Question

are the triangles below similar?
triangle cdb with angles at c: 41°, at b: 83°, at d: 56°, sides cd=6, db=4. triangle sqr with angles at s: 56°, at r: 41°, at q: 83°, sides sq=20, qr=25, sr=30.
options: yes, no

Explanation:

Step1: Check Angle Similarity

In triangle \(BCD\) (wait, actually triangle \(BCD\) is \( \triangle BCD\)? Wait, no, first triangle is \( \triangle BCD\)? Wait, labels: \(C\), \(B\), \(D\). Angles: \( \angle C = 41^\circ\), \( \angle B = 83^\circ\), \( \angle D = 56^\circ\) (since \(180 - 41 - 83 = 56\)). Second triangle \( \triangle QSR\) (wait, labels \(S\), \(Q\), \(R\)): \( \angle S = 56^\circ\), \( \angle R = 41^\circ\), \( \angle Q = 83^\circ\) (since \(180 - 56 - 41 = 83\)). So corresponding angles: \( \angle C = \angle R = 41^\circ\), \( \angle D = \angle S = 56^\circ\), \( \angle B = \angle Q = 83^\circ\). So angles are equal (AA similarity could apply, but let's check sides too).

Step2: Check Side Ratios

First triangle sides: \(CD = 6\), \(BD = 4\), \(BC\) (wait, maybe I mislabeled. Wait first triangle: \(C\) to \(D\) is 6, \(D\) to \(B\) is 4, \(B\) to \(C\) is... Wait second triangle: \(S\) to \(Q\) is 20, \(Q\) to \(R\) is 25, \(S\) to \(R\) is 30. Let's match sides to angles. In first triangle, angle \(C = 41^\circ\), angle \(D = 56^\circ\), angle \(B = 83^\circ\). In second triangle, angle \(R = 41^\circ\), angle \(S = 56^\circ\), angle \(Q = 83^\circ\). So side opposite \(41^\circ\): in first triangle, side opposite \( \angle C (41^\circ) \) is \(BD = 4\); in second triangle, side opposite \( \angle R (41^\circ) \) is \(QS = 20\). Ratio \(4/20 = 1/5\). Side opposite \(56^\circ\): in first triangle, side opposite \( \angle D (56^\circ) \) is \(BC\) (wait no, \( \angle D\) is at \(D\), so side opposite \( \angle D\) is \(BC\). Wait first triangle: \(CD = 6\) (between \(C\) and \(D\)), \(BD = 4\) (between \(D\) and \(B\)), \(BC\) (between \(B\) and \(C\)). Wait maybe better to list sides with angles. Let's assign:

First triangle (let's call it \( \triangle ABC\) no, labels: \(C\), \(B\), \(D\)):

• \( \angle C = 41^\circ\), side opposite: \(BD = 4\)
• \( \angle D = 56^\circ\), side opposite: \(BC\) (wait, no, in triangle \(CBD\), vertices \(C\), \(B\), \(D\). So sides: \(CB\) (length?), \(BD = 4\), \(CD = 6\). Angles: \( \angle C = 41^\circ\), \( \angle B = 83^\circ\), \( \angle D = 56^\circ\).

Second triangle \( \triangle SQR\):

• \( \angle S = 56^\circ\), side opposite: \(QR = 25\)
• \( \angle R = 41^\circ\), side opposite: \(QS = 20\)
• \( \angle Q = 83^\circ\), side opposite: \(SR = 30\)

Now, let's match angles:

\( \angle C (41^\circ) \) corresponds to \( \angle R (41^\circ) \)

\( \angle D (56^\circ) \) corresponds to \( \angle S (56^\circ) \)

\( \angle B (83^\circ) \) corresponds to \( \angle Q (83^\circ) \)

Now, sides:

• Side opposite \(41^\circ\): \(BD = 4\) (first triangle) and \(QS = 20\) (second triangle). Ratio \(4/20 = 1/5\)
• Side opposite \(56^\circ\): \(BC\) (first triangle) – wait no, in first triangle, \( \angle D = 56^\circ\), so side opposite is \(BC\). Wait first triangle: \(CD = 6\) (between \(C\) and \(D\)), \(BD = 4\) (between \(D\) and \(B\)), so \(BC\) can be found? Wait no, maybe I made a mistake. Wait the first triangle has sides 6 (CD), 4 (BD), and BC (unknown). The second triangle has sides 20 (QS), 25 (QR), 30 (SR). Let's check the ratios of the given sides. Wait CD is 6, SR is 30: 6/30 = 1/5. BD is 4, QS is 20: 4/20 = 1/5. Wait maybe CD corresponds to SR, BD corresponds to QS, and BC corresponds to QR. Let's check:

CD = 6, SR = 30: 6/30 = 1/5

BD = 4, QS = 20: 4/20 = 1/5

Now, QR is 25, so BC should be 5 (since 5/25 = 1/5). Let's check angles. Since angles are equal (AA: two angles equal, so triangles are similar by AA, and sides are proportional (ratio 1/5)). So the triangl…

Answer:

yes