposite the legs are equal (the base angles), and the angle opposite the base is the vertex angle. So in this case, the legs are 3.5 (left and right), and the base is 4 (bottom)? No, the diagram shows the bottom side as 3.5, left side as 3.5, right side as 4. So the legs are left (3.5) and bottom (3.5), and the base is right (4). Then the vertex angle is at the top, and the base angles are at the bottom left (x) and bottom right (55°). Wait, but if the legs are left (3.5) and bottom (3.5), then the base is right (4), so the base angles are at the bottom left (x) and top? No, this is confusing. Wait, maybe the angle at the bottom right is 55°, and the two sides adjacent to it are 3.5 (bottom) and 4 (right). The side opposite to angle x is 4, and the side opposite to 55° is 3.5. Wait, let's use the Law of Sines correctly. Let's label the triangle:

- Vertex B: bottom left, angle \(x\), side opposite: AC = 4
- Vertex C: bottom right, angle 55°, side opposite: AB = 3.5
- Vertex A: top, angle \(y\), side opposite: BC = 3.5

So by Law of Sines:

\(\frac{AB}{\sin C} = \frac{BC}{\sin A} = \frac{AC}{\sin B}\)

So \(AB = 3.5\), \(C = 55°\), \(BC = 3.5\), \(AC = 4\), \(B = x\)

So \(\frac{3.5}{\sin 55°} = \frac{3.5}{\sin A}\) ⇒ \(\sin A = \sin 55°\) ⇒ \(A = 55°\) or \(A = 180° - 55° = 125°\). But since the sum of angles in a triangle is 180°, if \(A = 125°\), then \(x + 55° + 125° = 180°\) ⇒ \(x = 0°\), which is impossible. So \(A = 55°\). Then \(x = 180° - 55° - 55° = 70°\)? Wait, no, that can't be. Wait, no, let's recalculate. If \(A = 55°\), \(C = 55°\), then \(x = B = 180° - 55° - 55° = 70°\). Wait, but that contradicts the Law of Sines. Wait, no, let's check the sides. AB = 3.5 (opposite C = 55°), BC = 3.5 (opposite A = 55°), so AB = BC, so angles opposite them (C and A) are equal, so A = C = 55°, then B = 180 - 55 - 55 = 70°, so x = 70°. Wait, that makes sense. So the angle at B (bottom left) is 70°, so \(x = 70\).

Wait, let's verify with Law of Sines. \(\frac{AC}{\sin B} = \frac{4}{\sin 70°} \approx \frac{4}{0.9397} \approx 4.25\)

\(\frac{AB}{\sin C} = \frac{3.5}{\sin 55°} \approx \frac{3.5}{0.8192} \approx 4.27\), which is close (considering rounding errors). So that works. So the angle \(x\) is 70 degrees.

Step2: Calculate the sum of angles

In any triangle, the sum of the interior angles is \(180^\circ\). We have two angles: one is \(55^\circ\), and the other (opposite the equal side) is also \(55^\circ\) (since the sides opposite them are equal, so the angles are equal). Wait, no, earlier we saw that the two sides of length 3.5 are opposite angles \(55^\circ\) and \(x\)? No, wait, the two sides of length 3.5 are AB and BC, so angles opposite them are C and A, which are equal (55° each). Then angle B (x) is \(180 - 55 - 55 = 70\).

So:

\(x + 55^\circ + 55^\circ = 180^\circ\)

\(x = 180^\circ - 55^\circ - 55^\circ\)

\(x = 70^\circ\)

Answer:

\(70\)